cells in series and parallel experiment conclusion

unifyphysics

Combination of Cells in Series and Parallel

In 1800, Italian physicist Alessandro Volta created the first practical chemical cell, known as the voltaic pile . This groundbreaking invention consisted of alternating layers of zinc and copper discs, separated by cardboard soaked in saltwater. When connected, the voltaic pile produced a steady electric current, demonstrating the conversion of chemical energy into electrical energy. This innovation marked the birth of the first true battery and laid the foundation for future developments in electrochemical cells.

Building on Volta’s pioneering work, British chemist John Frederic Daniell introduced the Daniell cell in 1836. Daniell’s design improved upon the voltaic pile by replacing the saltwater-soaked cardboard with a porous pot containing a copper sulfate solution. The zinc electrode was dipped into dilute sulfuric acid. This configuration provided a more stable electromotive force (EMF) and significantly reduced the polarization effects that plagued earlier designs. The Daniell cell was a major step forward, offering a more reliable and efficient power source.

The mid-19th century saw further advancements in cell technology with the contributions of William Grove and Robert Bunsen. Grove introduced the Grove cell , which used nitric acid and zinc, while Bunsen developed the Bunsen cell , employing sulfuric acid and zinc. Both cells improved the efficiency and reliability of electrochemical power sources, paving the way for more sophisticated applications.

In 1866, French engineer Georges Leclanché invented the Leclanché cell , a design known for its simplicity and practicality. This cell featured a zinc anode immersed in ammonium chloride (NH₄Cl) solution, with a cathode composed of a carbon rod surrounded by manganese dioxide (MnO₂). The Leclanché cell was widely used in telegraphs and doorbells due to its reliability, low cost, and ease of use. Its success demonstrated the potential for electrochemical cells to be integrated into everyday devices.

Around 1887, Carl Gassner introduced the dry cell , a significant advancement over previous designs. Unlike earlier cells that used liquid electrolytes, the dry cell utilized a paste electrolyte, making it more portable and less prone to leakage. The zinc container served as both the anode and casing, while the manganese dioxide and carbon rod formed the cathode. Dry cells became ubiquitous in devices like flashlights, radios, and toys, providing a convenient and reliable source of portable power.

In the 1950s, Lewis Urry developed the alkaline battery , which replaced the acidic electrolyte with an alkaline paste, typically potassium hydroxide. Alkaline batteries offered several advantages over earlier designs, including higher capacity, longer shelf life, and better performance under various conditions. These batteries quickly became the standard for many portable electronic devices, from remote controls to portable radios.

The late 20th century witnessed a revolution in battery technology with the advent of lithium-ion batteries . These batteries use lithium compounds as the anode material and are known for being lightweight and rechargeable. Lithium-ion batteries have become the powerhouse behind modern portable electronics, including smartphones, laptops, and electric vehicles. Their high energy density, long cycle life, and ability to maintain performance over numerous charge-discharge cycles make them indispensable in today’s digital age.

Table of Contents

When you think of electrical circuits, imagine a network of interconnected cells—tiny powerhouses that drive our devices. But what exactly is the Electromotive Force (EMF) of a cell, and how does it impact our circuits?

What Is EMF? EMF is like the cell’s energy passport. It measures how much energy the cell transfers per unit of charge passing through it.

Think of it as the cell’s motivation to push charges around the circuit. Without EMF, our gadgets would stay silent.

Cell Connections: Series and Parallel Imagine you have multiple cells. How do you connect them?

Series connection : Cells line up like a string of pearls. The positive terminal of one cell connects to the negative terminal of the next. In series, EMFs add up. If you connect two 1.5V cells in series, you get a total of 3V. Example: Old-style Christmas lights—they glow in sequence.

Parallel connection: Cells sit side by side, like stars in the night sky. All positive terminals connect, and all negative terminals connect. Each cell gets the full voltage across it. Example: Modern LED fairy lights—they shine independently.

In series, voltage divides among cells. Bulbs may appear dimmer. In parallel, each cell gets full voltage, maintaining brightness.

Series circuits are sensitive. One faulty cell affects the whole chain. Parallel circuits are robust. One cell’s hiccup doesn’t bother the others.

Internal Resistance

When you think of a battery or a cell, you might picture a compact energy source, ready to power your devices. But there’s more to the story—something hidden within the cell itself.

What Is Internal Resistance? Imagine a cell as a tiny power station. It generates an electromotive force (EMF) that pushes charges through a circuit. But wait! Inside the cell, there’s a subtle resistance—a roadblock for those charges. This is the internal resistance . Think of it like friction in a machine—it opposes the flow of current.

Inside the cell, chemical reactions occur. These reactions involve ions moving through the electrolyte. The electrolyte and electrodes (like zinc and copper) have inherent resistance. Even the wires connecting the cell contribute a bit of resistance.

As current flows, some energy is lost overcoming internal resistance. The actual voltage available to the external circuit is less than the cell’s nominal EMF. The energy lost due to internal resistance appears as heat within the cell.

The total voltage across the cell is:

\(\displaystyle\text{Total Voltage} = \text{EMF} – I \cdot r\)

So, the effective EMF is reduced by the internal voltage drop.

In everyday situations, finding batteries with exactly the voltage you need is not always possible. There are only a limited number of battery types available in the market. So, when you require a specific voltage that isn’t readily available, you can combine two or more batteries in different ways to achieve the desired voltage and current.

There are two fundamental ways to combine batteries: series combination and parallel combination. These two types form the basis for all other combinations.

• Series Combination: When batteries are connected in series, the positive terminal of one battery is linked to the negative terminal of the next. This arrangement adds up the voltages of each battery to produce a higher total voltage. However, the current remains the same across all batteries in the series.
• Parallel Combination: In a parallel combination, the positive terminals of all batteries are connected, and the negative terminals are also connected together. This setup keeps the voltage the same as that of a single battery but increases the total current capacity. It’s like having multiple paths for the current to flow, which collectively can supply more power.

Series Combination

• When cells are connected in series, the positive terminal of one cell is connected to the negative terminal of the next cell, and so on. The same current flows through each cell.
• The total voltage of the series combination is equal to the sum of the voltages of each cell. This is because the potential difference between the terminals of each cell adds up along the circuit.
• The total internal resistance of the series combination is equal to the sum of the internal resistances of each cell. This is because each cell’s internal resistance opposes the current flow along the circuit.
• A single equivalent cell with the same total voltage and internal resistance can replace the series combination of cells. The equivalent cell can provide the same current and power to the external load as the series combination.

When cells (batteries) are connected in series, the positive terminal of one cell is connected to the negative terminal of the next cell. The overall voltage is the sum of the individual cell voltages, but the current through each cell is the same.

Voltage in Series: The total voltage across cells connected in series is the sum of the voltages of each cell.

Current in Series: The same current flows through each cell in a series of connections.

Let’s consider (n) cells connected in series. Assume that each cell has an electromotive force (emf) (E) and an internal resistance (r).

For cells in series, the total emf (E total ) is the sum of the EMFs of each cell.

\(\displaystyle E_{\text{total}} = E_1 + E_2 + \cdots + E_n\)

If each cell has the same emf (E), then:

\(\displaystyle E_{\text{total}} = nE\)

The total internal resistance (R internal ) is the sum of the internal resistances of each cell.

\(\displaystyle R_{\text{internal}} = r_1 + r_2 + \cdots + r_n\)

If each cell has the same internal resistance (r), then:

\(\displaystyle R_{\text{internal}} = nr\)

When a load with resistance (R) is connected across the series combination of cells, the total voltage (V) across the load is given by:

\(\displaystyle V = E_{\text{total}} – I \cdot R_{\text{internal}}\)

where (I) is the current through the circuit. The current (I) in the circuit can be found using Ohm’s Law. The total resistance in the circuit is the sum of the external resistance (R) and the total internal resistance (R internal ).

\(\displaystyle I = \frac{E_{\text{total}}}{R + R_{\text{internal}}}\)

Substituting (\(\displaystyle E_{\text{total}} = nE\)) and (\(\displaystyle R_{\text{internal}} = nr\)):

\(\displaystyle I = \frac{nE}{R + nr}\)

The internal resistance of a series combination of cells reduces the circuit’s effective voltage and power output. The advantage of the series combination of cells is that it can increase the voltage and power output of the circuit. The disadvantage of the series combination of cells is that it can increase the internal resistance and energy loss of the circuit.

Parallel Combination

A combination of cells in parallel is a connection of cells where the positive terminals of all the cells are connected, and the negative terminals of all the cells are connected. The same potential difference is applied across each cell in the parallel combination.

When cells (batteries) are connected in parallel, all the positive terminals are connected together, and all the negative terminals are connected together. The overall voltage remains the same as the voltage of a single cell, but the total current is the sum of the currents from each cell.

Voltage in Parallel: The total voltage across cells connected in parallel is the same as the voltage of each individual cell.

Current in Parallel: The total current is the sum of the currents supplied by each cell.

Let’s consider (n) cells connected in parallel. Assume that each cell has an electromotive force (emf) (E) and an internal resistance (r).

For cells in parallel, the total voltage (E total ) is equal to the voltage of any single cell, because all cells have the same voltage:

\(\displaystyle E_{\text{total}} = E\)

The total internal resistance (R internal ) of cells in parallel is found by the formula for resistances in parallel:

\(\displaystyle\frac{1}{R_{\text{internal}}} = \frac{1}{r_1} + \frac{1}{r_2} + \cdots + \frac{1}{r_n}\)

\(\displaystyle\frac{1}{R_{\text{internal}}} = \frac{1}{r} + \frac{1}{r} + \cdots + \frac{1}{r} = \frac{n}{r}\)

Therefore, the total internal resistance is:

\(\displaystyle R_{\text{internal}} = \frac{r}{n}\)

When a load with resistance (R) is connected across the parallel combination of cells, the total current (I) supplied by the combination is the sum of the currents from each cell. Ohm’s law gives us:

Substituting (\(\displaystyle E_{\text{total}} = E\)) and (\(\displaystyle R_{\text{internal}} = \frac{r}{n})\):

\(\displaystyle I = \frac{E}{R + \frac{r}{n}}\)

The current from each cell (I cell ) is given by:

\(\displaystyle I_{\text{cell}} = \frac{E}{R + r}\)

The total current (I) is then:

\(\displaystyle I = n \times I_{\text{cell}} = n \times \frac{E}{R + r}\)

Also Read: Electrical Cell

Solved Examples

Problem 1: Three cells of EMF 1.5V, 2.0V, and 2.5V are connected in series. Calculate the equivalent EMF of the combination.

Solution: When cells are connected in series, their EMFs add up directly.

\(\displaystyle \text{Equivalent EMF} (E_{\text{eq}}) = E_1 + E_2 + E_3 \)

Given: \(\displaystyle E_1 = 1.5 \, \text{V} \); \(\displaystyle E_2 = 2.0 \, \text{V} \); \(\displaystyle E_3 = 2.5 \, \text{V} \)

\(\displaystyle E_{\text{eq}} = 1.5 \, \text{V} + 2.0 \, \text{V} + 2.5 \, \text{V} = 6.0 \, \text{V} \)

The equivalent EMF of the combination of cells in series is 6.0V.

Problem 2: Three cells with internal resistances 0.5Ω, 1.0Ω, and 1.5Ω are connected in series. Calculate the equivalent internal resistance of the combination.

Solution: When cells are connected in series, their internal resistances add up directly.

\(\displaystyle \text{Equivalent Internal Resistance} (r_{\text{eq}}) = r_1 + r_2 + r_3 \)

Given: \(\displaystyle r_1 = 0.5 \, \Omega \); \(\displaystyle r_2 = 1.0 \, \Omega \); \(\displaystyle r_3 = 1.5 \, \Omega \)

\(\displaystyle r_{\text{eq}} = 0.5 \, \Omega + 1.0 \, \Omega + 1.5 \, \Omega = 3.0 \, \Omega \)

The equivalent internal resistance of the combination of cells in series is 3.0Ω.

Problem 3: Two cells of EMF 1.5V and 2.0V with internal resistances 0.5Ω and 1.0Ω, respectively, are connected in parallel. Calculate the equivalent EMF and internal resistance of the combination.

Solution: For cells in parallel, the equivalent EMF (E eq ) is given by:

\(\displaystyle E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \)

And the equivalent internal resistance (\(\displaystyle r_{\text{eq}}\)) is given by:

\(\displaystyle \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} \)

Given: \(\displaystyle E_1 = 1.5 \), \text{V}, \(\displaystyle\quad r_1 = 0.5 \), \(\displaystyle\Omega \) \(\displaystyle E_2 = 2.0 \), \text{V}, \(\displaystyle\quad r_2 = 1.0 \), \(\displaystyle\Omega \)

Calculate (\(\displaystyle E_{\text{eq}}\)):

\(\displaystyle E_{\text{eq}} = \frac{(1.5 \times 1.0) + (2.0 \times 0.5)}{0.5 + 1.0} = \frac{1.5 + 1.0}{1.5} = \frac{2.5}{1.5} = 1.67 \, \text{V} \)

Calculate (\(\displaystyle r_{\text{eq}}\)):

\(\displaystyle \frac{1}{r_{\text{eq}}} = \frac{1}{0.5} + \frac{1}{1.0} = 2 + 1 = 3 \) \(\displaystyle r_{\text{eq}} = \frac{1}{3} = 0.33 \, \Omega \)

The equivalent EMF of the combination of cells in parallel is 1.67V and the equivalent internal resistance is 0.33Ω.

Problem 4: Two cells of EMF 1.5V and 2.5V with internal resistances 0.4Ω and 0.6Ω, respectively, are connected in parallel. Determine the maximum current that can be drawn from the combination.

Solution: The maximum current that can be drawn from the combination is determined by the equivalent EMF and internal resistance, using Ohm’s Law:

\(\displaystyle I_{\text{max}} = \frac{E_{\text{eq}}}{r_{\text{eq}}} \)

Using the formulas from Problem 3:

\(\displaystyle E_{\text{eq}} = \frac{(1.5 \times 0.6) + (2.5 \times 0.4)}{0.4 + 0.6} = \frac{0.9 + 1.0}{1.0} = 1.9 \, \text{V} \)

\(\displaystyle \frac{1}{r_{\text{eq}}} = \frac{1}{0.4} + \frac{1}{0.6} = 2.5 + 1.67 = 4.17 \)

\(\displaystyle r_{\text{eq}} = \frac{1}{4.17} \approx 0.24 \, \Omega \) \(\displaystyle I_{\text{max}} = \frac{1.9}{0.24} \approx 7.92 \, \text{A} \)

The maximum current that can be drawn from the combination of cells in parallel is approximately 7.92A.

Problem 5: Four cells each with EMF 1.5V and internal resistance 0.5Ω are connected in two parallel pairs, and these pairs are connected in series. Calculate the equivalent EMF and internal resistance of the combination.

Solution: For two cells in parallel:

\(\displaystyle E_{\text{parallel}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} = \frac{(1.5 \times 0.5) + (1.5 \times 0.5)}{0.5 + 0.5} = 1.5 \, \text{V} \)

\(\displaystyle r_{\text{parallel}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{0.5 \times 0.5}{0.5 + 0.5} = 0.25 \, \Omega \)

Each parallel combination has:

\(\displaystyle E_{\text{parallel}} = 1.5 \, \text{V} \) \(\displaystyle r_{\text{parallel}} = 0.25 \, \Omega \)

These two parallel combinations are then connected in series:

\(\displaystyle E_{\text{series}} = E_{\text{parallel}} + E_{\text{parallel}} = 1.5 \, \text{V} + 1.5 \, \text{V} = 3.0 \, \text{V} \)

\(\displaystyle r_{\text{series}} = r_{\text{parallel}} + r_{\text{parallel}} = 0.25 \, \Omega + 0.25 \, \Omega = 0.5 \, \Omega \)

The equivalent EMF of the mixed series-parallel combination is 3.0V and the equivalent internal resistance is 0.5Ω.

Problem 6: A load of 5Ω is connected across two cells in series, each with an EMF of 1.5V and internal resistance of 0.2Ω. Calculate the power delivered to the load.

Solution: Equivalent EMF and internal resistance for cells in series:

\(\displaystyle E_{\text{eq}} = 1.5 \, \text{V} + 1.5 \, \text{V} = 3.0 \, \text{V} \) \(\displaystyle r_{\text{eq}} = 0.2 \, \Omega + 0.2 \, \Omega = 0.4 \, \Omega \)

Total resistance in the circuit:

\(\displaystyle R_{\text{total}} = R_{\text{load}} + r_{\text{eq}} = 5 \, \Omega + 0.4 \, \Omega = 5.4 \, \Omega \)

Current through the circuit:

\(\displaystyle I = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{3.0 \, \text{V}}{5.4 \, \Omega} = 0.556 \, \text{A} \)

Power delivered to the load:

\(\displaystyle P = I^2 \times R_{\text{load}} = (0.556 \, \text{A})^2 \times 5 \, \Omega \)

\(\displaystyle P = 0.309 \, \text{A}^2 \times 5 \, \Omega = 1.545 \, \text {W} \)

The power delivered to the load is 1.545W.

What is the difference between cells connected in series and cells connected in parallel?

In a series connection, the cells are connected end-to-end, so the total voltage is the sum of the individual voltages, but the current remains the same through all cells. In a parallel connection, the cells are connected with all positive terminals together and all negative terminals together, resulting in the total current being the sum of the individual currents, but the voltage remains the same across all cells.

What are the advantages of connecting cells in series?

Connecting cells in series increases the total voltage output, which is useful when a higher voltage is needed to power a device. This setup is often used in applications where the voltage requirement exceeds what a single cell can provide, such as in flashlights or certain electronic devices.

What are the benefits of connecting cells in parallel?

Connecting cells in parallel increases the total current capacity and extends the overall battery life. This configuration is beneficial when a device requires more current than a single cell can provide or when longer operating time is needed, such as in large-capacity battery packs for electric vehicles or backup power supplies.

How does the internal resistance of cells affect their performance in series and parallel combinations?

In a series combination, the total internal resistance is the sum of the internal resistances of all cells, which can lead to a significant voltage drop under load and reduce efficiency. In a parallel combination, the effective internal resistance decreases, improving current delivery and reducing the voltage drop, which makes the parallel combination more efficient for high-current applications.

Can you combine cells in both series and parallel configurations?

Yes, cells can be combined in both series and parallel configurations to achieve a desired voltage and current output. This is often done in battery packs for applications requiring both high voltage and high current, such as in electric vehicles or large-scale energy storage systems.

What are some common applications of series and parallel cell combinations?

Series combinations are commonly used in applications like flashlights, remote controls, and certain electronic devices where higher voltage is needed. Parallel combinations are used in applications such as power banks, electric vehicles, and backup power supplies where higher current capacity and longer battery life are essential.

How can you determine the total voltage and current of a combination of cells?

For cells in series, the total voltage is the sum of the individual cell voltages, while the current remains the same as that of a single cell. For cells in parallel, the total current is the sum of the individual cell currents, while the voltage remains the same as that of a single cell. By combining these principles, you can calculate the overall voltage and current for any combination of series and parallel connections.

Leave a Comment Cancel reply

Save my name, email, and website in this browser for the next time I comment.

• Cells in Series and Parallel

As we know the most frequently used method to connect electrical components is Series Connection and Parallel Connection. Since the cell is an important part of an electric circuit. To know more about Cells, Series Connection and parallel Connection explore the article!

Suggested Videos

Cells generate electricity  and also derives chemical reactions.  One or more electrochemical cells are batteries . Every cell has two terminals namely:

• Anode: Anode is the terminal from where the current flows in from out i.e. it provides an incoming channel for the current to enter the circuit or the device.
• Cathode: Cathode is the terminal from where the current flows out i.e. it provides an outgoing current flow from the circuit or the device.

Learn more about  Electric Charge here in detail

There are two simplest ways for cell connectivity  are as follows:

• Series Connection: Series connection is the connectivity of the components in a sequential array of components.
• Parallel Connection: Parallel connection is the connectivity of the components alongside to other components.

Cells in Series Connection

In series, cells are joined end to end so that the same current flows through each cell.  In case if the cells are connected in series the emf of the battery is connected to the sum of the emf of the individual cells. Suppose we have multiple cells and they are arranged in such a way that the positive terminal of one cell is connected to the negative terminal of the another and then again the negative terminal is connected to the positive terminal and so on, then we can that the cell is connected in series.

Browse more Topics under Current Electricity

• Electric Current
• Electrical Energy and Power
• Resistivity of Various Materials
• Temperature Dependence of Resistivity
• Drift of Electrons and the Origin of Resistivity
• Combination of Resistors – Series and Parallel
• Atmospheric Electricity and Kirchhoff’s Law
• Wheatstone Bridge, Meter Bridge and Potentiometer
• Cells, EMF, Internal Resistance

Equivalent EMF/Resistance of Cells in Series

If E is the overall emf of the battery combined with  n number cells and E 1,  E 2,  E 3  , E n  are the emfs of individual cells.

Then   E 1  + E 2  +   E 3  + …….E n 

Similarly, if r 1 , r 2 , r 3 , r n  are the internal resistances of individual cells, then the internal resistance of the battery will be equal to the sum of the internal resistance of the individual cells i.e.

r = r 1  + r 2 +  r 3  +  r n

Cells in Parallel Connection

Cells are in parallel combination if the current is divided among various cells. In a parallel combination, all the positive terminal are connected together and all the negative terminal are connected together.

Equivalent EMF/Resistance of Cells in Parallel

If emf of each cell is identical, then the emf of the battery combined with n numbers of cells connected in parallel is equal to the emf of each cell. The resultant internal resistance of the combination is,

r =  \(( \frac{1}{r_1} \) + \( \frac{1}{r_2 } \) +  \( \frac{1}{ r_3} \) +……..  \( \frac{1}{r_n} \) ) -1

Equivalent EMF/Resistance of Cells in Series and Parallel

Assume the emf of each cell is E and internal resistance of each cell is r. As n  numbers of cells are connected in each series, the emf of each series, as well as the battery, will be nE.  The equivalent resistance of the series is nr. As, m the number of series connected in parallel equivalent internal resistance of that series and parallel battery is nr/m.

Solved Questions For You

Q. The internal resistance of a cell of emf 1.5 V, if it can deliver a maximum current of 3 A is,

Solution: A. For maximum amount, load resistance = 0

r = \( \frac{E}{I} \)

= \( \frac{1.5}{3} \)

Q.2 For a given cell, its terminal voltage depends on

• External resistance, Internal Resistance
• External resistance
• Internal Resistance
• None of these

Solution: A. Inside the cell, the energy is put into the circuit by the cell, but some of this energy is out by the internal resistor. So the potential difference available to the rest of the circuit is the emf minus the potential difference lost inside the cell.

Customize your course in 30 seconds

Which class are you in.

Current Electricity

• Electrical Insulators
• Electric Car
• Electrical Conductors
• Electric Power
• Cells, EMF and Internal Resistance
• Atmospheric Electricity and Kirchhoff’s Law
• Combination of Resistors in Series and Parallel

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Download the App

We think you are located in South Africa . Is this correct?

• Yes, I reside in South Africa
• Change country/curriculum

We use this information to present the correct curriculum and to personalise content to better meet the needs of our users.

11.1 Series circuits

Series and parallel circuits.

Chapter overview

This chapter builds on the Gr 6 and 7 electric circuits work, and the previous chapter of this book. Up until now, we have only been looking at simple circuits. We will now examine the concept of series and parallel circuits. We will look at the difference between these two set-ups in circuits, specifically looking at the effects of adding resistors in series or in parallel and observing the change in brightness of bulbs. The use of ammeters has also been included in this chapter. However, if you do not have these instruments, you can simply do a qualitative study, using the brightness of the bulbs.

You can also use the PhET simulations where learners can build their own circuits and test them out, observing the effects when they add or remove various components. These simulations will run directly within your browser from our website, www.curious.org.za . Here is a link to a guide (in pdf format) written by PhET in the use of some of the electric circuit simulations: phet.colorado.edu/files/teachers-guide/circuit-construction-kit-dc-guide.pdf

3.1 Series circuits (2.5 hours)

3.2 Parallel circuits (3 hours)

3.3 Other output devices (0.5 hours)

• Are there different types of electric circuits?
• If all the light bulbs in a house are part of the same circuit, how can you switch one light off without the rest also turning off?
• What is a series circuit?
• What is a parallel circuit?
• What happens when you connect more components in series or in parallel?

In the last chapter, and in Gr 6 and 7, we have been looking at electric circuits. These have mostly been series circuits. What does this mean? And how else can a circuit be arranged?

Series circuits

A series circuit is one in which there is only one pathway for the electric current to follow. The components are arranged one after another in a single pathway. When we connect the components we say that they are connected in series . We have already seen examples of series circuits in the last chapter.

An ammeter is a measuring device used to measure the electric current in the circuit. It is connected into the circuit in series. The current is measured in amperes (A).

The ampere is named after André-Marie Ampère (1775-1836), a French mathematician and physicist. He is considered the father of electrodynamics, which is the study of the effect of electromagnetic forces between electric charges and currents.

What is the symbol for an ammeter? Draw it here.

Do you think that an ammeter would have a high resistance or a low resistance to the current? Explain your choice.

Ammeters have an extremely low resistance because they must not alter the current they are measuring in any way.

The ampere is often shortened to 'amp'.

A series circuit only provides one pathway for the electrons to follow. Let's investigate what happens when we increase the resistance in a series circuit.

What happens when we add more resistors in series?

The aim of this investigation is to show the learners that adding more resistors in series causes the overall resistance of the circuit to increase and that this reduces the current strength.

AIM: To investigate the effect of adding resistors to a series circuit.

This is a good opportunity for group work if you have enough equipment, but make sure that each learner is able to connect an ammeter correctly and is able to read the ammeter scale accurately. If you do not have sufficient equipment for all the learners, you can do this experiment as a demonstration. Perhaps give several learners an opportunity to come up to the front and help to connect the ammeters. If you do not have any ammeters then you can use the brightness of the bulbs to indicate current strength. The larger the current, the brighter the bulb will glow. This means that if the bulb glows brightly, it must have a large current moving through it. If the bulb is dimmer, it means that there is a smaller current flowing through it.

If you do not have the physical apparatus for this investigation but you do have internet access, use the PhET simulations found here: http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc

The simulations are also useful because the ammeters (and voltmeters) commonly used in school laboratories are often not calibrated correctly or not serviced regularly and so often give slightly inaccurate results.

HYPOTHESIS: Write a hypothesis for this investigation.

This is a learner-dependent answer. The hypothesis should relate the dependent and independent variables and make a prediction. The dependent variable will change as the independent variable is changed. Here is an example of a possible answer:

As the number of resistors increases, the current strength decreases.

MATERIALS AND APPARATUS:

• 1,5 V cells
• 3 torch bulbs
• insulated copper conducting wires

It is important that the torch bulbs have the same resistance and are not randomly selected. The switch is not an essential part of this investigation. It can be left out of the circuit.

Construct the circuit with the cell, the ammeter, 1 bulb and the switch in series.

Note how brightly the bulb is shining and write down the ammeter reading. Draw a circuit diagram.

Note how brightly the bulbs are shining and write down the ammeter reading. Draw a circuit diagram.

Note how brightly the bulbs are shining and write down the ammeter reading. Draw a circuit diagram for the last circuit you built.

Complete the table:

The brightness of the bulbs is a qualitative comparison. Learners should use "bright, brighter, brightest" as a way to describe the glowing bulbs. The graph should show the quantitative data of the ammeter reading and the number of bulbs. If you do not have an ammeter to take readings, either do not draw a graph, or change the graph to a bar graph which has bright, brighter, brightest as the values on the y-axis. This is not a particularly useful graph but will give the learners a chance to practice drawing a bar graph and give them a visual representation of the decrease in current strength as the number of bulbs increases.

Draw a graph to show the relationship between the number of bulbs and the current.

These results are an example of possible results. The actual results obtained by the learners will differ but the trend should be similar. As the number of bulbs in series increases, both the ammeter reading and bulb brightness should decrease.

Using standard ammeters may not give perfect results and if the bulbs are allowed to heat up too much in between adding more bulbs, their resistance will be higher. It is important that the learners see a downward trend.

What happened to the brightness of the bulbs as the number of bulbs increased?

The bulbs got dimmer as more bulbs were added.

When you had two bulbs, did they glow with the same brightness, or was one brighter than the other?

The bulbs glowed with the same brightness.

When you had three bulbs, did they glow the same as each other or was one brighter than the others?

What do your answers to the previous questions tell you about the current in the series circuit?

If all the bulbs glow the same, it means that they all experience the same current. This means that the current is the same everywhere in a series circuit.

What happened to the reading on the ammeter as you added more bulbs in series?

The ammeter reading decreased.

CONCLUSION:

Based on your answers, what happened to the current when more bulbs were added in series?

As more bulbs were added, the current decreased.

Is your hypothesis accepted or rejected?

This answer will depend on the hypothesis written by the learner at the start of the investigation.

As more resistors are added in series, the total resistance of the circuit increases. As the total resistance increases, the current strength decreases. What would happen if we increased the number of cells connected in series? Would the current become larger or smaller? Let's investigate.

How does adding more cells in series affect the current?

This investigation will show that adding more cells in series increases the current strength. Be careful with this activity because if you do not have enough resistance in your circuit, you can damage the torch light bulbs. Use at least two torch light bulbs or a torch light bulb and a resistor in order to keep the resistance high enough. If you have ammeters, you can use quantitative data to show that adding more cells in series increases the current strength. If you do not have ammeters, then use the brightness of the bulbs as qualitative data. Use terms such as dim, bright, brightest. The learners will not be able to draw effective graphs with the qualitative data but you could give them the example data in the teacher's guide and ask them to draw a line graph if they need practice.

AIM: To investigate the effect of increasing the number of cells connected in series on the electric current strength.

HYPOTHESIS: Write a hypothesis for this investigation. Remember to mention how the increase in the number of cells will affect the current strength.

This answer is learner-dependant. They must mention how the dependent variable will be affected by the independent variable. Remember that the hypothesis does not need to be factually correct. They will prove or disprove it by completing the investigation. Here is an example of a possible hypothesis: As the number of cells connected in series increases, the current strength increases.

MATERIALS AND APPARATUS

• three 1,5 V cells
• 2 torch light bulbs (or 1 torch light bulb and one resistor)

Observe the brightness of the bulbs and record the ammeter reading in the table of results. Draw a circuit diagram.

Record the ammeter reading in the table of results. Draw a circuit diagram.

These results are example results. The actual results obtained by the learners will differ but the trend should be similar. As the number of cells increases, both the ammeter reading and the bulb brightness should increase.

As the number of cells connected in series increases, so does the current strength.

This answer depends on the learner's original hypothesis.

We have seen that increasing the number of cells in series increases the current, but increasing the number of resistors decreases the current.

We will now investigate the current strength at different points in a series circuit.

Testing the current strength

The first investigation looked at the decrease in current strength when more resistors were connected in series. This investigation confirms that the current strength is the same at all points in a series circuit. This is an optional investigation. This can be a demonstration if your equipment is limited. This is a good opportunity for group work, but make sure that each learner is able to connect an ammeter correctly and understands the ammeter scale.

INVESTIGATIVE QUESTION: Is the current strength the same at all points in a series circuit?

HYPOTHESIS: Write a hypothesis for this investigation. What do you think will happen in this investigation?

This is a learner-dependant answer. Learners need to mention the independent and dependent variables. The dependent variable will change as the independent variable is changed.

Here are two examples of an acceptable hypothesis:

• The current will be different at different points in the circuit OR
• The current will be the same at different points in the circuit.
• insulated copper connecting wires.
• two 1,5V cells
• two torch light bulbs

Measure the current strength using the ammeter. Draw a circuit diagram of this set up.

Complete the following table:

The ammeter readings should be the same at any point in the series circuit.

CONCLUSIONS:

Write a conclusion based on your results.

The current strength is the same at any point in a series circuit.

Is your hypothesis true or false?

In a series circuit, there is only one pathway for the electrons to move through. The current strength is the same everywhere in that pathway.

What have we learned about series circuits?

There is only one pathway for the electrons to follow.

The current flows at the samestrength everywhere in a series circuit, because there is only one pathway. We say that the current is the same at all points in the circuit.

If you add more resistors in series, the current in the whole circuit decreases .

Why does the current stay the same at all points? Let's think about how electric current moves through a circuit. Do you remember that we spoke about the delocalised electrons in metals in the last chapter?

Animation showing the movement of electrons. http://www.schoolphysics.co.uk/animations/Electric_current/index.html

The electrons in a conductor normally drift in various different directions within a metal, as shown in the diagram.

When we build a closed circuit with a cell as an energy source, the electrons will all begin to move towards the positive side of the cell. The rate at which the electrons move, is determined by the resistance of the conductor.

There are electrons everywhere in the conducting wires and electrical components. When the circuit is closed, all the electrons start moving in the same general direction at the same time . This is why a light bulb turns on immediately when you close the switch.

Flip the switch and watch the electrons with this simulation. [link] http://phet.colorado.edu/en/simulation/signal-circuit

The simulation identified in the visit box helps to demonstrate how a light bulb turns on immediately when the switch is turned on.

In a series circuit, all the electrons travel through every component and wire as they travel through the circuit. All the electrons experience the same resistance and so they all move at the same rate.

This means that in the diagram below, the readings on all three ammeters will be the same, so: A 1 = A 2 = A 3

Parallel circuits

• parallel circuit

Parallel circuits offer more than one pathway for the electrons to follow. When constructing a parallel circuit, we say that components are connected in parallel .

Look at the diagram which shows how two light bulbs are connected in parallel.

How can you tell whether or not a circuit is connected in series or in parallel? Let's look at some circuit diagrams to tell the difference.

Watch a video that explains the difference between series and parallel circuits

Series or parallel?

INSTRUCTIONS:

Look at the following circuits and decide which are in series and which are in parallel. The series circuits will only offer one pathway, but the parallel circuits will have more than one pathway for the electrons to follow.

Let's investigate how parallel circuits work.

How does adding resistors in parallel affect the current strength?

This investigation will show the learners that increasing the number of resistors in parallel to each other, causes the overall resistance of the circuit to decrease and the current strength to increase. There is no need to discuss how to calculate the effective resistance of a parallel circuit. The learners just need a qualitative understanding.

AIM: To investigate the effect of adding resistors in parallel on the current strength.

If you do not have physical apparatus for this investigation but you do have internet access, use the PhET simulations found here: http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc

This is a learner dependant answer. Learners need to mention the independent and dependent variables. The dependent variable will change as the independent variable is changed.

• As more bulbs are added in parallel, the current strength will decrease OR
• As more bulbs are added in parallel, the current strength will increase.
• three identical torch bulbs

It is important that the torch bulbs are the same resistance and not randomly selected. The switch and ammeter are not strictly necessary for this experiment. They can be left out if you don't have enough switches or ammeters.

Note how brightly the bulb is shining and record the ammeter reading. Draw a diagram of your circuit.

The brightness of the bulbs is a qualitative description. The learners should use "bright, brighter, brightest" in order to describe the glowing bulbs.

The graph will show the relationship between the main current (reading on the ammeter) and the number of bulbs connected in parallel. As more bulbs are connected in parallel, the current strength should increase because the overall resistance of the circuit decreases. This means that the graph should be a straight line with an increasing trend. Standard ammeters may not be accurate enough to produce a perfectly straight line. This is not as important as seeing the upward trend.

These results are just an example. The actual results will depend on the circuit set up by the learner.

The bulbs got brighter as more bulbs were added.

When you had two bulbs, did they glow with the same brightness or was one brighter than the other?

When you had three bulbs, did they glow the same brightness or was one brighter than the others?

What do your answers to the previous questions tell you about the current in the parallel branches of the circuit?

As all the bulbs are identical, if they all glow the same brightness, then they all experience the same current. This means that the current is the same in each branch.

What happened to the reading on the ammeter as you added more bulbs in parallel?

The ammeter reading increased.

Based on your answers, what happened to the current when more bulbs were added in parallel?

As more bulbs were added, the current increased.

As more resistors are added in parallel, the total current strength increases. The overall resistance of the circuit must therefore have decreased. The current in each light bulb was the same because all the bulbs glowed with the same brightness. This tells us that the current of electrons must have split up and moved through each of the branches.

We can also connect cells in parallel. What would happen if we increased the number of cells connected in parallel? Would the current get stronger or weaker?

What happens to the current strength when cells are connected in parallel?

AIM : To investigate how increasing the number of cells connected in parallel affects the current strength in a circuit.

This is a learner-dependent answer. Learners need to identify the independent and dependent variables. The dependent variable will change as the independent variable is changed.

• As more cells are added in parallel, the current strength will decrease OR
• As more cells are added in parallel, the current strength will increase.
• three 1,5V cells
• one torch light bulb

The ammeter is not essential to the experiment. The brightness of the bulb can serve as a qualitative measure.

Set up a circuit which has one cell, the ammeter and the torch light bulb in series with each other. Draw a circuit diagram of your circuit.

Connect another cell in parallel with the first cell. To connect the second cell in parallel, connect a wire from the positive terminal of the first cell to the positive terminal of the second cell. Connect another wire between the negative terminal of the first battery and the negative terminal of the second battery. Draw a circuit diagram of your circuit.

Connect a third cell in parallel to the other two cells. Draw a circuit diagram of your circuit.

The brightness of the bulbs is a qualitative description. The learners should use "bright, brighter, brightest" in order to describe the glowing bulbs. The ammeter readings should stay the same.

What did you notice about the brightness of the bulbs?

The brightness of the bulbs should not change.

What did you notice about the ammeter readings?

The ammeter readings are the same.

What conclusion can you draw from your results?

Adding cells in parallel does not change the overall current strength.

Adding cells in parallel has no overall effect on the current strength. The current strength stays the same if you add cells in parallel.

We saw that the current strength increased when bulbs were connected in parallel. However, we were only testing the current strength at one point in the parallel circuit. How does the current compare in the different pathways of the circuit? Let's do an investigation to find out.

The first investigation looked at the increase in current strength when more resistors were connected in parallel. This investigation confirms that the current strength is not the same at all points in a parallel circuit. This is a good opportunity for group work, but make sure that each learner is able to connect and read an ammeter correctly. If you do not have enough equipment to allow for small groups to build the circuits, you can rather use this investigation as a demonstration. Perhaps give several learners an opportunity to come up to the front and help to connect the ammeters.

INVESTIGATIVE QUESTION: Is the current strength equal at all points in a parallel circuit?

• three identical torch light bulbs
• Set up a parallel circuit with two cells in series with each other and three torch light bulbs in parallel with each other.
• Insert an ammeter in series between the cells and the first pathway, as shown in the diagram.

• Measure the current strength using the ammeter.
• Remove the ammeter and close the circuit again.
• Insert the ammeter in series in the first pathway.

• Insert the ammeter in series in the second pathway.

• Insert the ammeter, in series, in the third pathway.

• Insert the ammeter in series between the first pathway and the cells on the opposite side to the first reading.

These are some example readings to show the trend:

If you do not use identical bulbs, then the readings in each of the branches will not be identical, but they will add up to reading in the main branch. If possible, it is worthwhile to demonstrate this to learners.

The current strength is not the same at all points in a parallel circuit. If the bulbs are identical, then the current is the same in the three branches, however the current in the main part of the circuit is greater than that in the individual pathways. The current in the main part of the circuit is the sum of the currents in the pathways.

What have we learned about parallel circuits?

There is more than one pathway for the current to follow.

• The current divides between the different branches so that each branch gets some of the current. As the torch bulbs in each branch in our example were identical, the current divided equally between them.

If you add more resistors in parallel, the total current supplied by the cell in the circuit increases .

Why does the current divide when offered an alternative pathway?

Imagine that you are sitting in a school hall during assembly. You are bored and waiting for it to end so that you can go out to break to chat to your friends. There is only one exit from the hall. When you are dismissed, everyone has to exit through the same door. It takes a while because only some learners can leave at a time.

Now imagine that there is a second door that is the same as the first door. Now you and your friends have a choice of which door to go through. The speed at which the learners exit the hall will increase and some of you will exit through the first door while others will exit through the second door. No one can go through both doors at the same time.

This is similar to the way current behaves when in a parallel circuit. As the electrons approach the branch in the circuit, some electrons will take the first path and others will take the other path. The current is divided between the two pathways.

In the following circuit A1 = A4 and A1 = A2 + A3 and A4 = A2 + A3

We have looked at how resistors and cells behave in series and parallel circuits. Let's look at how different metals conduct electricity. All conductors have some resistance in a circuit. Are some metals better conductors of electricity than others?

Let's have a look at which metals offer more resistance than others to the flow of charge (current) through an electric circuit .

Which metals offer the most resistance?

This activity only compares the effect of the type of material on resistance. The other factors that affect resistance will be covered in the Grade 9 Energy and Change syllabus.

Each metal will have a particular resistance based on the resistivity. You do not need to measure the resistance of each metal, all that is required is a qualitative description of the light bulb. The brighter the light bulb, the higher the current. If there is a high current it means that there is little resistance. So the brighter the bulb glows, the less resistance offered by the metal wire. The learners may make small mistakes if the brightness of the bulbs is difficult to distinguish.

Use whichever metal wires you have available. Try to get copper and nickel. You could twist aluminium foil into a wire (just make sure it is the same length and approximate thickness as the other metals). Aluminium wire will often ignite if placed in a circuit so test it beforehand and make sure that it does not get too hot. If you use the materials listed below, then nichrome will have the highest resistance, followed by zinc, then aluminium and copper has the lowest resistance of the four.

• torch light bulb
• insulated copper wires
• lengths of copper, aluminium, zinc and nichrome wire
• crocodile clips (if available)

The actual length of wire that you use is not important, but they should all be the same length and thickness. If you cannot find these metals, any other combination of metals can be successfully used.

INSTRUCTIONS

• Build a circuit with the cell and the torch light bulb and leave a gap for the metal to be tested. You can use crocodile clips at the end of each piece of metal for easy insertion.
• Insert each metal into the circuit (one at a time).

Observe the brightness of the bulb.

Draw a circuit diagram of your apparatus.

An example circuit diagram with the break in the circuit where metals are to be tested shown on the left.

Why can we use the brightness of the bulb to qualitatively measure resistance?

High resistance opposes the movement of electrons, decreasing the current so there is less energy for the light bulb. The higher resistance wire will cause the bulb to be dimmer than the lower resistance wire.

List the metals in order of increasing resistance.

Copper, aluminium, zinc and nichrome.

Why do you think copper is used for connecting wires in electrical circuits?

Copper has an extremely low resistance, and so has a minimal effect on the overall resistance of the circuit. Other materials would add to the overall resistance of the circuit, decreasing the maximum possible current in that circuit.

There are several factors which influence the amount of resistance a material offers to an electric current. We have seen that the type of material is one of those factors.

In Gr 9 we will look at the other factors that influence resistance. If you want to see the content in other grades, remember that you can visit www.curious.org.za

Other output devices

Light bulbs are not the only devices used in electrical circuits. Devices that use electrical energy to function, including light bulbs, are called output devices. Let's look at some other common examples of output devices.

LEDs (Light-Emitting Diodes)

LEDs are widely used electronic devices. They are small lights but they do not have a filament like an incandescent bulb has. They therefore cannot burn out, as there is no filament to wear out, and they do not get as hot. LEDs are used in electronic timepieces, high definition televisions and many other applications. Larger LEDs are also replacing traditional light bulbs in many homes because they do not use as much electricity. They last longer than incandescent bulbs and are more efficient.

Watch this video about the history of the LED

In the last chapter, we looked at the energy transfers in an electrical system. We will now represent energy transfer within electrical systems in a different way. We will apply this new representation to the difference between energy outputs in an LED and an incandescent light bulb.

Video on drawing a basic Sankey diagram.

Sankey diagrams

Sankey diagrams were first introduced in the Gr 7 CAPS workbook as a way of representing the transfers of energy within a system, specifically focusing on the transfer of input energy to useful and wasted output energy. They provide a very clear illustration of the process. This links back to the previous chapter to reinforce learning.

You might have drawn Sankey diagrams in Grade 7. If not, here is some quick revision.

In an energy system, input energy is transferred to useful output energy and wasted output energy. A Sankey diagram is a visual and proportional representation of the energy transfers that happen in a system.

For example, a kettle uses about 2000 J of input energy, but only about 1400 J is used to heat the water. The remaining 600 J is wasted as sound. Here is the Sankey diagram to represent the energy transfer.

Remember that energy is measured in joules (J).

We will now compare an LED with an incandescent light bulb.

Draw a Sankey diagram for an LED if the input energy is 100 J, 75 J of energy is used to produce light and the rest is lost as heat.

Draw a Sankey diagram for a filament light bulb if the input energy is 100 J, the wasted heat energy is 80 J and the rest produces light.

Which bulb do you think is more efficient? Explain your answer.

The LED bulb is more efficient as more of the input energy is transferred to useful output (light) than is wasted as heat. In the filament light bulb, much more energy is wasted as heat.

Can you think of any other output devices? Make a list of as many as you can.

Some are: motors, buzzers, beepers.

History of electricity production

• Work in groups of three or four.
• Research the history of electricity production: How was electricity discovered and how did electricity become widely used?
• Create a basic timeline for the discovery of electricity and it's production.

The timeline does not need to be too specific. We want learners to realise that this was not an overnight discovery, but involved many people over a significant time. Here are some pertinent facts. This list is not complete and not all of the dates are necessary. Another useful resource is available here: http://www.timetoast.com/timelines/118814

• 600 BC - Discovery that amber, rubbed with silk, would attract light objects such as feathers
• 1600 AD - William Gerbert coined the term electricity. He was the first to make a link between magnetism and electricity
• 1700s - Wimshurst machine, used to generate static electricity
• 1752 - Benjamin Franklin proved that lightning was a form of electricity
• 1800s - Sir Humphrey Davey discovered electrolysis; Volta created the first simple cell
• 1831 - Michael Faraday demonstrated electromagnetic induction
• 1825 - Ampere published his theories on electricity and magnetism. The unit of current, the ampere, is named after him
• 1827 - George Ohm published his study of electricity. The unit of resistance, the ohm, is named after him
• 1831 - Charles Wheatstone and William Fothergill created the telegraph machine
• 1870 - Thomas Edison built a DC generator
• 1876 - Alexander Graham Bell invented the telephone which uses electricity to transfer speech
• 1878 - Joseph Swan demonstrated an electric light bulb
• 1880s - Nikola Tesla developed an AC generator
• 1881 - The first British public electricity generator was built in Surrey
• 1883 - Magnus Volk built the first electric train line
• 1896 - Nikola Tesla established hydroelectric power plants in America
• 1905 - Albert Einstein demonstrated the photoelectric effect which led to the production of photovoltaic cells

An electricity timeline animation. http://resources.schoolscience.co.uk/britishenergy/14-16/index.htm

Write a short paragraph describing the career. Include information on how one can study or prepare for your chosen career.

The Eskom website has information regarding various careers and the internet has many different sources.

• A series circuit has only one pathway for the electrons to travel through.
• A parallel circuit has more than one pathway for the electrons to travel through.
• In a series circuit, the current is the same at all points in the circuit.
• In a series circuit, the resistance increases as more resistors are added in series.
• In a parallel circuit, the current splits between the available paths.
• In a parallel circuit, the resistance decreases as more resistors are added in parallel.

Revision questions

Look at the three circuit diagrams. Rank the circuits from brightest bulb to dimmest bulbs. [3 marks]

Brightest, bright, dim

Explain your choices in the previous question. [5 marks]

The first circuit has the brightest bulb because it has the least resistance and so it has the highest current. The third circuit has the highest resistance because it has two resistors connected in series with the light bulb. The more resistors connected in series, the higher the resistance and the lower the current.

Look at the three circuit diagrams. Rank the circuits from brightest bulb(s) to dimmest bulb(s). [3 marks]

Dimmest, bright, brightest

The third circuit will have the brightest bulb because adding resistors in parallel lowers the overall resistance in the circuit. The current is therefore greater and the bulb shines brighter. The first circuit is the dimmest because it has no parallel branches, and so offers the highest resistance.

Look at the circuit diagram below. Each light bulb is identical.

Is this a series or parallel circuit? Explain your answer. [2 mark]

How do the brightness of bulbs A, B and C compare? (which is the brightest?) [3 marks]

What would happen to the brightness of the bulbs if the switch was opened? Explain your answer. [5 marks]

This circuit has both series components (the cell and bulb A are in series) and a parallel branch consisting of bulb B and C.

Bulb A is the brightest, Bulbs B and C would have the same brightness as each other.

If switch S is opened, then bulb C will not glow. Bulbs A and B would now have equal brightness but they would be dimmer than when the switch was closed. A and B would now be in series with each other and there is no parallel branch. The overall resistance of the circuit would therefore be higher, resulting in a smaller current.

Study the following diagram.

What is the relationship between the ammeter readings on A1 and A4? In other words, how do the current strengths compare at these points in the circuit? Explain your answer. [3 marks]

What is the relationship between the ammeter readings on A1, A2 and A3? In other words, how do the current strengths compare at these points in the circuit? Explain your answer. [3 marks]

A1 = A4. The total current flows through the circuit at both of these points.

A1 = A2 + A3. The current splits between parallel branches in a circuit.

Total [38 marks]

Series And Parallel Combinations Of Cells

• Current and Electricity - An introduction

Learning Objectives

• Understand the concept of series and parallel combinations of cells
• Learn how current flows in series and parallel circuits
• Understand the calculated results of voltage and current in different configurations
• Apply Ohm’s Law to solve circuit problems involving series and parallel combinations of cells

Introduction

• In electronic circuits, multiple cells or batteries are often connected together to provide the required voltage and current.
• There are two common methods of connecting cells in circuits: series and parallel combinations.
• In this lecture, we will explore the behavior of cells connected in series and parallel, and analyze their effect on voltage and current.

Series Combination of Cells

• The positive terminal of one cell is connected to the negative terminal of the next cell.
• The total voltage of the combination is the sum of individual cell voltages.
• The total emf (electromotive force) of the series combination is given by: Emf_total = Emf_1 + Emf_2 + Emf_3 + ...

Series Combination of Cells (contd.)

• The current flowing through each cell in a series combination is the same.
• The total internal resistance of the series combination is the sum of individual internal resistances.
• The total internal resistance (R_total) is given by: R_total = R_1 + R_2 + R_3 + ...
• The total resistance of the series combination is also equal to the sum of individual resistances of the cells.

Example - Series Combination of Cells

• Consider three cells with emf values of 1.5V, 2V, and 3V connected in series.
• The internal resistances of the cells are 0.1Ω, 0.2Ω, and 0.3Ω respectively.
• What is the total emf and internal resistance of the series combination?
• Emf_total = 1.5V + 2V + 3V = 6.5V
• R_total = 0.1Ω + 0.2Ω + 0.3Ω = 0.6Ω

Parallel Combination of Cells

• The positive terminals of all cells are connected together and similarly, the negative terminals are connected together.
• The total emf (Emf_total) of the parallel combination is the same as the emf of each individual cell.
• The total internal resistance (R_total) of the parallel combination is given by: 1 / R_total = 1 / R_1 + 1 / R_2 + 1 / R_3 + ...

Parallel Combination of Cells (contd.)

• The current divides among the cells in a parallel combination.
• The sum of currents through each cell is equal to the total current.
• The resistance of the total parallel combination (R_total) is less than the smallest resistance of any individual cell.

Example - Parallel Combination of Cells

• Consider three cells with emf values of 1.5V, 2V, and 3V connected in parallel.
• Calculate the total emf and internal resistance of the parallel combination.
• Emf_total = Emf_1 = Emf_2 = Emf_3 = 1.5V
• 1 / R_total = 1 / R_1 + 1 / R_2 + 1 / R_3
• 1 / R_total = 1 / 0.1Ω + 1 / 0.2Ω + 1 / 0.3Ω
• R_total = 0.1667Ω

Voltage and Current in Series and Parallel Combinations

• In series combination, the current is the same through all cells, while the voltage divides among the cells.
• In parallel combination, the voltage is the same across all cells, while the current divides among the cells.
• Ohm’s Law can be applied to calculate the current, voltage, and resistance in the series and parallel combinations.
• The total resistance in series combination is the sum of individual resistances, while in parallel combination it is less than the smallest resistance.
• Emf_total = Emf_1 + Emf_2 + Emf_3 + ...
• R_total = R_1 + R_2 + R_3 + ...
• 1 / R_total = 1 / R_1 + 1 / R_2 + 1 / R_3 + ...
• If the current flowing through each cell is 0.5A, what is the total current in the series combination?
• The current flowing through each cell is the same, so the total current is also 0.5A.
• If the current flowing through each cell is 1A, what is the total current in the parallel combination?
• The total current is the sum of currents flowing through each cell, which is 1A + 1A + 1A = 3A.
• If the internal resistances of the cells are 0.1Ω, 0.2Ω, and 0.3Ω respectively, what is the total resistance of the series combination?
• The total resistance is the sum of individual resistances, which is 0.1Ω + 0.2Ω + 0.3Ω = 0.6Ω.
• If the internal resistances of the cells are 0.1Ω, 0.2Ω, and 0.3Ω respectively, what is the total resistance of the parallel combination?
• The total resistance is given by the formula: 1 / R_total = 1 / R_1 + 1 / R_2 + 1 / R_3
• Substituting the values, we get: 1 / R_total = 1 / 0.1Ω + 1 / 0.2Ω + 1 / 0.3Ω 1 / R_total = 10Ω + 5Ω + 3.333Ω 1 / R_total = 18.333Ω
• Therefore, the total resistance is: R_total = 1 / 18.333Ω ≈ 0.0545Ω
• Cells can be connected in series or parallel combinations to provide the required voltage and current in electronic circuits.
• In series combination, the voltage adds up and the current remains the same.
• In parallel combination, the voltage remains the same and the current adds up.
• The total internal resistance of series combination is the sum of individual internal resistances, while in parallel combination it is less than the smallest resistance.
• Ohm’s Law can be applied to calculate the current, voltage, and resistance in series and parallel combinations.