in a potentiometer experiment the balancing with a cell

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Chapter 14: Problem 163

In a potentiometer experiment, the balancing with a cell is at length $240 \mathrm{~cm}$. On shunting the cell with a resistance of $2 \Omega$, the balancing length becomes 120 $\mathrm{cm} .$ The internal resistance of the cell is (A) $1 \Omega$ (B) $0.5 \Omega$ (C) $4 \Omega$ (D) $2 \Omega$

Short answer, step by step solution, understand the formula for the potential difference in a potentiometer circuit, calculate the potential difference in each case, find the relation between the potential differences, understand the equivalent resistance in a parallel circuit, apply the formula for equivalent resistance in the second case, relate the currents in both cases, solve for the internal resistance, key concepts.

These are the key concepts you need to understand to accurately answer the question.

Understanding Internal Resistance

Balancing length in potentiometers.

In the original experiment, the initial balancing length is noted as 240 cm without any shunt resistor in place.
Once a resistor is connected in parallel, the balancing length changes to 120 cm, indicating a difference in potential drop due to the additional load introduced.

Potential Gradient Essentials

Parallel circuit resistance.

The parallel connection effectively reduces the total resistance.
This changes the potential difference across the cell, thereby altering the balancing length in the potentiometer setup.

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in a potentiometer experiment the balancing with a cell

Most popular questions from this chapter

Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are $3 \%$ each, then error in the value of resistance of the wire is (A) $6 \%$ (B) Zero (C) $1 \%$ (D) $3 \%$

The resistance of a bulb filament is $100 \Omega$ at a temperature $100^{\circ} \mathrm{C}$. If its temperature coefficient of resistance be $0.005 /{ }^{\circ} \mathrm{C}$, its resistance will become $200 \Omega$ at a temperature of (A) $300^{\circ} \mathrm{C}$ (B) $400^{\circ} \mathrm{C}$ (C) $500^{\circ} \mathrm{C}$ (D) $200^{\circ} \mathrm{C}$c

Two conductors have the same resistance at $0^{\circ} \mathrm{C}$ but their temperature coefficients of resistance are $\alpha_{1}$ and $\alpha_{2} .$ The respective temperature coefficients of their series and parallel combinations are nearly (A) $\frac{\alpha_{1}+\alpha_{2}}{2}, \alpha_{1}+\alpha_{2}$ (B) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$ (C) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}$ (D) $\frac{\alpha_{1}+\alpha_{2}}{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of $4 / 3$ and $2 / 3$, then the ratio of the currents passing through the wire will be (A) 3 (B) $1 / 3$ (C) $\underline{8 / 9}$ (D) 2

Figure $14.47$ shows the circuit of a potentiometer. The length of the potentiometer wire $A B$ is $50 \mathrm{~cm}$. The EMF $E_{1}$ of the battery is $4 \mathrm{~V}$, having negligible internal resistance. Value of $R_{1}$ and $R_{2}$ are $15 \Omega$ and $5 \Omega$, respectively. When both the keys are open, the null point is obtained at a distance of $31.25 \mathrm{~cm}$ from $A$, but when both the keys are closed, the balance length reduces to $5 \mathrm{~cm}$ only. Given $R_{A B}=10 \Omega$ The balance length when key $K_{2}$ is open and $K_{1}$ is closed is given by (A) $10.5 \mathrm{~cm}$ (B) $11.5 \mathrm{~cm}$ (C) $12.5 \mathrm{~cm}$ (D) $13.5 \mathrm{~cm}$

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Potentiometer - Principle And Applications

A potentiometer is an adaptable tool used to find out and also compare electricity potentials- mostly in DC currents. Basically, it operates under the assumption that voltage drop on a constant length of wire varies uniformly with its length provided there is a constant current flowing through it, therefore, allowing accurate measurement and not requiring withdrawal of power from the test setup.

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In this article, we will discuss the concept of Potentiometer - Principle And Applications. It is an essential gadget in the learning of current electricity which is important for Class 12, NEET and JEE Main exams. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of fifteen questions have been asked on this concept. And for NEET seven questions were asked from this concept.

What is a Potentiometer?

The potentiometer is a device which does not draw current from the given circuit and still measures the potential difference. It is a device used to measure the e.m.f of a given cell and to compare the e.m.f of cells. It is also used to measure the internal resistance of a given cell

The potentiometer consists of wires of length 5 to 10 meters arranged on a wooden block as parallel strips of wires with 1-meter length each and the ends of wires are joined by thick coppers. The wire has a uniform cross-section and is made up of the same material. A driver circuit that contains a rheostat, key, and a voltage source with internal resistance r. The driver circuit sends a constant current (I) through the wire.

The potential across the wire AB having length L is given as V=IR, Where R is the resistance of the wire AB

Since the driver circuit sends a constant current (I) through the wire So

$V \propto R$

Using $R=\frac{\rho L}{A}$ we can say that $R \propto L$ since area and resistivity are constant.

Therefore we get V is proportional to length. I.e $V \propto L$

The secondary circuit contains cells/resistors whose potential is to be measured. One end is connected to a galvanometer and another end of the galvanometer is connected to a jockey which is moved along the wire to obtain a point where there is no current through the galvanometer. So the potential of the secondary circuit is proportional to the length at which there is no current through the galvanometer. This is how the potential of a circuit is measured using the potentiometer.

Calibration of Potentiometer

In the potentiometer, a battery of known emf E is connected to the secondary circuit. A constant current I is flowing through AB from the driver circuit (that is the circuit above AB). The jockey is slide on potentiometer wire AB to obtain null deflection in the galvanometer. Let l be the length at which the galvanometer shows null deflection.

Since the potential of wire AB (V) is proportional to the length AB(L).

$\text { Similarly } E \quad \propto \quad l$

$\begin{aligned} & \frac{V}{E}=\frac{L}{l} \\ & V=E \frac{L}{l} \end{aligned}$

Thus we obtained the potential of wire AB when a constant current is passing through it. This is known as calibration.

Potential Gradient

The potential difference per unit length of wire i.e $x=\frac{V}{L}$

$ \begin{aligned} & V=i R=\left(\frac{e}{R+R_h+r}\right) R \\ & x=\frac{V}{L}=\frac{e}{\left(R+R_h+r\right)} \frac{R}{L} \end{aligned}$

For the above figure

$l_1$ is the balancing length obtained when a cell with emf $E_1$ is included in the secondary circuit. That is key is at position $1 . l_2$ is the balancing length obtained when cell with emf $E_2$ is included in the secondary circuit. That is key is at position 2.

So since $E \alpha l$ we get $ \frac{E_1}{E_2}=\frac{l_1}{l_2} $

With the help of the above ratio, we can compare the emf of these cells.

Determine the Internal Resistance of a Cell

Note- The cell in the secondary circuit has emf E and internal resistance r

Here $l_1$ is the balancing length obtained when key $\mathrm{K}^{\prime}$ is open that is we include only the cell in the secondary circuit. So corresponding potentials of wire of balancing length $l_1$ is $\mathrm{E}$. And we know that $E \propto l_1 \ldots$ (1)

Similarly, $l_2$ is the balancing length obtained when key $\mathrm{K}^{\prime}$ is closed that is both cell and $R^{\prime}$ is connected in the secondary circuit. So corresponding potentials of wire of balancing length $l_2$ is $\mathrm{V}$.

And we know that

$ V \propto l_2 \ldots \text { (2) } $ or we can say that $ \begin{aligned} & I R^{\prime} \propto l_2 \\ \Rightarrow & \frac{E}{r+R^{\prime}} * R^{\prime} \propto l_2 \ldots \end{aligned} $

So taking the ratio of equation (1) to equation (2), we get

$ \begin{aligned} & \frac{E}{V}= \frac{l_1}{l_2} \\ & \frac{E}{\frac{E R^{\prime}}{r+R^{\prime}}}=\frac{l_1}{l_2} \\ & \Rightarrow \frac{r+R^{\prime}}{R^{\prime}}=\frac{l_1}{l_2} \end{aligned} $

Then the internal resistance is given by $ \begin{aligned} & r=\left(\frac{l_1-l_2}{l_2}\right) R^{\prime} \\ & r=\left(\frac{E}{V}-1\right) R^{\prime} \end{aligned} $

Comparison of Resistances

The balance point is at a length l 1 cm from A when jockey J is plugged in between Y and X, while the balance point is at a length l 2 cm from A when jockey J is plugged in between Y and Z.

Then we get a ratio of resistances as

$\frac{R_2}{R_1}=\frac{l_2-l_1}{l_1}$

With the help of this ratio, we can compare these resistances.

Solved Examples Based on Potentiometer - Principle And Applications

Example 1: In the given circuit of the potentiometer, the potential difference E across AB (10 m length) is larger than E 1 and E 2 as well. For key K 1 (closed), the jockey is adjusted to touch the wire at J 1 so that there is no deflection in the galvanometer. Now the first battery (E 1 ) is replaced by the second battery (E 2 ) for working by making K 1 open and K 2 closed. The galvanometer gives them a null deflection at J 2 . The value of $\frac{E_1}{E_2} \text { is } \frac{a}{b}$where a = ______

Length of AB=10 m

For battery $\mathrm{E}_1$, balancing length is $l_1$ and $l_1=380 \mathrm{~cm}$ [from end $\mathrm{A}$ ] For battery $\mathrm{E}_2$, balancing length is $l_2$ and $l_1=760 \mathrm{~cm}$ [from end $\mathrm{A}$ ]

Now, we know that

$\begin{aligned} & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1}{l_2} \\ & \Rightarrow \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{380}{760}=\frac{1}{2}=\frac{\mathrm{a}}{\mathrm{b}} \\ & \therefore a=1 \& b=2 \\ & \Rightarrow \mathrm{a}=1 \end{aligned}$

Hence, the answer is option (1).

Example 2: In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells \frac{\varepsilon_1}{\varepsilon_2} is :

1) $\frac{5}{3}$ 2) $\frac{8}{5}$ 3) $\frac{4}{3}$ 4) $\frac{3}{2}$

When the galvanometer is connected to the point (1) $ \varepsilon_1=\phi l_1 \rightarrow(1) $

When the galvanometer is connected to point (2) $ \varepsilon_1+\varepsilon_2=\phi l_2 \rightarrow(2) $

Where $l_1=250 \mathrm{~cm} \& l_2=400 \mathrm{~cm}$ $ \begin{aligned} & \frac{\varepsilon_1}{\varepsilon_1+\varepsilon_2}=\frac{250}{400}=\frac{5}{8} \\ & \frac{\varepsilon_1+\varepsilon_2}{\varepsilon_1}=\frac{8}{5} \\ & 1+\frac{\varepsilon_2}{\varepsilon_1}=\frac{8}{5} \\ & \frac{\varepsilon_2}{\varepsilon_1}=\frac{3}{5} \\ & \Rightarrow \frac{\varepsilon_1}{\varepsilon_2}=\frac{5}{3} \end{aligned} $

Example 3: A DC main supply of e.m.f. $220 \mathrm{~V}$ is connected across a storage battery of e.m.f. $200 \mathrm{~V}$ through a resistance of $1 \Omega$. The battery terminals are connected to an external resistance ' $R$ '. The minimum value of ' $R$ '(in ohm), so that a current passes through the battery to charge it is :

Given $E=200 \mathrm{~V}$ $ \begin{aligned} & \mathrm{r}=1 \Omega \\ & \mathrm{V}=220 \mathrm{~V} \\ & \therefore r=\left(\frac{E-V}{V}\right) R=1=\left(\frac{20}{220}\right) R \\ & \therefore R=11 \Omega \end{aligned} $

Hence, the answer is (1).

Example 4: In a potentiometer experiment the balancing with a cell is at a length $240 \mathrm{~cm}$. On shunting the cell with a resistance of $2 \Omega$ the balancing length becomes $120 \mathrm{~cm}$ The internal resistance (in $\Omega$ ) of the cell is:

Determine the internal resistance

$\begin{aligned} & r=\left(\frac{l_1-l_2}{l_2}\right) R \\ & r=\left(\frac{E}{V}-1\right) R \\ & \frac{E}{V}=\frac{l_1}{l_2} \end{aligned}$

The internal resistance of a cell is given by

$\begin{aligned} & r=R\left(\frac{l_1}{l_2}-1\right)=R\left(\frac{l_1-l_2}{l_2}\right) \\ & \therefore \quad r=2\left[\frac{240-120}{120}\right]=2 \Omega \end{aligned}$ Hence, the answer is option (2).

Example 5: The balancing length for a cell is $560 \mathrm{~cm}$ in a potentiometer experiment. When an external resistance of $10 \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \mathrm{~cm}$. If the internal $ \frac{N}{10} \Omega $ resistance of the cell is 10 , where $\mathrm{N}$ is an integer then the value of $\mathrm{N}$ is:

$\begin{aligned} & E \propto 560 \\ & \frac{E \times 10}{10+r} \propto 500 \\ & \frac{10+r}{10}=\frac{56}{50} \\ & \Rightarrow r=1.2=\frac{n}{10} \\ & \Rightarrow n=12 \\ & \end{aligned}$

Hence, the answer is option (2).

A potentiometer is a gadget customarily utilized for ascertaining the potential difference (voltage) over a component in a circuit without sourcing any current from that particular element. It works based on the fact the loss of potential over a length of wire is directly proportional to that length. By employing a sliding contact along the wire with the intention of comparing an unknown voltage to a given reference electrical potential point the unknown voltage could be found precisely.

Frequently Asked Questions (FAQs)

It works on the principle that the voltage drop across a uniform wire is directly proportional to its length when a constant current flows through it.

The main components include a long uniform resistance wire, a calibrated scale, a sliding contact (jockey), and a reference voltage source.

By balancing the potential drop across the potentiometer wire with the emf of the cell and finding the null point, the emf can be determined accurately.

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