Question

AB is a uniform potentiometer wire of resistance 6 $\Omega$. $C_1$ and $C_2$ are two cells; the emf of one is double that of the other. Each cell has internal resistance equal to that of 10 cm of the potentiometer wire. The null point appears at 50 cm from A. Then, length of AB in cm is.

• A. 100 cm
• B. 180 cm
• C. 90 cm
• D. 80 cm

Answer 1

Answer: (D)

80 cm

Answer 2

We assume that the two cells have emfs in the ratio 2 : 1. For consistency with the direction of current (from A to B along the potentiometer), take

$$\mathcal{E}_1=2E\quad\text{(cell }C_1\text{)}\quad\text{and}\quad \mathcal{E}_2=E\quad\text{(cell }C_2\text{)}.$$

Each cell has an internal resistance equal to that of 10 cm of the wire. Since the whole wire of length $L$ (in cm) has a resistance of 6 Ω, the resistance per cm is

$$\frac{6}{L}\, \Omega/\text{cm}.$$

Thus, each cell’s internal resistance is

$$r=\frac{6}{L}\times 10=\frac{60}{L}\,\Omega.$$

The circuit is arranged in a loop as follows:

• $A$ is connected to the positive of $C_1$.
• $C_1$ (emf $2E$, internal $60/L$) is connected to the positive of $C_2$ at point $X$.
• $C_2$ (emf $E$, internal $60/L$) is connected to point $B$.
• $B$ is connected back to $A$ through the potentiometer wire (total resistance 6 Ω).

Let the loop current be $I$. Writing the KVL in the loop (taking the direction of $I$) we have:

$$\underbrace{-(2E + I\cdot\frac{60}{L})}_{\text{across }C_1} \;-\; \underbrace{(E + I\cdot\frac{60}{L})}_{\text{across }C_2} \;+\; \underbrace{I\cdot 6}_{\text{wire from B to A (against }I\text{)}}=0.$$

Thus,

$$I\cdot 6 = 3E + I\cdot\frac{120}{L}\quad\Longrightarrow\quad I\Bigl(6-\frac{120}{L}\Bigr)=3E,$$

or

$$I=\frac{3E}{6-\frac{120}{L}}\quad\quad (1).$$

Now, in the potentiometer method the galvanometer is connected between the tap at a point $P$ (50 cm from $A$) and the junction $X$ (between $C_1$ and $C_2$). At the null point, the potential at $P$ equals that at $X$.

Taking the potential of $A$ as $V_A$, the potential drop along the wire is proportional to the resistance up to that point. The resistance of the wire from $A$ to $P$ is:

$$R_{AP}=\frac{6}{L}\times 50=\frac{300}{L}\,\Omega.$$

Thus the potential at $P$ is

$$V_P=V_A - I\cdot\frac{300}{L}.$$

On the other hand, dropping from $A$ to $X$ through $C_1$ gives:

$$V_X=V_A - \Bigl(2E+ I\cdot\frac{60}{L}\Bigr).$$

At null, $V_P=V_X$; hence,

$$V_A - I\frac{300}{L}=V_A- \Bigl(2E+ I\frac{60}{L}\Bigr).$$

Cancel $V_A$ and rearrange:

$$I\frac{300}{L}=2E+ I\frac{60}{L}\quad\Longrightarrow\quad I\Bigl(\frac{300-60}{L}\Bigr)=2E,$$

i.e.,

$$I\Bigl(\frac{240}{L}\Bigr)=2E\quad\Longrightarrow\quad I=\frac{2E\,L}{240}=\frac{E\,L}{120}\quad\quad (2).$$

Equate (1) and (2):

$$\frac{E\,L}{120}=\frac{3E}{6-\frac{120}{L}}.$$

Cancel $E$ (nonzero) and cross–multiply:

$$\frac{L}{120}\Bigl(6-\frac{120}{L}\Bigr)=3.$$

Simplify the bracket:

$$\frac{L}{120}\Bigl(\frac{6L-120}{L}\Bigr)=\frac{6L-120}{120}=3.$$

Then,

$$6L-120=360\quad\Longrightarrow\quad6L=480\quad\Longrightarrow\quad L=80\,\text{cm}.$$

Thus, the length of the potentiometer wire AB is 80 cm.

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Summary of the Core Solution:

1. Find resistance per cm: $6/L$; so internal resistance of each cell is $60/L$.
2. Write loop KVL for the circuit including cells and potentiometer wire: $I(6-\frac{120}{L})=3E$.
3. Use the null condition in the potentiometer: $I(300/L)=2E+ I(60/L)$ ⇒ $I(240/L)=2E$ which gives $I=\frac{E\,L}{120}$.
4. Equate both expressions for $I$ and solve to get $L=80\,\text{cm}$.