Current Electricity

JEE Main level, single correct, medium

312 of 380

Question 312 of 380

In the potentiometer, when the cell in the secondary circuit is shunted with 4Ω4 \Omega resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12Ω12 \Omega resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ____\_\_\_\_ Ω\Omega

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Answer: (A) 4

In a potentiometer, the internal resistance rr of a cell is given by the formula r=R(l0l−1)r = R \left( \frac{l_0}{l} - 1 \right), where l0l_0 is the balancing length in open circuit, ll is the balancing length when shunted with resistance RR.
This can be rewritten as l0l=1+rR\frac{l_0}{l} = 1 + \frac{r}{R} or l=l01+r/Rl = \frac{l_0}{1 + r/R}.
Case 1: R1=4ΩR_1 = 4 \Omega and l1=120l_1 = 120 cm.
120=l01+r/4=4l04+r120 = \frac{l_0}{1 + r/4} = \frac{4l_0}{4 + r} ... (i)
Case 2: R2=12ΩR_2 = 12 \Omega and l2=180l_2 = 180 cm.
180=l01+r/12=12l012+r180 = \frac{l_0}{1 + r/12} = \frac{12l_0}{12 + r} ... (ii)
Dividing equation (ii) by (i):
180120=12l012+r×4+r4l0\frac{180}{120} = \frac{12l_0}{12 + r} \times \frac{4 + r}{4l_0}
32=3(4+r)12+r\frac{3}{2} = \frac{3(4 + r)}{12 + r}
12=4+r12+r\frac{1}{2} = \frac{4 + r}{12 + r}
12+r=8+2r12 + r = 8 + 2r
r=4Ωr = 4 \Omega
The internal resistance of the cell is 4Ω4 \Omega.

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