In the potentiometer, when the cell in the secondary circuit is shunted with $4\Omega$ resistance, t

Question

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

Accepted Answer

Let E is emf and r is internal resistance of cell.

$${\frac{E \cdot 4}{r + 4} = 120\text{ }K }{\frac{E \cdot 12}{r + 12} = 180\text{ }K }{\Rightarrow \frac{1}{3}\frac{r + 12}{r + 4} = \frac{2}{3} }{r + 12 = 2(r + 4) }{\Rightarrow r = 4}$$

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