Current Electricity - Result Question 96
101. A potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is
balanced at a length $l$ of the potentiometer wire. The e.m.f. E will be given by:
[2015 RS]
(a) $\frac{E_0 r}{(r+r_1)} \cdot \frac{l}{L}$
(b) $\frac{E_0 l}{L}$
(c) $\frac{LE_0 r}{(r+r_1) l}$
(d) $\frac{LE_0 r}{r_1}$
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Answer:
Correct Answer: 101. (a)
Solution:
- (a) $EMF, E=K /$ where $K=\frac{V}{L}$ potential gradient $K=\frac{V}{L}=\frac{iR}{L}=(\frac{E_0 r}{r+r_1}) \frac{l}{L}$
So, $E=K l=\frac{E_0 r l}{(r+r_1) L}$
