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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:
A
$\frac{LE_0r}{(r+r_1)l}$
B
$\frac{LE_0r}{lr_1}$
C
$\frac{E_0r}{(r+r_1)}\cdot\frac{l}{L}$
D
$\frac{E_0l}{L}$
Detailed Solution
Current in the wire: $i = \frac{E_0}{r + r_1}$
Potential gradient: $x = \frac{ir}{L} = \frac{E_0r}{(r + r_1)L}$
$E = xl = \frac{E_0r}{(r + r_1)}\cdot\frac{l}{L}$
Potential gradient: $x = \frac{ir}{L} = \frac{E_0r}{(r + r_1)L}$
$E = xl = \frac{E_0r}{(r + r_1)}\cdot\frac{l}{L}$