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# In the circuit the galvanometer $G$ shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor $R$ will be

$\begin {array} {1 1} (A)\;200\Omega & \quad (B)\;100\Omega \\ (C)\;500\Omega & \quad (D)\;1000\Omega \end {array}$

The potential drop across the resistanceR is 2V.
Now,  no current flows through the galvanometer .and if $V_1$ is the potential drop across the resistor R and V is the total potentialthen
$\large\frac{V R}{500+R}$ $= V_1$
$\large\frac{12R}{500+R}$ $= 2$
Ans : (B)

edited Sep 10, 2014