Question

Analyze the given circuit in the steady state condition. Charge on the capacitor is $q_o=16\mu C$.
Find the e.m.f. of the battery.

Answer 1

Given : $q_o=16\mu$ C $C=4\mu F$

Thus potential across the capacitor $V_{OB}=\frac{q_o}{C}=\frac{16\mu }{4\mu }=4$ volts ........(1)

From figure, $V_{AB}=\frac{(1+2)}{(1+2+3+3)}E=\frac{E}{3}$

Also $V_{AO}=\frac{\left(4\right)}{(4+4)}E=\frac{E}{2}$

$\therefore$ $V_{OB}=V_{AB}-V_{AO}=\frac{E}{2}-\frac{E}{3}=\frac{E}{6}$ .............(2)

From (1) and (2), $\frac{E}{6}=4$ $⟹E=24$ volts