Question

Two capacitors of capacity $6\mu F$ and $3\mu F$ are charged to
$100V$ and $50V$ separately and connected as shown in figure. Now
all the three switches $S_1,S_2$ and $S_3$ are closed. Charge on the $6\mu F$ capacitor in steady state will be :

• A. $400\mu C$
• B. $700\mu C$
• C. $800\mu C$
• D. $250\mu C$

Answer 1

Answer: (B)

$700\mu C$

Total charge on plate 1 and 2 is $q=CV=600\mu C$ and total charge on plate 3 and 4 is $50\times 3=150\mu C$
Now charge $x\mu C$ is flowing in the circuit
$\frac{(150-x)}{3}+\frac{(600-x)}{6}-200=0$
$x=-100\mu C$
Hence, final charge on $6\mu F$ capacitor is
$q_1=600-x=600+100=700\mu C$