Question

The following question refers to the circuit shown. Assume that the capacitors are initially uncharged
After the switch has been closed for a long time, how much charge is on the positive plate of the $3\mu F$ capacitor?

• A. $24\mu C$
• B. $32\mu C$
• C. $56\mu C$
• D. $72\mu C$

Answer 1

Answer: (C)

$56\mu C$

Initially capacitors act as conducting wire. So initially all the three resistors will be in parallel. After long time capacitors are fully charged and no current flow through them. Now all the resistances will be in series.
The equivalent resistance $R_{eq}=2+3+4=9\Omega$
Current in the circuit is $I=V/R_{eq}=24/9=8/3A$
Here, potential across $3\mu F=$potential across resistor $(3+4)\Omega$
or $V_3=7I=7\times \left(8/3\right)=\frac{56}{3}V$
Charge on $3\mu F$ will be $Q=3\times \frac{56}{3}=56\mu C$