The correct option is
C 200 μC
In steady state capacitor behaves as open circuit
The circuit can be reduced as following:
Hence,
Req=9 Ω
Given
E=72 V
The current supplied by the battery
i=72 V9 Ω=8 A. (ohm's law )
Thus, the current supplied by the battery is
8 A.
assume zero potential at lower potential terminal of cell
Now distributing currents using
KCL and writing potentials at different nodes using ohms law .
ohm's law for
6Ω:
VA−VBR=8A
⇒72V−VB6Ω=8A
⇒VB=24V
ohm's law for
2Ω:
VB−VCR=2A
⇒24V−VC2Ω=2A
⇒VC=20 V
We can observe the potential difference across the capacitor is
20V.
Now, voltage across capacitor ,
V=20V,
C=10μF
∴ charge on capacitor
q=CV
=10×10−6F×20V=200 μC
Hence, option C is correct.