Correct option is (1) 8 µC
In the steady state, there will be no current in the upper branch of the circuit. Hence, the upper branch will behave as the fully capacitive circuit. Therefore, we can neglect the 6Ω resistance in the upper circuit when the steady state is reached.
The equivalent circuit of the upper branch in steady state can be drawn as:
Here, two 2 μF capacitors are in parallel with each other and their combination is in series with the 4 μF capacitor. Hence, the equivalent capacitance of the circuit is :
Ceq = \(\frac{4\times 4}{4 + 4}\) = 2 μF
Now the current I is:
I = \( \frac V {R + r}\)
\(= \frac 5{4 + 1}\)
= 1 A
Where, r is the internal resistance of the battery.
Hence, the voltage applied across the combination of the capacitors is :
ΔV = E − Ir
= 5 − 1 × 1
= 4 V
Therefore, the charge sent by th battery or charge on the 4 μF capacitor is :
Q = C ΔV
= 2 × 4
= 8 μC
