Question : A capacitor of 4µF is connected as shown in the circuit.
The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be
a) 0
b) 4 µC
c) 16µC
d) 8 µC
Doubt by Eshani
Ans : d) is the correct option.
Key Concept : A capacitor acts like a perfect conductor while charging, offering zero resistance. However, when fully charged, it behaves like an open circuit with very high resistance, reaching a stable state called steady state.
Detailed Solution :
Here we are assuming that the capacitor is fully charged i.e. steady stage is reached so it offers a high resistance to the flow of current through it. So there would be no current in the top arm of the circuit.
Now, the current in the remaining circuit would be
I = ε/R+r
I = 2.5/(2+0.5)
I = 2.5/2.5
I = 1 A
Now the TPD of the Cell
V=ε-Ir
V=2.5-(1)(0.5)
V=2.5-0.5
V=2 V
Here we are assuming that the capacitor is fully charged i.e. steady stage is reached so it offers a high resistance to the flow of current through it. So there would be no current in the top arm of the circuit.
Now, the current in the remaining circuit would be
I = ε/R+r
I = 2.5/(2+0.5)
I = 2.5/2.5
I = 1 A
Now the TPD of the Cell
V=ε-Ir
V=2.5-(1)(0.5)
V=2.5-0.5
V=2 V
Now this voltage also applied across the plates of the capacitor (because the top arm is connected in parallel).
So using
Q=CV
Q=4µF×2
Q=8µC
So using
Q=CV
Q=4µF×2
Q=8µC
Hence, d) 8 µC, would be the correct option.
