Question
A capacitor of capacitance 5µF is connected as shown in the figure. The internal resistance of the cell is 0.5 Ω . The amount of charge on the capacitor plates is
80 μC
40 μC
20 μC
10 μC
Solution
D.
10 μC
In steady state, there will be no current in the capacitor branch
Net resistance of the circuit
R = 1 + 1 + 0.5
R = 2.5 Ω
Current drawn from the cell
i =
=
i = 1 A
Potential drop across two parallel branches
V = E - i r
= 2.5 - 1 × 0.5
V = 2.0 V
So charge on the capacitor plates
q = CV
= 5 × 2
q = 10 μC
