In the given circuit diagram, when the current reaches a steady-state in the circuit, the charge on the capacitor of capacitance $C$ will be:

Question

In the given circuit diagram, when the current reaches a steady-state in the circuit, the charge on the capacitor of capacitance C will be:

TestHub · Accepted Answer

In steady state, current through capacitor will be zero, so current through $r_{1}$ will be zero.

Hence, the net current flowing in the circuit will be,
$I = \frac{E}{r + r_{2}}$
The potential difference across the capacitor will be,
$V_{2} = I \times r_{2} = \frac{E r_{2}}{r + r_{2}}$

Therefore, charge on the capacitor plate will be,
$q = C V_{2} = \frac{C E r_{2}}{r + r_{2}}$

Physics - Capacitor Question with Solution | TestHub

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