A capacitor of capacitance \( 5\,\mu\mathrm{F} \) is connected as shown in the figure. The internal resistance of the cell is \( 0.5\,\Omega \). The amount of charge on the capacitor plate is
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- 10 \( \mu \) C
- 5 \( \mu \) C
- 6 \( \mu \) C
- 10.2 \( \mu \) C
The Correct Option is A
Solution and Explanation
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A circuit consisting of a capacitor C, a resistor of resistance R and an ideal battery of emf V, as shown in figure is known as RC series circuit.
As soon as the circuit is completed by closing key S₁ (keeping S₂ open) charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference Vc (= q/C) across the capacitor also increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (Q = VC). During this process of charging, the charge q on the capacitor changes with time t as
\(q = Q[1 - e^{-t/RC}]\)
The charging current can be obtained by differentiating it and using
\(\frac{d}{dx} (e^{mx}) = me^{mx}\)
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