Two condensers one of capacity $\frac{C}{2}$ and other capacity $C$ are connected to a battery of voltage $V$ as shown. The work done in charging fully both the condensers is
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- \frac{1}{2} CV^2
- \frac{3}{4} CV^2
- \frac{3}{2} CV^2
- 2 CV^2
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
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The charging current can be obtained by differentiating it and using
\(\frac{d}{dx} (e^{mx}) = me^{mx}\)
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