The potential difference between the plate of a capacitor is increased by $20$ %. The energy stored on the capacitor increases by:

20

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22

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40

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44

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Solution

The correct option is B $44$ %
Energy stored in a capacitor with potential difference $V$ applied across it is $\frac{1}{2} C V^{2} = E_{0}$ .
When potential difference increases by 20%, $V^{'} = 1.2 V$
$⟹ E^{'} = \frac{1}{2} C V^{' 2} = \frac{1}{2} C ({1.2}^{2} V^{2})$
$⟹ E^{'} = {1.2}^{2} E_{0} = 1.44 E_{0}$
Thus the energy stored increases by $44$ %.

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The potential difference between the plate of a capacitor is increased by 20%. The energy stored on the capacitor increases by:

The correct option is B 44%Energy stored in a capacitor with potential difference V applied across it is 12CV2=E0.When potential difference increases by 20%, V′=1.2V⟹E′=12CV′2=12C(1.22V2)⟹E′=1.22E0=1.44E0Thus the energy stored increases by 44%.

The potential difference between the plate of a capacitor is increased by $20$ %. The energy stored on the capacitor increases by: