Question:
In a circuit consisting of a capacitance and a generator with alternating emf \(E_g = E_{g_0}\sin\omega t\), \(V_C\) and \(I_C\) are the voltage and current. Correct phasor diagram for such circuit is
In a circuit consisting of a capacitance and a generator with alternating emf \(E_g = E_{g_0}\sin\omega t\), \(V_C\) and \(I_C\) are the voltage and current. Correct phasor diagram for such circuit is
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In a purely capacitive AC circuit the current leads the voltage by \(90^\circ\); pick the phasor diagram where \(I_C\) is \(90^\circ\) ahead (anticlockwise) of \(V_C\).
Updated On: Nov 4, 2026
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The Correct Option is C
Solution and Explanation
For a capacitor, \(V_C = V_0\sin\omega t\) and \(I_C = C\frac{dV_C}{dt} = \omega C V_0\cos\omega t = I_0\sin\left(\omega t + \frac{\pi}{2}\right)\).
So the current leads the voltage by \(90^\circ\). In the phasor diagram the phasors rotate anticlockwise, so \(I_C\) must be drawn \(90^\circ\) ahead (anticlockwise) of \(V_C\).
Option (1): \(I_C\) lags \(V_C\) by \(90^\circ\) (inductor). Option (2): in phase (resistor). Option (4): opposite phase (\(180^\circ\)). Option (3): \(I_C\) is \(90^\circ\) ahead of \(V_C\).
Answer: (3) the diagram with \(I_C\) leading \(V_C\) by \(90^\circ\)
So the current leads the voltage by \(90^\circ\). In the phasor diagram the phasors rotate anticlockwise, so \(I_C\) must be drawn \(90^\circ\) ahead (anticlockwise) of \(V_C\).
Option (1): \(I_C\) lags \(V_C\) by \(90^\circ\) (inductor). Option (2): in phase (resistor). Option (4): opposite phase (\(180^\circ\)). Option (3): \(I_C\) is \(90^\circ\) ahead of \(V_C\).
Answer: (3) the diagram with \(I_C\) leading \(V_C\) by \(90^\circ\)
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