Discussion
If a 50 Hz sinusoidal voltage having an rms value of V volt is applied to a capacitance of C farad, the current in the circuit, with the applied voltage as the reference phasor, is given by
Answer: A
The current is \(jV\omega C\).
The impedance of a capacitor is \(Z_{C} = \frac{1}{j\omega C} = -\frac{j}{\omega C}\). Taking the voltage V as the reference phasor:
\[ I = \frac{V}{1/j\omega C} = j\omega C V \]
The factor j shows the current leading the voltage by 90°, as it does in a capacitor, and its magnitude is \(\omega C V\). \(-jV\omega C\) would lag (an inductor's behaviour), \(jV/\omega C\) has the wrong magnitude, and \(V/j\omega C\) multiplies by the impedance instead of dividing.
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