Question

Show that the current leads the voltage in phase by \[\pi /2\] in an AC circuit containing an ideal capacitor.

Answer 1

Hint: The voltage in the AC supply oscillates in the sine wave pattern. Recall the formula for charge stored in the capacitor and then express the current flowing through the capacitor. The current through the capacitor is the rate of flow of charges in the capacitor. Differentiate the equation of charge and identify the phase angle between current and voltage.

Formula used:
Charge stored in the capacitor, \[q = CV\]
Here, C is the capacitance and V is the voltage.

Complete step by step answer:
We know that in an AC circuit is given the voltage oscillates in sine wave pattern given by the expression,
\[V = {V_0}\sin \omega t\] …… (1)
Here, \[{V_0}\] is the maximum voltage and \[\omega \] is the angular frequency.

We know that the charge on the capacitor is expressed as,
\[q = CV\]
Here, C is the capacitance and V is the voltage.

Using equation (1) in the above equation, we have,
\[q = C{V_0}\sin \omega t\] …… (2)
We know that the current is the rate of flow of charges per unit time. Therefore, we can express the current through the capacitor as.
\[I = \dfrac{{dq}}{{dt}}\]

Using equation (2) in the above equation, we get,
\[I = \dfrac{d}{{dt}}\left( {C{V_0}\sin \omega t} \right)\]
\[ \Rightarrow I = C{V_0}\dfrac{d}{{dt}}\left( {\sin \omega t} \right)\]
\[ \Rightarrow I = C{V_0}\omega \cos \omega t\]
We know that, \[\sin \left( {\phi + \dfrac{\pi }{2}} \right) = \cos \phi \]. Therefore, we can write the above equation as,
\[\therefore I = \omega C{V_0}\sin \left( {\omega t + \dfrac{\pi }{2}} \right)\] …… (3)

From equation (1) and (2), we can conclude that the current leads the voltage by phase \[\dfrac{\pi }{2}\].

Note:To determine the charge on the capacitor, you can apply Kirchhoff’s voltage law in the circuit where the capacitor is the only component in the circuit. Applying Kirchhoff’s law, you will get the expression, \[{V_0}\sin \omega t - \dfrac{q}{C} = 0\]. Note that when the circuit contains only the capacitor as the active component, the current always leads the voltage by \[\dfrac{\pi }{2}\].