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AC Voltage Applied To A Capacitor MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • AC voltage applied to a capacitor is considered one of the most asked concept.

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In an a.c. circuit the voltage applied is E=E_{0}\, sin\, \omega t. The resulting current in the circuit is I=I_{0}sin\left ( \omega t-\frac{\pi }{2} \right ). The power consumption in the circuit is given by

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AC voltage applied to a capacitor

AC voltage applied to a capacitor:

 The circuit containing alternating voltage source V=V0sinωt connected to a capacitor of capacitance C.

Suppose at any time t, q be the charge on the capacitor and i be the current in the circuit. Since there is no resistance in the circuit, so the instantaneous potential drop q/C across the capacitor must be equal to applied alternating voltage so,

\frac{q}{C}=V_0sin\omega t

Since i=dq/dt  is the instantaneous current in the circuit so, 

\begin{aligned} i=& \frac{d q}{d t}=\frac{d}{d t}\left(C V_{0} \sin \omega t\right) \\ &=C V_{0} \omega \cos \omega t \\ &=\frac{V_{0}}{(1 / \omega C)} \cos \omega t \\ &=i_{0} \cos \omega t=i_{0} \sin \left(\omega t+\frac{\pi}{2}\right) \end{aligned}

Where,   i_{0}=\frac{V_{0}}{(1 / \omega C)} is the peak value of current.

Comparing equation of current  with V=V0sinωt ,we see that in a perfect capacitor current leads the emf by a phase angle of π/2.

Again comparing peak value of current with ohm's law ,we find that quantity 1/ωC has the dimension of the resistance. 

Thus the quantity X_C=\frac{1}{\omega C}=\frac{1}{2 \pi fC} is known as capacitive reactance. 

Phase difference (between voltage and current):

\phi = -\frac{\pi }{2}

Power :

P= 0 

Power factor :

\cos(\phi ) = 0

Time difference:

T.D =   \frac{T}{4}

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AC voltage applied to a capacitor

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