Question

what is the phase difference between voltage and current in an AC circuit containing pure capacitor?​

Answer 1

capacitor is a device for storing charging. It turns out that there is a 90° phase difference between the current and voltage, with the current reaching its peak 90° (1/4 cycle) before the voltage reaches its peak.

Answer 2

In a capacitive circuit, the phase between the current and voltage containing only pure capacitor the current leads the applied voltage by 90 degree i.e π/2.

Explanation:

Ac circuit containing only capacitor

Consider a circuit containing a capacitor of capacitance C connected across an alternating voltage source.

The alternating voltage is given by

Vm sin ωt        (1)

Let q be the instantaneous charge on the capacitor.

The emf across the capacitor at that instant  $\frac{q}{c}$ is

$v = \frac{q}{c} =0$  (According to Kirchoff’s loop rule,).

⇒  $CV_{m}$sin ωt

By the definition of current

$i -= \frac{dq}{dt} = \frac{d}{dt} CV_{m} \frac{d}{dt}$(sin ωt)

= $CV_{m}$sin ωt

And it also written as

$i = \frac{\frac{V_{m} }{I} }{\frac{I}{C_{w} } } sin(wt + \frac{\pi }{2} )$

$i = l_{m} sin(wt + \frac{\pi }{2} )$        (2)

where $\frac{\frac{V_{m} }{I} }{\frac{I}{C_{w} } } = l_{m}$

It is the peak value of the alternating current.

Final answer:

In a capacitive circuit, the current leads the applied voltage by π/2

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