- #1

Kaushik

- 282

- 17

Consider a circuit with a witch, capacitor and an AC voltage source.

The

We know that, ##Q = CV##

##\frac{dQ}{dt} = C \frac{dV}{dt}##

##i = C\frac{dV}{dt} \tag{1}##

So from the graph, the voltage increases rapidly around ## t = 0##.

So, as ##\frac{dV}{dt}## is high (but ##V = 0##), ##i## must also be high.

Similarly, when ##V## attains a maximum, ##\frac{dV}{dt}=0##. Hence, ##i = 0##.

So we can see that current is ahead of voltage by ##\pi/2##.

But, I can't

The

**sinusoidal AC voltage**source is depicted in the following graph:We know that, ##Q = CV##

##\frac{dQ}{dt} = C \frac{dV}{dt}##

##i = C\frac{dV}{dt} \tag{1}##

So from the graph, the voltage increases rapidly around ## t = 0##.

So, as ##\frac{dV}{dt}## is high (but ##V = 0##), ##i## must also be high.

Similarly, when ##V## attains a maximum, ##\frac{dV}{dt}=0##. Hence, ##i = 0##.

So we can see that current is ahead of voltage by ##\pi/2##.

But, I can't

*feel*it. Could someone please give an intuitive explanation to this?
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