I can see that how current leads voltage while capacitor is charging. Looking at any capacitor charging diagram will explain this: (e.g figures in http://en.wikipedia.org/wiki/RC_circuit)

However, I don't see how the current leads voltage while discharging? Because for discharging, both current and voltage look alike in the same descending format. And, it doesn't seem that there is a phase difference between current and voltage curves during the discharge! Can someone please explain what is happening?


Talk about "current leading voltage" or "phase difference" only applies to AC analysis. In the more general case, one could say what a capacitor really does is differentiate voltage, according to:

$$ i = C\frac{dv}{dt} $$

From this, you can derive all sorts of well-known things about capacitors. Such as, if you want a linearly changing voltage across a capacitor, you must apply a constant-current source to it. As an example, consider a 1 ampere current source connected to a 1 farad capacitor:

$$ \require{cancel} \begin{align} 1A &= 1F \frac{dv}{dt} \\ 1A &= \frac{1 A \cdot s}{V} \frac{dv}{dt} \\ \frac{1\cancel{A}\cdot V}{1\cancel{A}\cdot s} &= \frac{dv}{dt} \\ \frac{1V}{s} &= \frac{dv}{dt} \end{align} $$

If you consider the case where the applied voltage is sinusoidal, then so too is the current:

$$ \begin{align} i &= C\frac{dv}{dt} \\ i &= C\frac{d\sin(t)}{dt} \\ i &= C\cos(t) \end{align} $$

because \$\cos\$ is the derivative of \$\sin\$.

You will also see if you graph these functions, that \$\cos\$ (current) leads \$\sin\$ (voltage) by 90 degrees, as an electrical engineer would put it:

plot of sin and cos


You can only realistically talk about phase angles when sine waves are applied and if you apply a sinewave voltage, the cap current will lead voltage by 90 degrees all the time.

Current = \$ C\dfrac{dV}{dt}\$ and the differential of a voltage sinewave is a cosinewave of magnitude C. Cosine lead sine by 90 degrees.


Generally speaking, we can only meaningfully speak of a relative phase difference between waveforms if the two waveforms have the same form but are displaced in time.

Now, as others have pointed out, the current through a capacitor is proportional to the rate of change of the voltage across so, in general, the current and voltage associated with a capacitor do not have the same form.

For example, if the capacitor voltage is a ramp, the capacitor current is a constant. If the capacitor voltage is parabolic, the capacitor current is a ramp.

How can we meaningfully talk about the relative phase between a parabolic voltage and a current ramp?

Thus, for it to be possible to meaningfully speak of a phase difference, we need a very special type of waveform; a waveform that has the same form as its rate of change.

An example of such a waveform is

$$v_C(t) = \sin( \omega t)$$

The rate of change (the time derivative) of this is

$$\dot v_C(t) = \omega \cdot \cos (\omega t) = \omega \cdot \sin(\omega t + 90^\circ)$$


$$i_C(t) = C \,\dot v_C(t) = \omega C \cdot \sin(\omega t + 90^\circ)$$

Now, it's easy to see that, in this case, the voltage across and current through a capacitor have the same form and that there is a relative phase of \$90^\circ\$.

In the case of the RC circuit charge and discharge waveforms, note that the solutions are, for DC excitation:

$$v_C(t) = V_{DC}(1 - e^{-t/RC}) + v_C(0)\cdot e^{-t/RC}$$

$$i_C(t) = \dfrac{V_{DC} - v_C(0)}{R}e^{-t/RC}$$

For zero initial condition (the capacitor is charging), these are:

$$v_C(t) = V_{DC}(1 - e^{-t/RC})$$

$$i_C(t) = \dfrac{V_{DC}}{R}e^{-t/RC}$$

For zero DC excitation (the capacitor is discharging), these are:

$$v_C(t) = v_C(0)\cdot e^{-t/RC}$$

$$i_C(t) = - \dfrac{v_C(0)}{R}e^{-t/RC}$$

As you can see, in either case, there isn't any apparent relative phase parameter we can identify in the above voltage and current waveforms.

There is a subtle reason for this. In the case of a sinusoidal waveform, we can add a constant to the argument which has the effect of displacing the waveform in time; adding this constant changes the phase of the sine waveform:

$$\sin(\omega t + \phi)$$

is a sine waveform shifted in time by \$\frac{\phi}{\omega}\$ seconds.

However, if we add a constant to the argument for the exponential, the result is not a displacement in time but a scaling (multiplication by a constant).

$$e^{-t/RC + \phi} = e^{-t/RC}e^\phi = Ke^{-t/RC}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.