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Why is there a phase difference between resistor and capacitor? I am looking for the conceptual reason behind this. What does a phase difference of \$\pi/2\$ show conceptually? According to me, the one with positive phase difference would reach to its peak value earlier than the another, but how do we come to know the time difference between the two quantities to reach at its peak? What's the reason that there is a difference in reaching at its peak value? Please illustrate in as much detail as possible.

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    \$\begingroup\$ time. When you go from the time domain to the frequency domain (using the laplace transform) you end up with some phasey goodness! \$\endgroup\$ – KyranF Mar 31 '15 at 5:27
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    \$\begingroup\$ in the frequency domain, there is no "time = 0" state but there also cannot be knowledge of the future, so you cannot "lead" a signal, you can only ever lag behind by certain number of phases etc etc. Deep stuff. need more alcohol \$\endgroup\$ – KyranF Mar 31 '15 at 5:29
  • \$\begingroup\$ The most basic reason is that the very definition of capacitance is that it resists a change in voltage by absorbing or releasing charge, thus creating the delays and phase differences. \$\endgroup\$ – Sean Boddy Mar 31 '15 at 6:34
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Q = CV in a capacitor and \$\dfrac{dQ}{dt}\$ = current therefore: -

I = \$C\dfrac{dV}{dt}\$

This means that current is proportional to the derivative of voltage.

If that voltage is a sine wave then the derivative is a cosine wave hence a phase difference of pi/2 (90 degrees).

In a resistor, V = IR i.e. the relationship between voltage and current is that they are in-phase.

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The current through the resistor is proportional to the voltage across it. So the voltage and current will have the same shape and hence they will have maxima and minima together.

In case of capacitor, the current through the capacitor is proportional to the rate of change of voltage across it. So for a sine-voltage source, the current will be cosine wave as shown below.

So looking at the image, one can say that the current through a capacitor leads the voltage across it leads by 90 degrees or it lags by 270 degrees.

Why is there a phase difference between resistor and capacitor?

Current through resistor is proportional to the voltage across it so both current and voltage will have same shape.

Where as in case of a capacitor, the current through capacitor is proportional to the derivative of voltage across it. And in case of a sinusoidally varying voltage source, it appears as if they have a constant phase shift between them.

how do we come to know the time difference between the two quantities to reach at its peak?

We can measure. Or we can calculate mathematically. The relationship between time difference and phase difference is:

$$\Delta \phi = \frac{\Delta t}{T} \times 2\pi$$

where, T is the total time period of one cycle.

What's the reason that there is a difference in reaching at its peak value?

For any continuous signal, its derivative will have a value zero at its peaks. So a signal continuous in time and its derivative can not have the peaks at same point in time.

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  • \$\begingroup\$ But in case of RC differentiator circuit when output is taken across resistor how do we get a differentiated output of the input waveform for a sine waveform(i.e. a cosine with a phase difference why is there a phase difference) or any other.Should we not get the same output as the input? \$\endgroup\$ – Curious Apr 2 '15 at 14:16
  • \$\begingroup\$ What I explained above corresponds to a capacitor only circuit. In an RC circuit, the phase shift between current and voltage will will less than 90 degree. In a differentiator, the phase shift is almost 90 deg. so for \$\sin\theta\$ voltage source, the current through the resistor will be having a shape of \$\cos\theta\$. and voltage across resistor will be \$R\times\cos\theta\$. So the output waveform is cosine. \$\endgroup\$ – nidhin Apr 2 '15 at 14:54
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One simplified answer would be:

A capacitor is a simple device that when a greater voltage than what its current voltage is will charge and result in its own voltage increasing. If you put a load ( or aply a smaller voltage) on the capacitor then it will discharge. This is all fine. But if we start using an AC waveform to rapidly charge and discharge the capacitor we get this interesting phase shift. What happens is in a sinus wave the voltage does not change at a constant rate. And the faster you try to charge or discharge a capacitor , the more current you would need. It just so happens that the place in a sine wave where the current changes the fastest is \$90^\circ\$ (\$0.5 \times \pi\$ or \$0.25 \times \tau\$) offset from where the peak is.

So a capacitors current is offset by \$90^\circ\$ because that's when the voltage changes the fastest.

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