# AC circuit with resistor

As you can see from the image that "potential drop must be equal to the applied emf". This means after that point there is no flow of current as potential difference is zero.But how is this possible?It would be great if anybody explain me in detail.

• I don't understand the question. "This means that after that point" ... after what point? The diagram shows one complete cycle of AC current, but that cycle continues repeating indefinitely, as long as the circuit is closed. Commented Jan 15, 2019 at 16:05
• I am talking about net voltage. Commented Jan 16, 2019 at 3:20
• I STILL don't understand the question. What is "net voltage", and please be very specific. Commented Jan 16, 2019 at 3:25
• "Potential drop must be equal to the applied emf" why it is true? Sorry for inconvenience. Commented Jan 16, 2019 at 3:27
• The total potential drop across a circuit must equal the applied emf, because emf is directly related to electric potential energy, and energy must be conserved. Commented Jan 16, 2019 at 4:40

This means after that point there is no flow of current as potential difference is zero. But how is this possible?

In the example you gave, potential drop across the resistor is not always zero. The voltage drop on the resistor and voltage drop on the voltage source add up to zero, but this sum is not some "net voltage" or so.

The confusion here may be caused by unlucky terminology in that example: it uses the concept of "source of emf" of magnitude $$\epsilon$$. It takes a while to explain various kinds of emf and the difference between electromotive force and voltage of source here. In the end, what this is meant to mean here is that we have a voltage source that has voltage drop $$-\epsilon$$.

The rest is just an application of Kirchhoff's voltage law (KVL):

Sum of voltage drops in a closed circuit is zero.

To apply KVL, we have to choose one "direction of circling" as positive, say, the counter-clockwise one. Voltage drop on an element is positive when electric potential decreases in this direction. Current is positive when it runs in the same direction.

Let us use this at a time when both current and voltage drop on the resistor are positive, and $$\epsilon$$ is positive. The KVL says

$$\text{voltage drop across the resistor} + \text{voltage drop across the voltage source} = 0$$ or

$$(IR) + (-\epsilon) = 0.$$

$$(IR) + (-E_0\sin \omega t) = 0.$$ $$IR = E_0\sin \omega t.$$ The same equation generally holds also at times when $$I$$ is negative, so we have only one universal equation.