An AC circuit consists of a combination of circuit elements like resistors, inductors and capacitors fed by an AC generator which provides sinusoidal voltage. To analyse an AC circuit is to find the current through the given circuit and the phase difference between the current and the applied voltage.

## RC Circuit

Consider a resistor of resistance R and a capacitor of capacitance C connected in series across an AC voltage source. The capacitor in the circuit will store charges, and the resistor connected to the circuit controls the rate of charging and discharging. The charging and discharging of the capacitor is not an instantaneous process but a time-consuming process. The capacitor charges gradually. In an RC circuit, a fully discharged capacitor is connected to the capacitor, and the switch is left open. When the switch is closed, the current starts flowing through the capacitor and the resistor.

The current that starts flowing is called the charging current. It is given by the expression

I = I0 = ε/R

Here, ε is the emf of the cell

R is the resistance of the resistor.

Potential difference is developed across the capacitor as time progresses. The potential difference across the capacitor is given by q/C, and across the resistor, the potential difference is iR. Applying Kirchoff’s voltage law, we get the equation of the transient current, and it depends on the time.

I = (ε/R) e-t/RC

## AC Circuit Containing a Resistor

Consider a resistor of resistance R connected in series with an AC source. Let the source supply a sinusoidal voltage given by

V = V0 sin ωt ———–(1)

Let V be the instantaneous voltage

V0 is the peak value

ω = 2πf , f is the frequency of AC

Then V = VR = V0sinωt

Instantaneous current through the circuit is I = V/R = V0sinωt/R = I0sinωt

Here I0 = V0/R represents the peak value of current

i.e. I = I0sinωt ——–(2)

I is also the current through the resistor R.

From equations (1) and (2), it can be seen that the current through the resistor is in phase with the applied voltage.

## AC Circuit Containing a Capacitor

Consider a capacitor of capacitance C connected across an AC source. Let the source supply a sinusoidal voltage given by

V = V0 sin ωt ———–(1)

Let V be the instantaneous voltage

V0 is the peak value

ω = 2πf, f is the frequency of AC

Then V = Vc = V0sinωt

The instantaneous charge on the capacitor is given by q = CV = CV0sinωt

The instantaneous value of current through the circuit is given by

I = dq/dt = d(CV)/dt = d(CV0sinωt)/dt

⇒ I = ωCV0cosωt

Let ωCV0 = I0

Therefore, I = I0cosωt

I = I0 sin (ωt + π/2) ———-(2)

Comparing equations (1) and (2) it can be observed that current leads across the capacitor by π/2 or 900.

## Voltmeter

Check BYJU’S pages to learn more about other Physics topics like voltmeter, ammeter etc. The voltmeter is the instrument used to measure voltage or potential difference among two points of a circuit. It has a high internal resistance. The high resistance of the voltmeter will impede the flow of the current through it. This allows the device to take correct readings of the voltage.

The main principle of a voltmeter is that it must be connected in parallel across the two points of the circuit in which we want to measure the voltage. For an ideal voltmeter, the resistance will be infinity and hence the current drawn by the voltmeter will be zero and there will not be any power loss in the circuit.