The current flowing through a d.c. circuit is given by the voltage applied across the circuit divided by the resistance of the circuit. But in an a.c. circuit this simple relationship does not hold good. Alternating currents set up magnetic effects and alternating e.m.fs. set up electrostatic effects, and these effects must be taken into account. With comparatively low voltage and high current, magnetic effects are larger and electrostatic effects are negligible, but with high voltage, electrostatic effects are appreciable.

**A.C. Circuits Containing Resistance only:**

An a.c. circuit containing a non-inductive resistance of R ohm is shown in fig. 24(a). When current flows through the circuit, the voltage drop in the resistance is equal to the product of current and resistance of the circuit, i.e.

V=IR volt,

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where V is the voltage applied across the circuit and I is the current in ampere flowing through the circuit.

When the circuit contains only resistance, the applied voltage and the circuit current are in phase, i.e. the current and voltage vectors always work in the same direction and the angle of phase difference between them in zero. This is shown in fig. 24(b) and 24(c).

**Power absorbed by the circuit:**

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The power absorbed by a d.c. circuit in watt is given by the product of voltage applied across the circuit and the current in ampere flowing through the circuit. In an a.c. circuit this relation holds good only for instantaneous values. In practice average power in watt absorbed by an a.c. circuit over a period Is taken into consideration. This is the value indicated by a wattmeter connected in the circuit.

The average power absorbed by an a.c. circuit is given by the product of three factors,—r.m.s. voltage applied across the circuit, r.m.s. current flowing through the circuit and the power factor of the circuit.

P = VI cos θ watt,

where,

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P = average power absorbed by the circuit over a period,

V = r.m.s. voltage applied across the circuit,

I = r.m.s. current flowing through the circuit, and

θ = phase angle between V and I.

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For an a.c. circuit containing non-inductive resistance (ohmic resistance) only, θ = 0° and cos θ = 1.

.** ^{.}**. P = VI watt.

Again V = IR volt.

.** ^{.}**. P = IR x I = I

^{2}R watt.

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Thus, an a.c. circuit containing resistance only behaves as a d.c. circuit. The power is sinusoidal, and it has a frequency twice that of voltage or current.

**A. C. Circuits Containing Inductance only:**

When an a.c. circuit contains a coil, it has inductance. The effect of the presence of inductance in the circuit is to make the current lagging behind the applied voltage. If the circuit has only inductance (and negligible resistance), the current lags behind the voltage by 90° or the voltage leads the current by 90°. This is shown in fig. 25.

Let an a.c. circuit contains negligible resistance and an inductance of L henry. This is shown in fig. 25(a). Since the current lags behind the applied voltage by 90°, current in the circuit is zero when the voltage is maximum, and current is maximum when voltage is zero. This is shown in the wave diagram in fig. 25(b).

As soon as the voltage is applied across the circuit, current does not flow through the circuit. Current starts growing in the circuit gradually after the voltage reaches its maximum value. Similarly, when the circuit is switched off, there is no voltage across its terminals. But the current still remains in the circuit and gradually dies out.

Due to change in current, an e.m.f. of self induction is set up in the coil. This induced e.m.f. acts in opposition to the applied voltage, i.e, applied voltage neutralizes the self-induced e.m.f. at every instant. The wave of self-induced e.m.f. is shown by the dotted line in the wave diagram. The vector diagram representing the voltage and the current vectors is shown in fig. 25(c).

Inductance, in conjunction with supply frequency, offers an opposition to the flow of current through the coil. This opposition is known as inductive reactance. It is usually denoted by X_{L} and is expressed in ohm. If the coil has an inductance of L henry and if the supply frequency is f hertz, inductive reactance

X_{L} = 2πfL ohm.

The voltage applied across the coil divided by the inductive reactance gives the current flowing through the coil. If the applied voltage be V volt and the current through the coil be I ampere, then

I = V/X_{L} ampere,

or V = I x X_{L} volt.

**Power absorbed by the circuit:**

The average value of power absorbed by an inductance over a period is zero. Since the current lags behind the voltage by 90°, the phase angle between voltage and current, i.e, θ=90°. Therefore the power factor cosθ = cos 90° = 0.

**Hence, the average power absorbed by the coil: **

P=VI cos θ = 0.

For quarter cycle when current in the circuit gradually increases, the magnetic field set up in and around the coil expands. During this time inductance draws power from the supply and stores the same in the magnetic field. But for the next quarter cycle when current gradually decreases, magnetic field collapses with the current and the stored energy is returned back to the supply. Therefore, the average power is zero.

Since inductance does not absorb any power, it cannot do any useful work in the circuit. All useful work is done by the resistance.

**Ex. 1:**

Find the current which will flow through a coil of negligible resistance and inductance of 0.02 henry when connected to a 100-volt, 50-hertz supply. What will be the current when the frequency is doubled?

**Solution:**

Here, L = 0.02 henry, V = 100 volts, and f = 50 hertz.

Inductive reactance,

X_{L} = 2πfL = 2 x 3.14 x 50 x 0.02 = 6.28 ohms.

Current flowing through the circuit,

I = V/X_{L} = 100/6.28 = 15.92 amperes.

When frequency is doubled, f = 100 hertz.

.** ^{.}**. X

_{L}= 2 x 3.14 x 100 x 0.02 = 12.56 ohms.

Current flowing through the circuit,

I = 100/12.56 = 7.96 amperes

**A.C. Circuits Containing Resistance and Inductance in Series:**

No a.c. circuit contains only inductance. In practice inductance is always associated with certain amount of resistance. Some devices, such as choking coil or choke coil, has small resistance and comparatively large inductance. But no coil has inductance without resistance.

Resistance and inductance do not exist as two separate items in the circuit. They remain associated together. But for the sake of study, they are shown separately. Since the same current flows through resistance and inductance at every instant and since the sum of voltage drops in resistance and in inductance gives the total applied voltage across the coil, these two items are considered to be connected in series. Thus, an inductive coil may be considered to be consisting of two a.c. circuits connected in series, one containing resistance only and the other inductance only.

An a.c. circuit containing a resistance of R ohm in series with an inductance of L henry is shown in fig. 26(a). Let,

I = r.m.s. value of current flowing through the circuit,

V_{R} = r.m.s. voltage dropped across resistance

= IR volt in phase with I,

V_{L} = r.m.s. voltage dropped across inductance

= IX_{L} volt leading I by 90°, and

V = r.m.s. voltage applied across the whole circuit and is the vector sum of V_{R} and V_{L}.

The vector diagram of the circuit is shown in fig. 26(c). From this diagram we find that the current lags behind the applied voltage by an angle θ (or the applied voltage leads the circuit current by an angle θ) such that tan θ = X_{L}/R. This is also shown in the wave diagram in fig. 26(b).

In the vector diagram V_{R}, V_{L} and V have formed a right-angled triangle in which V_{R} represents the adjacent side, V_{L} represents the opposite side or the perpendicular and V represents the hypotenuse.

√R^{2} + X_{L}^{2} is the opposition to the flow of current offered jointly by the resistance and reactance of the circuit. This opposition is known as Impedance of the circuit. It is usually denoted by Z and is expressed in ohm. Thus,

**Power absorbed by the circuit:**

Since inductance does not absorb any power, in the circuit only resistance absorbs power. Again, power absorbed by resistance is equal to the product of voltage across resistance and the current flowing through it. Therefore the power absorbed by the whole circuit is given by-

P = V_{R}I watt.

Also V_{R} = IR volt,

.** ^{.}**. P = IR x I = I

^{2}R watt.

From the vector diagram,

cosθ = V_{R}/V

or, V_{R} = Vcos θ volt.

.** ^{.}**. P = Vcos θ x I = VI cos θ watt.

cos θ is the power factor of the circuit. In this circuit since the current lags behind the applied voltage, the power factor of the circuit is a lagging power factor, and the phase angle θ is called a lagging angle.