Power Consumed in AC Circuits : Some Important AC Circuits

Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.

What is Power Factor? What is Wattless Current?

Power consumed in ac circuits: In any dc circuit, the potential difference (voltage) and the current are always in phase. But this is not so for ac circuits—in general, a phase difference is evolved between the voltage and the current. Mathematically the voltage V and the current I are expressed as,
V = V₀sinωt and I = I₀sin(ωt – θ)
where, θ = phase difference between the voltage and the current. θ always lies between -90° and +90°, i.e., -90° ≤ θ ≤ 90° . Then, power consumed (or power dissipated) in the circuit is,
P = VI = V₀I₀sinωtsin(ωt – θ)
= V₀I₀sinωt[sinωtcosθ – cosωtsin]
= V₀I₀[sin²ωtcosθ – sinωtcosωt sinθ]
= V₀I₀[sin²ωtcosθ – \(\frac{1}{2}\)sin2ωtsinθ]

Over a complete cycle, the average of sin²ωt = \(\frac{1}{2}\) and that of sin²ωt = 0. So the average power of the circuit is,
\(\bar{P}\) = \(\frac{1}{2}\)V₀I₀cosθ = \(\frac{V_0}{\sqrt{2}} \frac{I_0}{\sqrt{2}} \cos \theta\) = V_rmsI_rmscosθ
No experiment can measure the instantaneous power P; every measurement leads to the average power \(\bar{P}\). This \(\bar{P}\) is referred to as the effective power, and is usually denoted by the simple symbol P. So the effective power (or true power) of an ac circuit is,
P = \(\frac{1}{2}\)V₀I₀cosθ = V_rmsI_rmscosθ …. (1)

Power factor: The factor cosθ in equation (1) is vitally important; this factor arises, clearly, due to the voltage-current phase difference θ. This cosθ is called the power factor of an ac circuit. If θ ≠ o, cosθ < 1; so the power factor, in general, reduces the power consumed in an ac circuit to some extent.
Also, as -90° ≤ θ ≤ 90°, the power factor cosθ is never negative; as a result, the power consumed in the circuit can never be negative.

The unit of power P is watt (W); to distinguish between P and the product V_rmsI_rms, watt is never used as the unit of V_rmsI_rms —the usual unit of this product is volt, ampere or V ᐧ A. Incidentally, from the similarity with the dc expression P = VI, the product V_rmsI_rms is often called the apparent power of an ac circuit. Clearly, true power = apparent power × power factor.

1. If voltage and current are in the same phase, then θ = 0 and cosθ = 1. In this condition, an ac circuit consumes the maximum power.
2. If the voltage and the current are either -90° or +90° out of phase, then cosθ = 0 and P = 0. An ac circuit of this type consumes no power. The corresponding current is referred to as wattless current.

Time interval between the peak values of voltage and current : V = V₀sin ωt, I = I₀ sin(ωt – sinθ)
So, V = V₀, when ωt = \(\frac{\pi}{2}\), t = \(\frac{\pi}{2 \omega}\)
Similarly, I = I₀, when (ωt’ – θ) = \(\frac{\pi}{2}\) i.e., t’ = \(\frac{\pi}{2 \omega}+\frac{\theta}{\omega}\)
Thus, the minimum time interval between the peak values of ac voltage and current is,
t₀ = t’ – t = \(\frac{\theta}{\omega}\) …. (2)

Purely Resistive Circuit

Fig. shows this circuit. Alternating voltage applied in the circuit,
V= V₀sinωt
Here, the peak value of alternating voltage = V₀;

rms value, V_rms = \(\frac{V_0}{\sqrt{2}}\)frequency, f = \(\frac{\omega}{2 \pi}\)
I = \(\frac{V}{R}\) = \(\frac{V_0}{R} \sin \omega t\)
or I = I₀ sin ωt
According to Ohm’s law,
I = \(\frac{V}{R}\) = \(\frac{V_0}{R}\) sinωt

So, the peak value of ac, I₀ = \(\frac{V_0}{R}\);
rms value, Irms = \(\frac{I_0}{\sqrt{2}}\) = \(\frac{V_0}{\sqrt{2} R}\) = \(\frac{V_{\mathrm{rms}}}{R}\);
frequency, f = \(\frac{\omega}{2 \pi}\)

The factor sinwt in equations (1) and (2) indicates that

1. voltage and current are in phase [Fig.(a) and (b)];
2. there is no change in frequency in this type of circuit.

Power in the circuit: Phase difference between voltage and current, θ = 0; so, power factor, cosO = 1 . For this, maximum power is consumed in a purely resistive circuit, which is
P = V₀I₀cosθ = \(\frac{1}{2}\)(I₀R)I₀ ᐧ 1 = \(\frac{1}{2} I_0^2 R\) = \(\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)
i.e., P = I²_rms ᐧ R ……….. (3)

Purely Inductive Circuit

Fig. shows this circuit. Let the alternating voltage applied in the circuit be,
V= V₀sinωt ….. (1)

If the instantaneous change of current in the circuit be dl, the emf induced in the two ends of the inductor = -L\(\frac{d I}{d t}\).

This emf reduces the main voltage in the circuit. For this the effective voltage of the circuit V₀ sinωt – L\(\frac{d I}{d t}\)
A purely inductive circuit has no resistance i.e., R = 0. So according to Ohm’s law,
V₀sinωt – L\(\frac{d I}{d t}\) = 0
or, \(\int d I\) = \(\int \frac{V_0}{L} \sin \omega t d t\)
or, I = –\(\frac{V_0}{\omega L}\)cosωt + k [k = integration constant] …… (2)

It is clear that dimensionally k is the same as I. Also k is a time- independent quantity. As the source voltage oscillates symmetrically about zero, the current also oscillates symmetrically about zero. For this, no constant or time independent current can flow through the circuit, i.e., k = 0.
∴ I = \(-\frac{V_0}{\omega L}\)cosωt = +\(\frac{V_0}{\omega L} \sin \left(\omega t-90^{\circ}\right)\)
or, I = I₀ = sin (ωt – 90°) ….. (3)
where I₀ = \(\frac{V_0}{\omega L}\) = peak value of current.

From the equations (1) and (3) we come to the conclusions—

i) current lags behind the voltage by 90° [Fig.(a) and (b)];

ii) the quantity ωL plays the same role in an inductive circuit as the resistance R in a resistive circuit. This quantity (i.e., ωL) is known as the inductive reactance of an ac circuit.
Inductive reactance is the opposition offered by an inductor to the flow of alternating current through it. It is denoted by the symbol X_L.
Hence for the above circuit,
X_L = ωL = inductive reactance
Similar to that of R its unit is also ohm (H).

Power in the circuit: Phase difference between voltage and current, θ = 90°, so, power factor, cosθ = 0.
Therefore, power consumption, P = \(\frac{1}{2} V_0 I_0 \cos \theta\) = 0 Clearly, the current I in this circuit is wattless.

Purely Capacitive Circuit

Fig. shows the circuit. Alternating voltage applied in the circuit,
V = V₀ sin ωt ….. (1)

At any moment, if Q be the charge stored in the capacitor C, then its terminal potential difference is \(\frac{Q}{C}\).
This potential difference opposes the applied instantaneous voltage in the circuit.
So, the effective voltage of the circuit, V_e = V₀ sin ωt – \(\frac{Q}{C}\)
As there is no resistance in the circuit, according to Ohm’s law,
V_e = IR = 0
and hence, V₀sinωt – \(\frac{Q}{C}\) = 0
or, Q = CV₀sinωt ……. (2)
Therefore alternating current,
I = \(\frac{d Q}{d t}\) = ωCV₀cosωt = ωCV₀sin(ωt + 90°)
= I₀sin(ωt + 90°) …… (3)
where, I₀ = ωCV₀ = \(\frac{V_0}{1 / \omega C}\) = \(\frac{V_0}{X_C}\) = peak value of current
From the equations (1) and (3) we come to the conclusions—

i) current leads the applied voltage by 90° [Fig.(a) and (b)];

ii) the quantity \(\frac{1}{\omega C}\) plays the same role in a capacitive circuit as the resistance R in a resistive circuit. This quantity (i.e., \(\frac{1}{\omega C}\)) is known as capacitive reactance (X_C). Capacitive reactance is the opposition offered by a capacitor to the flow of alternating current through it. Hence for the above circuit,
X_C = \(\frac{1}{\omega C}\) = capacitive reactance
Similar to that of R or X_L, its unit is also ohm (Ω).

Power in the circuit: Phase difference between voltage and current, # = -90°; so, power factor, cos θ = 0. Therefore, power consumption P = \(\frac{1}{2} V_0 I_0 \cos \theta\) = 0.
Here again this current is wathers.

Wattless or Idle current: Power consumed in an ac circuit, P = \(\frac{1}{2} V_0 I_0 \cos \theta\). Now the phase difference between voltage and current in a pure inductive circuit, θ = 90°. So power in the circuit is zero. On the other hand, phase difference between voltage and current in a pure capacitive circuit, θ = -90°. So, again power is equal to zero. It means, a purely inductive or capacitive circuit does not dissipate any power; current through the inductor or capacitor is hence called wattless or idle current.