Power in AC Circuit

In a DC circuit, power is simply P = VI. But in an AC circuit, both voltage and current vary sinusoidally — and they are often out of phase. This means the power delivered to the circuit is not simply the product of voltage and current magnitudes.

Think of pushing a swing. If you push in sync with the swing's motion, you do maximum work (pure resistive circuit, φ = 0). If you push at the wrong time — say, when the swing is coming toward you — you do no net work at all (purely reactive circuit, cos φ = 0). The power factor is a measure of how "in sync" your push is.

In the expression Pav = V_rmsI_rms cos⁡ ϕ, the quantity cos φ is called the power factor.

OR

Power Factor is the cosine of the phase angle (φ) between the voltage and current in an AC circuit. It equals the ratio of true (average) power to apparent power.

The current flowing in a purely inductive or purely capacitive circuit for which cos φ = 0 and no power is dissipated even though the current is flowing is called wattless current.

OR

Wattless Current (also called reactive current) is the component of current in a purely inductive or purely capacitive circuit that flows without consuming any power.

P = VI = \[\frac{V_mI_m}{2}[\cos\phi-\cos(2\omega t+\phi)]\]

Where:

• V_m​ = Peak voltage
• I_m​ = Peak current
• ϕ = Phase angle between voltage and current
• ω = Angular frequency
• t = Time

P_av ​= VI cos ϕ = I²Z cos ϕ = \[\frac {V_m​I_m}{2}\]​​cos ϕ = V_rms​I_rms​ cos ϕ

Let voltage and current in a series LCR circuit be:

where φ is the phase angle by which the current lags the voltage.

The instantaneous power is the product of instantaneous voltage and current:

Using the trigonometric identity sin ⁡A sin ⁡B = \[\frac {1}{2}\][cos⁡(A − B) − cos⁡(A + B)]:

\[\cos\phi=\frac{P_{av}}{P_{apparent}}=\frac{P_{av}}{V_{rms}\cdot I_{rms}}\]

1. For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
2. Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.

Solution

1. We know that P = IV cos⁡ ϕ where cos⁡ ϕ is the power factor. To supply a given power at a given voltage, if cos⁡ ϕ is small, we have to increase the current accordingly. But this will lead to large power loss (I²R) in transmission.
2. Suppose in a circuit, current I lags the voltage by an angle φ. Then, the power factor cos ⁡ϕ = R/Z. We can improve the power factor (tending to 1) by making Z tend to R. Let us resolve I into two components. I_p​ along the applied voltage V and I_q perpendicular to the applied voltage. I_q​ is called the wattless component since, corresponding to this component of current, there is no power loss. I_p​ is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. It's clear from this analysis that if we want to improve power factor, we must completely neutralise the lagging wattless current I_q by an equal leading wattless current I_q′. This can be done by connecting a capacitor of appropriate value in parallel so that I_q​ and I_q′ cancel each other, and P is effectively I_pV.

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH, and C = 796 μF. Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.

Solution

(a) To find the impedance of the circuit, we first calculate X_L and X_C:

Therefore:

(b) Phase difference:

Since φ is negative, the current in the circuit lags the voltage across the source.

(c) The power dissipated in the circuit is P = I²R. Now:

Therefore: P = (40)² × 3 = 4800 W

(d) Power factor = cos⁡(−53.1°) = 0.6

Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition.

Solution

(a) The frequency at which resonance occurs is:

(b) The impedance Z at the resonant condition is equal to the resistance:

The rms current at resonance is:

The power dissipated at resonance is:

You can see that in the present case, the power dissipated at resonance is more than the power dissipated in Example 2.

At an airport, a person is required to walk through a metal detector doorway for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work?

Solution

The metal detector works on the principle of resonance in AC circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned to resonance. When you walk through with metal in your pocket, the circuit's impedance changes, resulting in a significant change in current. This change in current is detected, and the electronic circuitry emits a sound as an alarm.