Power in AC Circuits: The Power Factor

Average Power and the Power Factor

Drive a series LCR circuit with $v = v_m\sin\omega t$; the current is $i = i_m\sin(\omega t + \phi)$ with $i_m = v_m/Z$ and $\tan\phi = (X_C - X_L)/R$. The instantaneous power is

$$p = vi = \frac{v_m i_m}{2}\left[\cos\phi - \cos(2\omega t + \phi)\right]$$

The second term averages to zero over a cycle, leaving

$$\boxed{P = \frac{v_m i_m}{2}\cos\phi = VI\cos\phi = I^2Z\cos\phi}$$

The factor $\cos\phi$ is the power factor: average power depends not just on voltage and current but on the cosine of the phase angle between them.

The Four NCERT Cases

1. Resistive circuit (pure R): $\phi = 0$, $\cos\phi = 1$ — maximum power dissipation: $P = VI$.
2. Purely inductive or capacitive: $\phi = 90$ degrees, $\cos\phi = 0$ — no power dissipated even though current flows. This current is called the wattless current.
3. LCR series circuit: $P = VI\cos\phi$ with $\phi = \tan^{-1}[(X_C - X_L)/R]$. Crucially, power is dissipated only in the resistor — never in ideal L or C.
4. At resonance: $X_C - X_L = 0$, so $\phi = 0$, $\cos\phi = 1$: maximum power $P = I^2Z = I^2R$.

Key Point: since $\cos\phi = R/Z$, the power formula can always be rewritten $P = I^2 Z\cdot\frac{R}{Z} = I^2R$ — whatever the circuit, the resistor takes all the heat.

[NEET Important] 'Wattless current' = the current in a purely reactive circuit ($\phi = 90$ degrees): real amperes, zero average watts. The phrase itself is asked as a one-mark definition.

Why Engineers Fight Low Power Factor (NCERT Example 7.7)

(a) Why transmit at high voltage? Transmission cables carry power $P = VI\cos\phi$. For a given P, a higher V means a smaller I — and the line loss $I^2R_{line}$ falls as the square. So power is transported at the highest practical voltages (and stepped down near the consumer — Section 7).

(b) Why is low power factor expensive? For fixed required power P and supply voltage V, the current is $I = \frac{P}{V\cos\phi}$. A small $\cos\phi$ forces a large current, and the line loss $I^2R$ balloons. Suppliers therefore want loads with power factor close to 1.

The fix: most industrial loads (motors!) are inductive, with current lagging. Installing a capacitor in parallel supplies a leading component that cancels the lagging (wattless) component, raising $\cos\phi$ towards 1 without changing the useful power.

[JEE Tip] Keep the two power formulas straight: the instantaneous power oscillates at $2\omega$ (twice the supply frequency); the average is $VI\cos\phi$. Questions about 'frequency of power oscillation' want $2\nu$ — 100 Hz on the 50 Hz mains.

Solved Examples

Example 1: The 283 V LCR circuit (NCERT Example 7.8)

A sinusoidal voltage of peak 283 V, 50 Hz, drives a series circuit with R = 3 $\Omega$, L = 25.48 mH, C = 796 $\mu$F. Find (a) Z, (b) the phase, (c) the power dissipated, (d) the power factor.

Solution:

1. (a) $X_L = 2\pi(50)(0.02548) = 8\ \Omega$; $X_C = \frac{1}{2\pi(50)(796\times10^{-6})} = 4\ \Omega$; $Z = \sqrt{3^2 + (4-8)^2} = 5\ \Omega$.
2. (b) $\phi = \tan^{-1}\frac{X_C - X_L}{R} = \tan^{-1}\left(-\frac{4}{3}\right) = -53.1^{\circ}$ — net inductive, current lags.
3. (c) $i_m = 283/5 = 56.6$ A; $I = 40$ A; $P = I^2R = 1600 \times 3 = 4800$ W.
4. (d) $\cos\phi = \cos 53.1^{\circ} = 0.6$. Check: $P = VI\cos\phi = 200 \times 40 \times 0.6 = 4800$ W. Consistent.

Example 2: The same circuit at resonance (NCERT Example 7.9)

If the source frequency is changed to the resonant frequency, find the impedance, current and power.

Solution:

1. At resonance $Z = R = 3\ \Omega$ and $\phi = 0$.
2. $i_m = 283/3 = 94.3$ A, so $I = 66.7$ A.
3. $P = I^2R = 66.7^2 \times 3 \approx 13.35$ kW — nearly three times the off-resonance power. Maximum power at resonance.

Example 3: A straight power-factor computation [NEET Numerical]

An AC circuit draws 2 A (rms) from a 220 V source with the current lagging the voltage by 60 degrees. Find the power consumed.

Solution:

1. $P = VI\cos\phi = 220 \times 2 \times \cos 60^{\circ}$.
2. $= 440 \times 0.5 = 220$ W.
3. The rest of the apparent power (VI = 440 V A) is the reactive, sloshing share — only $\cos\phi$'s share becomes heat.

Example 4: Wattless in numbers [NEET Numerical]

A pure inductor on a 220 V source carries 5 A (rms). How much average power does the source deliver?

Solution:

1. Pure inductor: $\phi = 90^{\circ}$, $\cos\phi = 0$.
2. $P = VI\cos\phi = 220 \times 5 \times 0 = 0$ W.
3. Five real amperes, zero average watts — the wattless current in person.

Example 5: Power factor from the impedance triangle [JEE Numerical]

A series circuit has R = 30 $\Omega$ and Z = 50 $\Omega$, drawing rms current 2 A. Find the power factor and the power dissipated.

Solution:

1. $\cos\phi = R/Z = 30/50 = 0.6$.
2. $P = I^2R = 4 \times 30 = 120$ W. (Equivalently $I^2Z\cos\phi = 4 \times 50 \times 0.6 = 120$ W.)
3. All 120 W appear in R; the reactance handles only the borrowed energy.

Example 6: Resonance vs off-resonance power [JEE Numerical]

A series LCR circuit (R = 3 $\Omega$) is driven by a fixed 200 V rms source. Compare the power at resonance with the power when Z = 5 $\Omega$.

Solution:

1. At resonance: $I = 200/3 = 66.7$ A; $P = I^2R = 13.3$ kW.
2. Off resonance: $I = 200/5 = 40$ A; $P = I^2R = 4.8$ kW.
3. Ratio $= (Z/R)^2 = (5/3)^2 \approx 2.8$ — power falls off the peak as the square of the impedance ratio.

Example 7: Why high-voltage transmission? (NCERT Example 7.7a)

For circuits transporting electric power, why is a low power factor implied to be wasteful, and why is power transmitted at high voltages?

Solution:

1. The cable must deliver $P = VI\cos\phi$. For fixed P, raising V lowers I proportionally.
2. Line loss is $I^2R_{line}$ — quartering the current cuts the loss sixteen-fold.
3. Hence: generate, step UP to very high voltage, transmit thin-current power across the country, step DOWN at the destination.

Example 8: The cost of low power factor (NCERT Example 7.7b)

A factory needs power P from the V-volt mains at power factor 0.5. Compare its line current (and line losses) with a unity-power-factor load of the same P.

Solution:

1. $I = \frac{P}{V\cos\phi}$: at $\cos\phi = 0.5$ the current is twice the unity-pf current.
2. Line losses $I^2R$ become four times larger.
3. That's why suppliers penalise low power factor — and why factories install capacitor banks to cancel the lagging wattless component.

Example 9: Where exactly is power dissipated?

In a series LCR circuit carrying current, identify where the average power goes, and why L and C take none of it.

Solution:

1. Across L and C the voltage is 90 degrees out of phase with the current — their average power $VI\cos 90^{\circ}$ is zero; they only borrow and return energy.
2. Across R, voltage and current are in phase: average power $I^2R > 0$.
3. Hence $P = VI\cos\phi = I^2R$ — the source's net output all lands in the resistor as heat.

Example 10: The frequency of power oscillation

On the 50 Hz mains, at what frequency does the instantaneous power delivered to a resistor oscillate?

Solution:

1. $p = i_m^2R\sin^2\omega t = \frac{i_m^2R}{2}(1 - \cos 2\omega t)$.
2. The oscillating part has angular frequency $2\omega$ — twice the supply's.
3. Answer: 100 Hz. (This is why some lamps flicker at 100 Hz, not 50 Hz.)