Alternating Current — AC Circuits, Resonance, Transformers

Why AC Matters

Alternating current reverses direction periodically, unlike DC which flows one way. Nearly all power generation and transmission uses AC because transformers can step voltage up or down — reducing transmission losses dramatically. This chapter covers AC circuit analysis, phasor diagrams, resonance, and power.

CBSE Class 12 boards give 4-6 marks. JEE Main tests 1 question on average, often involving LCR circuits or resonance.

graph TD
    A[AC Circuit] --> B{Components?}
    B -->|R only| C[V and I in phase]
    B -->|L only| D[V leads I by 90°]
    B -->|C only| E[I leads V by 90°]
    B -->|R + L + C Series| F[LCR Analysis]
    F --> G[Find Z = √R² + XL-XC²]
    F --> H{XL = XC?}
    H -->|Yes| I[Resonance! Z = R, I max]
    H -->|No| J[Phase angle φ = tan⁻¹ XL-XC /R]
    I --> K[ω₀ = 1/√LC]

Essential Formulas

v = V_0 \sin(\omega t), \quad i = I_0 \sin(\omega t + \phi)

RMS values: $V_{rms} = \frac{V_0}{\sqrt{2}}, \quad I_{rms} = \frac{I_0}{\sqrt{2}}$

Component	Reactance/Resistance
Resistor	$R$
Inductor	$X_L = \omega L$
Capacitor	$X_C = \frac{1}{\omega C}$
Series LCR	$Z = \sqrt{R^2 + (X_L - X_C)^2}$

\omega_0 = \frac{1}{\sqrt{LC}}, \quad f_0 = \frac{1}{2\pi\sqrt{LC}}

At resonance: $X_L = X_C$ , $Z = R$ (minimum), current is maximum.

Quality factor: $Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR}$

P_{avg} = V_{rms} I_{rms} \cos\phi

$\cos\phi$ is the power factor. At resonance, $\cos\phi = 1$ (maximum power transfer).

\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}

Step-up: $N_s > N_p$ . Step-down: $N_s < N_p$ . Ideal: $V_p I_p = V_s I_s$ .

Solved Examples

Example 1 (Easy — CBSE)

An AC source of 220 V rms is connected to a 100 $\Omega$ resistor. Find rms current and peak current.

$I_{rms} = 220/100 = \mathbf{2.2 \text{ A}}$ . $I_0 = I_{rms}\sqrt{2} = \mathbf{3.11 \text{ A}}$ .

Example 2 (Medium — JEE Main)

A series LCR circuit has $R = 100\ \Omega$ , $L = 0.5$ H, $C = 20\ \mu$ F. Find resonant frequency and current at resonance if $V = 200$ V rms.

$f_0 = \frac{1}{2\pi\sqrt{0.5 \times 20 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-5}}} = \frac{1}{2\pi \times 3.16 \times 10^{-3}} \approx \mathbf{50.3 \text{ Hz}}$

At resonance: $Z = R = 100\ \Omega$ . $I = 200/100 = \mathbf{2 \text{ A}}$ .

Example 3 (Hard — JEE Main)

In the above circuit, find the voltage across the inductor at resonance.

$V_L = IX_L = I \times \omega_0 L = 2 \times (2\pi \times 50.3) \times 0.5 = 2 \times 158 = \mathbf{316 \text{ V}}$

Notice: the voltage across the inductor (316 V) exceeds the source voltage (200 V). This is the voltage amplification effect at resonance, characterized by the Q factor.

Common Mistakes to Avoid

Mistake 1 — Adding AC voltages algebraically. In series LCR, $V_R$ , $V_L$ , $V_C$ are not in phase. $V_{total} \neq V_R + V_L + V_C$ . Use phasor (vector) addition.

Mistake 2 — Confusing peak and rms values. Household “220 V” is rms. Peak is $220\sqrt{2} \approx 311$ V. Always check which one the problem asks for.

Mistake 3 — Applying DC formulas to AC circuits. Power is $P = V_{rms}I_{rms}\cos\phi$ , not simply $VI$ .

Mistake 4 — Wrong sign for impedance. $Z = \sqrt{R^2 + (X_L - X_C)^2}$ . The subtraction is $X_L - X_C$ , not the other way. The sign determines whether the circuit is inductive or capacitive.

Practice Questions

Q1. Find the reactance of a 0.1 H inductor at 50 Hz.

$X_L = 2\pi fL = 2\pi \times 50 \times 0.1 = 31.4\ \Omega$ .

Q2. Find the reactance of a 100 $\mu$ F capacitor at 50 Hz.

$X_C = 1/(2\pi fC) = 1/(2\pi \times 50 \times 10^{-4}) = 31.8\ \Omega$ .

Q3. A transformer has 500 primary turns and 50 secondary turns. If input is 220 V, find output voltage.

$V_s = V_p \times N_s/N_p = 220 \times 50/500 = 22$ V.

Q4. Power factor of a circuit is 0.5. If $V_{rms} = 200$ V and $I_{rms} = 5$ A, find average power.

$P = 200 \times 5 \times 0.5 = 500$ W.

Q5. In a pure capacitive circuit, what is the power factor?

Current leads voltage by $90°$ . $\cos 90° = 0$ . Average power = 0. (Energy oscillates between source and capacitor.)

FAQs

Why is AC used for power transmission instead of DC?

AC voltage can be easily stepped up by transformers for long-distance transmission (high voltage, low current = low $I^2R$ losses) and stepped down for domestic use. DC doesn’t work with conventional transformers.

What is power factor and why does it matter?

Power factor $= \cos\phi$ tells us how much of the apparent power is actually doing useful work. Low power factor means more current for the same useful power, increasing transmission losses. Industries pay penalties for low power factor.

What happens at resonance in an LCR circuit?

Impedance is minimum ( $Z = R$ ), current is maximum, and the circuit is purely resistive (voltage and current are in phase). The voltages across $L$ and $C$ can individually exceed the source voltage but cancel each other.

What is the quality factor (Q)?

$Q = \omega_0 L/R$ measures the sharpness of resonance. High Q means a narrow, tall resonance peak (selective). Low Q means a broad, flat peak. Radio tuners use high-Q circuits to select one station.

Advanced Concepts

Phasor diagrams — the graphical tool

A phasor is a rotating vector that represents a sinusoidal quantity. The length represents amplitude, and the angle represents phase.

In a series LCR circuit:

$V_R$ is along the current direction (in phase)
$V_L$ leads current by 90° (points upward)
$V_C$ lags current by 90° (points downward)

The resultant voltage: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$

This is why $V_R + V_L + V_C \neq V_{source}$ — you must add them as phasors (vectors), not as scalars.

Bandwidth and selectivity

The bandwidth of a resonant circuit is the range of frequencies over which the current stays above $I_{max}/\sqrt{2}$ :

\Delta f = \frac{f_0}{Q} = \frac{R}{2\pi L}

A high Q-factor means narrow bandwidth (high selectivity). This is how a radio receiver picks one station out of hundreds — the LC circuit resonates at only the selected frequency.

Wattless current

When current and voltage are 90° out of phase (pure inductor or capacitor), the average power is zero. This current is called wattless current because it does no useful work — energy just oscillates back and forth.

In a general circuit: $I_{wattless} = I_{rms}\sin\phi$ , $I_{active} = I_{rms}\cos\phi$

CBSE boards love asking: “What is wattless current?” Answer: the component of current that is 90° out of phase with voltage, contributing zero average power. It equals $I\sin\phi$ , where $\phi$ is the phase angle.

LC oscillations

A charged capacitor connected to an inductor creates oscillations — energy bounces between the electric field of the capacitor and the magnetic field of the inductor.

q(t) = q_0 \cos(\omega_0 t), \quad \omega_0 = \frac{1}{\sqrt{LC}}

This is analogous to a mass-spring system in mechanics: $C$ is like the spring (stores potential energy), $L$ is like the mass (stores kinetic energy).

\frac{1}{2}\frac{q^2}{C} + \frac{1}{2}LI^2 = \frac{1}{2}\frac{q_0^2}{C} = \text{constant}

Electric field energy + magnetic field energy = total energy (conserved).

Additional Solved Examples

Example 4 (JEE Main): An LCR series circuit has $R = 40\,\Omega$ , $L = 5$ H, $C = 80\,\mu$ F connected to 200 V, 50 Hz. Find impedance, current, and power factor.

$X_L = 2\pi \times 50 \times 5 = 500\pi \approx 1571\,\Omega$

$X_C = \frac{1}{2\pi \times 50 \times 80 \times 10^{-6}} = \frac{1}{0.02513} \approx 39.8\,\Omega$

$Z = \sqrt{40^2 + (1571 - 39.8)^2} = \sqrt{1600 + 2345721} \approx 1531\,\Omega$

$I = 200/1531 \approx 0.131$ A

$\cos\phi = R/Z = 40/1531 \approx 0.026$ (very inductive circuit, almost all reactive)

Example 5 (CBSE): At what frequency does a capacitor of 10 $\mu$ F have a reactance of 100 $\Omega$ ?

$X_C = 1/(2\pi fC)$ . $f = 1/(2\pi \times 100 \times 10 \times 10^{-6}) = 1/(0.00628) \approx \mathbf{159}$ Hz.

Additional Practice Questions

Q6. What is the phase difference between voltage and current in a pure inductor?

Voltage leads current by 90° ( $\pi/2$ radians). Remember ELI: in an inductor (L), EMF (E) leads current (I).

Q7. An LC circuit has $L = 0.2$ H and $C = 5\,\mu$ F. Find the frequency of oscillation.

$f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.2 \times 5 \times 10^{-6}}} = \frac{1}{2\pi \times 10^{-3}} = \frac{1000}{2\pi} \approx 159$ Hz.

Q8. A series LCR circuit is at resonance. The voltage across C is 500 V while the source voltage is 50 V. Find Q.

At resonance, $V_C = QV_{source}$ . $Q = V_C/V_{source} = 500/50 = 10$ .

Exam Weightage

Exam	Typical weight	Key topics
CBSE Class 12	4–6 marks	LCR impedance, resonance, transformer
JEE Main	1–2 questions	Resonance, power factor, reactance
NEET	1 question	RMS values, basic AC concepts

Choke coil vs resistor

A choke coil (inductor with minimal resistance) limits AC current without wasting power. A resistor wastes energy as heat. Fluorescent tube lights use a choke (called ballast) for this reason.

The choke consumes zero average power (in the ideal case) because voltage and current are 90° out of phase. In practice, a small resistance causes some power loss.

Comparison of DC and AC

Property	DC	AC
Direction	One way	Reverses periodically
Frequency	0 Hz	50 Hz (India), 60 Hz (USA)
Transformable	Not easily	Yes (transformers)
Long-distance transmission	High losses	Low losses (step up voltage)
Storage	Batteries	Difficult to store directly
Skin effect	None	Present at high frequencies

This comparison is a frequent 3-mark CBSE board question.