Intermediate

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Resonance, Power in AC Circuits & Transformers

Master AC Resonance and Transformers for JEE & NEET. Covers ω0=1/√LC, Q-factor three forms, bandwidth, power factor cosφ=R/Z, wattless current, transformer turns ratio, losses.

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Resonance in a series LCR circuit is the specific frequency at which the circuit's impedance hits an absolute minimum and the current hits a maximum — a phenomenon with profound practical applications ranging from radio tuning to power engineering. The Power factor governs how efficiently an AC circuit converts apparent power into real, useful work — a critical concept for industrial power systems. Lastly, the Transformer — arguably the single most important application of electromagnetic induction — is what enables the long-distance transmission of electricity by stepping voltages up for transmission and down for safe distribution. For JEE and NEET, this topic provides high-yield numerical questions on resonant frequency, Q-factor, power factor, transformer turns ratios, and efficiency.

1. Resonance in a Series LCR Circuit

At resonance, the inductive and capacitive reactances perfectly cancel out ( $X_{L} = X_{C}$ ). Therefore, $Z = R$ (minimum impedance) and $I_{rms} = V_{rms} / R$ (maximum current).

Setting $X_{L} = X_{C}$ : $ω_{0} L = \frac{1}{ω_{0} C} ⟹$

$ω_{0} = \frac{1}{\sqrt{L C}} f_{0} = \frac{1}{2 π \sqrt{L C}}$

Key Features at Resonance

Impedance $Z = R$ (minimum); current $I_{max} = V / R$ (maximum).
Phase angle $ϕ = 0^{\circ}$ — voltage and current are perfectly in phase.
Power factor $= \cos (0^{\circ}) = 1$ — maximum power transfer occurs.
$V_{L} = V_{C}$ — the individual voltages across the inductor and capacitor are equal and opposite, cancelling each other out.
$V_{L}$ and $V_{C}$ can individually be much larger than the total supply voltage — a phenomenon known as voltage amplification.

Worked Example

An LCR circuit has $L = 0.5 H$ and $C = 200 μ F$ . Find the resonant frequency.

$ω_{0} = \frac{1}{\sqrt{0.5 \times 200 \times 10^{- 6}}} = \frac{1}{\sqrt{10^{- 4}}} = \frac{1}{0.01} = 100 rad/s$

$f_{0} = \frac{100}{2 π} \approx 15.9 Hz$

2. Quality Factor ( $Q$ -factor)

The $Q$ -factor measures the "sharpness" of resonance — essentially how highly selective the circuit is at tuning into $ω_{0}$ .

$Q = \frac{ω_{0} L}{R} = \frac{1}{ω_{0} C R} = \frac{1}{R} \sqrt{\frac{L}{C}}$

All three forms are mathematically equivalent. Typical values for practical radio circuits range from $Q = 5$ to $100$ .

Bandwidth and Half-Power Points

The bandwidth ( $Δ ω$ ) is the spread of frequencies over which the dissipated power is at least half of its maximum peak value:

$Δ ω = \frac{R}{L} = \frac{ω_{0}}{Q}$

At the exact half-power points ( $ω_{0} \pm Δ ω / 2$ ): Current $= \frac{I_{max}}{\sqrt{2}}$ , and Power $= \frac{P_{max}}{2}$ .

$Q$ Value	Bandwidth	Resonance Peak
High $Q$ (Low $R$ )	Narrow	Sharp and tall (Highly selective)
Low $Q$ (High $R$ )	Broad	Flat and wide (Non-selective)

Worked Example

If $L = 0.5 H$ , $C = 200 μ F$ , $R = 10 Ω$ , and $ω_{0} = 100 rad/s$ :

$Q = \frac{ω_{0} L}{R} = \frac{100 \times 0.5}{10} = 5$

Bandwidth $Δ ω = \frac{R}{L} = \frac{10}{0.5} = 20 rad/s$ (Verification: $ω_{0} / Q = 100 / 5 = 20 ✓$ )

3. Power in AC Circuits

Instantaneous and Average Power

Instantaneous power $p = v i$ . For $v = V_{0} \sin (ω t)$ and $i = I_{0} \sin (ω t - ϕ)$ , the average power over a full cycle evaluates to:

$P_{avg} = V_{rms} I_{rms} \cos ϕ = I_{rms}^{2} R$

Power Factor

$Power factor = \cos ϕ = \frac{R}{Z} = \frac{P_{avg}}{V_{rms} I_{rms}}$

Circuit Element	Power Factor ( $\cos ϕ$ )	Average Power
Pure Resistor	$1$ ( $ϕ = 0^{\circ}$ )	$P = V_{rms} I_{rms}$ (maximum)
Pure Inductor	$0$ ( $ϕ = 90^{\circ}$ )	$P = 0$ (wattless)
Pure Capacitor	$0$ ( $ϕ = 90^{\circ}$ )	$P = 0$ (wattless)
Series LCR at Resonance	$1$ ( $ϕ = 0^{\circ}$ )	$P = V_{rms} I_{rms}$ (maximum)
General LCR	$R / Z$	$P = I_{rms}^{2} R$

Wattless Current

The component of the current that is perfectly out of phase with voltage does no net work: $I_{wattless} = I_{rms} \sin ϕ$ .

The component that does the actual work: $I_{active} = I_{rms} \cos ϕ$ (in phase with voltage).

Proof: $P = V_{rms} \times I_{active} = V_{rms} \times I_{rms} \cos ϕ ✓$

4. The Transformer

A transformer uses mutual induction to effortlessly change AC voltage (and current) levels. An ideal transformer has absolutely no energy losses.

Transformer Equations (Ideal)

$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} = \frac{I_{p}}{I_{s}}$

where $p$ = primary coil, $s$ = secondary coil, and $N$ = number of turns.

Energy conservation constraint: $V_{p} I_{p} = V_{s} I_{s}$ (Input Power exactly equals Output Power).

Type	Turns Condition	Effect
Step-up	$N_{s} > N_{p}$	$V_{s} > V_{p}$ and $I_{s} < I_{p}$ (Boosts voltage, drops current)
Step-down	$N_{s} < N_{p}$	$V_{s} < V_{p}$ and $I_{s} > I_{p}$ (Drops voltage, boosts current)

Transformer Efficiency & Energy Losses

$η = \frac{P_{s}}{P_{p}} = \frac{V_{s} I_{s}}{V_{p} I_{p}} \times 100 %$

Real transformers have $η \approx 95$ – $99 %$ due to practical losses:

Loss Type	Cause	Mitigation Strategy
Copper loss	Ohmic resistance of windings ( $I^{2} R$ heating)	Use thick, low-resistance copper wire
Iron (Eddy) loss	Induced Eddy currents swirling in the iron core	Use a laminated core (thin insulated sheets)
Hysteresis loss	Energy spent reversing the core's magnetic domains	Use soft iron (which has a narrow hysteresis loop)
Flux leakage	Magnetic flux failing to fully link to the secondary coil	Wind the primary and secondary coils tightly over one another

Power Transmission — Why Step Up Voltage?

The power lost to heating in a long transmission line is $P_{loss} = I^{2} R_{line}$ . For a fixed power output ( $P = V I$ ), stepping up the voltage proportionately steps down the current. This dramatically slashes $I^{2} R$ losses.
Example: Stepping up voltage from 11 kV to 220 kV (a 20x increase) drops the current by a factor of 20, which reduces the $I^{2} R$ heating loss by a massive factor of 400.

Resonance, Q-factor, and Bandwidth — All in One

Quantity	Formula	Physical Meaning
Resonant frequency	$ω_{0} = 1 / \sqrt{L C}$	$X_{L} = X_{C}$ ; Min impedance $Z$ ; Max current $I$ .
Q factor (3 forms)	$Q = \frac{ω_{0} L}{R} = \frac{1}{ω_{0} C R} = \frac{1}{R} \sqrt{\frac{L}{C}}$	Curve Sharpness; acts as a direct voltage multiplier at resonance.
Bandwidth	$Δ ω = R / L = ω_{0} / Q$	The specific frequency range where Power $\geq P_{max} / 2$ .
Voltage magnification	$V_{L} / V = V_{C} / V = Q$ (at resonance only)	$V_{L}$ and $V_{C}$ individually blow up to equal $Q \times$ the supply voltage.

JEE / NEET Power Tips — Resonance, Power & Transformers

$ω_{0} = 1 / \sqrt{L C}$ is completely independent of $R$ . Resistance does not dictate the resonant frequency — it only controls the sharpness (Q-factor) and the height of the current peak. Increasing $R$ broadens the curve and lowers the max current without shifting the peak left or right.
At resonance: $V_{L} = Q \times V_{supply}$ and $V_{C} = Q \times V_{supply}$ . If a circuit has a Q-factor of 5 and a 100 V supply, the voltage across the inductor and capacitor will hit a staggering 500 V each. This selective voltage amplification is exactly how AM/FM radio tuners isolate weak signals.
Average power $= I_{rms}^{2} R$ always — only resistors dissipate true power. Whether or not you know the phase angle, $P = I_{rms}^{2} R$ is always correct. The general formula $P = V_{rms} I_{rms} \cos ϕ$ yields the exact same answer since $\cos ϕ = R / Z$ .
Transformer: The current ratio is the INVERSE of the turns ratio. $I_{p} / I_{s} = N_{s} / N_{p}$ . A step-up transformer steps up the voltage but forces the current to step down to conserve power. The most common student error is reversing this ratio.
Eddy current losses are fixed by lamination, NOT by soft iron. Lamination directly slices up the bulky core to break the conducting paths of Eddy currents. Soft iron is used solely to shrink the Hysteresis loop. These are two distinct physical loss mechanisms requiring distinct material solutions.

Practice Questions

Q1 (JEE Main / NEET): A series LCR circuit has $L = 0.5 H$ , $C = 200 μ F$ , and $R = 10 Ω$ . Find (i) the resonant angular frequency, (ii) the Q-factor, and (iii) the bandwidth.

Explanation:

(i) $ω_{0} = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{0.5 \times 200 \times 10^{- 6}}} = \frac{1}{\sqrt{10^{- 4}}} = \frac{1}{0.01} = 100 rad/s$

(ii) $Q = \frac{ω_{0} L}{R} = \frac{100 \times 0.5}{10} = 5$

(iii) $Δ ω = \frac{R}{L} = \frac{10}{0.5} = 20 rad/s$ (Verification: $ω_{0} / Q = 100 / 5 = 20 ✓$ )

Q2 (JEE Main): In a series LCR circuit at perfect resonance, $L = 2 H$ , $C = 50 μ F$ , $R = 10 Ω$ , and the supply voltage is 100 V. Find the voltage across the inductor.

Explanation:

First, find the resonant frequency and Q-factor:
$ω_{0} = 1 / \sqrt{2 \times 50 \times 10^{- 6}} = 1 / \sqrt{10^{- 4}} = 100 rad/s$
$Q = \frac{ω_{0} L}{R} = \frac{100 \times 2}{10} = 20$

At resonance, $I_{rms} = V / R = 100 / 10 = 10 A$
$V_{L} = I_{rms} \times X_{L} = 10 \times (ω_{0} L) = 10 \times 100 \times 2 = 2000 V$

(Shortcut check: $V_{L} = Q \times V_{supply} = 20 \times 100 = 2000 V ✓$ )

Q3 (NEET MCQ): The power factor of a series LCR circuit at exactly the resonant frequency is:

A) 0
B) 0.5
C) $1 / \sqrt{2}$
D) 1

Answer: D) 1.

Explanation: At resonance, $X_{L}$ perfectly cancels $X_{C}$ , meaning the net impedance $Z = R$ . Because there is no net reactance, the voltage and current are perfectly in phase ( $ϕ = 0^{\circ}$ ). The power factor is $\cos (0^{\circ}) = 1$ , yielding maximum power transfer.

Q4 (Board): A step-down transformer has 3000 turns in the primary and 150 turns in the secondary. If the primary voltage is 2200 V and the secondary circuit draws 10 A, find (i) secondary voltage, (ii) primary current, and (iii) power delivered.

Explanation:

(i) $V_{s} = V_{p} \times (\frac{N_{s}}{N_{p}}) = 2200 \times (\frac{150}{3000}) = 2200 \times \frac{1}{20} = 110 V$

(ii) $I_{p} = I_{s} \times (\frac{N_{s}}{N_{p}}) = 10 \times (\frac{150}{3000}) = 10 \times \frac{1}{20} = 0.5 A$

(iii) $P_{del} = V_{s} I_{s} = 110 \times 10 = 1100 W$
(Verification via Primary side: $V_{p} I_{p} = 2200 \times 0.5 = 1100 W ✓$ )

Q5 (JEE Main): An AC circuit is observed to have $V_{rms} =$ 200 V, $I_{rms} =$ 4 A, and a real average power $P =$ 400 W. Find the power factor and phase angle.

Explanation:

Use the average power formula $P = V_{rms} I_{rms} \cos ϕ$ :
$\cos ϕ = \frac{P}{V_{rms} I_{rms}} = \frac{400}{200 \times 4} = \frac{400}{800} = 0.5$ (This is the Power Factor)

To find the phase angle, take the inverse cosine:
$ϕ = \cos^{- 1} (0.5) = 60^{\circ}$

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Resonance, Power in AC Circuits & Transformers

1. Resonance in a Series LCR Circuit

Key Features at Resonance

Worked Example

2. Quality Factor ( $Q$ -factor)

Bandwidth and Half-Power Points

Worked Example

3. Power in AC Circuits

Instantaneous and Average Power

Power Factor

Wattless Current

4. The Transformer

Transformer Equations (Ideal)

Transformer Efficiency & Energy Losses

Power Transmission — Why Step Up Voltage?

Resonance, Q-factor, and Bandwidth — All in One

Practice Questions

Related Study Material

Practice Questions

Topic Information

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Target Subjects

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Subject Mastery

Resonance, Power in AC Circuits & Transformers

1. Resonance in a Series LCR Circuit

Key Features at Resonance

Worked Example

2. Quality Factor (Q-factor)

Bandwidth and Half-Power Points

Worked Example

3. Power in AC Circuits

Instantaneous and Average Power

Power Factor

Wattless Current

4. The Transformer

Transformer Equations (Ideal)

Transformer Efficiency & Energy Losses

Power Transmission — Why Step Up Voltage?

Resonance, Q-factor, and Bandwidth — All in One

Practice Questions

Related Study Material

Practice Questions

Topic Information

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Track Your Learning

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2. Quality Factor ( $Q$ -factor)