Is it correct to say: Using the peak value where the rms value is needed, or the other way round?

Meters and mains ratings give rms values. P = VI cos φ uses rms values; with peak values it becomes P = ½ v_m i_m cos φ.

Is it correct to say: Writing X_L = νL or X_C = 1/(νC)?

Use the angular frequency ω = 2πν: X_L = 2πνL and X_C = 1/(2πνC).

Is it correct to say: Adding voltages arithmetically in series LCR: V = V_R + V_L + V_C?

The voltages are out of phase and add as phasors: V = √(V_R² + (V_C − V_L)²).

Is it correct to say: Saying the impedance is zero at resonance?

At resonance X_C − X_L = 0, so Z = R, not zero. The current is then limited only by R.

Is it correct to say: Thinking a step-up transformer increases power?

It steps voltage up and current down. Ideally V_p I_p = V_s I_s; a real transformer delivers slightly less power than it takes in.

Is it correct to say: Calculating average power in a pure inductor as I²X_L?

A pure inductor or capacitor consumes zero average power. Reactance is not resistance — only R dissipates energy.

📄 Formula sheet (PDF)

Alternating Current

Peak and rms values, reactance and impedance, phasors, series LCR circuit, resonance, power factor and the transformer — NCERT Class 12 Physics Ch 7

📐 14 formulas✏️ 4 examples🎯 6 practice⚖️ Units III–IV: 17 marks (with Ch 4, 5 and 6)🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Always say whether a value is peak or rms. Mains ratings such as 220 V or 230 V are rms values; the peak is √2 times larger.
→Draw the phasor diagram for a series LCR circuit before writing Z: take current as reference, V_R along I, V_L 90° ahead, V_C 90° behind. It earns marks and fixes the sign of the phase angle.
→Use ω = 2πν. X_L = ωL rises with frequency and X_C = 1/ωC falls with frequency, which is why a capacitor blocks DC.
→At resonance X_L = X_C, Z = R, the current is maximum and the power factor is 1. The voltage across L or C alone can then be much larger than the supply voltage.
→For the transformer, learn the working principle, the turns ratio and the four energy losses (flux leakage, resistance of windings, eddy currents, hysteresis) with how each is reduced.

📐 Formulas(14)

Alternating Voltage

v = v_{m} sin ω t, ω = 2 π ν

Symbol	Meaning
$v$	Instantaneous voltage (V)
$v_{m}$	Peak voltage (V)
$ω$	Angular frequency (rad s⁻¹)
$ν$	Frequency (Hz)

AC Through a Resistor

i = i_{m} sin ω t, i_{m} = \frac{v _{m}}{R}

Symbol	Meaning
$i_{m}$	Peak current (A)
$R$	Resistance (Ω)

RMS (Effective) Current and Voltage★ Board fav

I = \frac{i _{m}}{2} = 0.707 i_{m}, V = \frac{v _{m}}{2} = 0.707 v_{m}

Symbol	Meaning
$I$	rms current (A)
$V$	rms voltage (V)

Average Power in a Resistor

P ˉ = \frac{1}{2} i_{m}^{2} R = I^{2} R = V I

Inductive Reactance

X_{L} = ω L = 2 π ν L

Symbol	Meaning
$X_{L}$	Inductive reactance (Ω)
$L$	Inductance (henry, H)

Capacitive Reactance

X_{C} = \frac{1}{ω C} = \frac{1}{2 π ν C}

Symbol	Meaning
$X_{C}$	Capacitive reactance (Ω)
$C$	Capacitance (farad, F)

Power in a Pure Inductor or Capacitor

P ˉ_{L} = P ˉ_{C} = 0

Impedance of a Series LCR Circuit★ Board fav

Z = R^{2} + (X_{C} - X_{L})^{2}, i_{m} = \frac{v _{m}}{Z}

Symbol	Meaning
$Z$	Impedance (Ω)

Phase Angle in a Series LCR Circuit

tan ϕ = \frac{X _{C} - X _{L}}{R}

Symbol	Meaning
$ϕ$	Phase angle between current and voltage

Resonant Frequency of a Series LCR Circuit★ Board fav

ω_{0} = \frac{1}{L C}, ν_{0} = \frac{1}{2 π L C}

Symbol	Meaning
$ω_{0}$	Resonant angular frequency (rad s⁻¹)
$ν_{0}$	Resonant frequency (Hz)

Average Power in an AC Circuit★ Board fav

P = V I cos ϕ = I^{2} Z cos ϕ = I^{2} R

Symbol	Meaning
$P$	Average power (W)
$cos ϕ$	Power factor

Power Factor

cos ϕ = \frac{R}{Z}

Transformer Voltage Ratio★ Board fav

\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}

Symbol	Meaning
$V_{p}, V_{s}$	Primary and secondary voltages (V)
$N_{p}, N_{s}$	Number of turns in primary and secondary

Ideal Transformer: Current Ratio

V_{p} I_{p} = V_{s} I_{s}, \frac{I _{p}}{I _{s}} = \frac{N _{s}}{N _{p}}

Symbol	Meaning
$I_{p}, I_{s}$	Primary and secondary currents (A)

✏️ Solved Examples

1Solved Exampleeasy3 steps

A 55 Ω resistor is connected to a 220 V, 50 Hz AC supply. Find (a) the rms current, (b) the peak current and (c) the average power dissipated.

220 V is the rms value, so Ohm's law gives the rms current

I = \frac{V}{R} = \frac{220}{55} = 4.0 A

2Solved Exampleboard8 steps

A series LCR circuit has R = 30 Ω, L = 80 mH and C = 25 μF. It is connected to a source v = 200√2 sin(1000t) volt. Find (a) the impedance, (b) the rms current, (c) the phase angle, (d) the power factor and average power, and (e) the resonant angular frequency.

Read off the rms voltage and angular frequency

V = \frac{200 2}{2} = 200 V, ω = 1000 rad s^{- 1}

3Solved Exampleboard3 steps

An ideal step-down transformer has 2000 turns in its primary and is connected to 220 V AC mains. It must deliver 11 V to a device that draws 4.0 A. Find (a) the number of turns in the secondary, (b) the primary current and (c) the power drawn from the mains.

Voltage ratio equals turns ratio

N_{s} = N_{p} \frac{V _{s}}{V _{p}} = 2000 \times \frac{11}{220} = 100

4Solved ExampleHOTS5 steps

A 0.50 H inductor, a capacitor C and a 22 Ω resistor are connected in series across a 220 V, 50 Hz supply. (a) Find C for resonance. (b) Find the rms current at resonance. (c) Find the rms voltage across the inductor at resonance and comment on the result.

At resonance ω = 1/√(LC), so C = 1/(ω²L) with ω = 2πν

C = \frac{1}{4 π ^{2} ν ^{2} L} = \frac{1}{4 π ^{2} \times 5 0 ^{2} \times 0.50} \approx 2.03 \times 1 0^{- 5} F

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Using the peak value where the rms value is needed, or the other way round
✓Meters and mains ratings give rms values. P = VI cos φ uses rms values; with peak values it becomes P = ½ v_m i_m cos φ.
2
Writing X_L = νL or X_C = 1/(νC)
✓Use the angular frequency ω = 2πν: X_L = 2πνL and X_C = 1/(2πνC).
3
Adding voltages arithmetically in series LCR: V = V_R + V_L + V_C
✓The voltages are out of phase and add as phasors: V = √(V_R² + (V_C − V_L)²).
4
Saying the impedance is zero at resonance
✓At resonance X_C − X_L = 0, so Z = R, not zero. The current is then limited only by R.
5
Thinking a step-up transformer increases power
✓It steps voltage up and current down. Ideally V_p I_p = V_s I_s; a real transformer delivers slightly less power than it takes in.
6
Calculating average power in a pure inductor as I²X_L
✓A pure inductor or capacitor consumes zero average power. Reactance is not resistance — only R dissipates energy.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
The rms voltage of a domestic supply is 230 V. What is its peak value?
Q2
Find the reactance of a 10 μF capacitor at 50 Hz.
Q3
Find the reactance of a 0.50 H inductor at 60 Hz.
Q4
Find the resonant frequency of a series LCR circuit with L = 10 mH and C = 1.0 μF.
Q5
In a series LCR circuit R = 6 Ω and Z = 10 Ω. If the rms supply voltage is 100 V, find the power factor and the average power.
Q6
What is the average power consumed by an ideal inductor connected to an AC source?

📝 Notes

Alternating Current

An AC source gives $v = v_m\sin\omega t$ . This chapter shows how a resistor, an inductor and a capacitor each respond, and what happens when all three are in series.

Peak versus rms

Every power calculation uses rms values: $I = i_m/\sqrt{2}$ and $V = v_m/\sqrt{2}$ . If a question gives $v = 311\sin(314t)$ , then $v_m = 311$ V, $V \approx 220$ V and $\nu = 314/2\pi \approx 50$ Hz.

How R, L and C respond

Resistor: current in phase with voltage; opposition $R$ , independent of frequency.
Inductor: current lags by $\pi/2$ ; $X_L = \omega L$ grows with frequency.
Capacitor: current leads by $\pi/2$ ; $X_C = 1/\omega C$ falls with frequency.

Pure L and pure C take zero average power.

Series LCR with phasors

Draw $V_R$ along the current, $V_L$ up and $V_C$ down. The resultant gives

$Z = \sqrt{R^2 + (X_C - X_L)^2}$ and $\tan\phi = (X_C - X_L)/R$ .
Resonance when $X_L = X_C$ : $\omega_0 = 1/\sqrt{LC}$ , $Z = R$ , current maximum.

Deriving $Z$ and $\phi$ from the phasor diagram is a common long-answer question, so practise drawing it neatly.

Power and the transformer

Average power is $P = VI\cos\phi$ with power factor $\cos\phi = R/Z$ . A low power factor means a large current for the same useful power, and so greater losses in transmission lines.

A transformer changes AC voltage in the ratio of turns, $V_s/V_p = N_s/N_p$ . Power is stepped up to high voltage (low current) for transmission to cut $I^2R$ losses, then stepped down near the consumer.