Is it correct to say: Treating the given mains voltage as the peak value?

Quoted AC values (230 V, 220 V) are rms. Peak = √2 × rms.

Is it correct to say: Adding R, X_L and X_C directly: Z = R + X_L + X_C?

They are out of phase. Z = √(R² + (X_L − X_C)²); X_L and X_C subtract because they are 180° apart.

Is it correct to say: Saying the current leads the emf in an inductor?

Inductor: current lags by π/2. Capacitor: current leads by π/2. Remember ‘L lags’.

Is it correct to say: Writing average power as e_rms × i_rms?

P_av = e_rms i_rms cos φ. The cos φ factor makes the power zero for pure L or C.

Is it correct to say: Using f instead of ω in X_L = ωL and X_C = 1/ωC?

ω = 2πf. With f = 50 Hz, ω ≈ 314 rad/s, not 50.

Is it correct to say: Stating that the mean value of AC over a full cycle is 2i₀/π?

2i₀/π is the mean over a half cycle. Over a full cycle the mean of a sinusoidal current is zero.

📄 Formula sheet (PDF)

AC Circuits

Mean and rms values, phasors, AC through R, L and C, series LCR impedance, power factor, wattless current, resonance, Q-factor, LC oscillations and choke coil — Maharashtra HSC Physics Ch 13

📐 14 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Learn the phase relations as one picture: in a resistor e and i are in phase, in an inductor current lags emf by π/2, in a capacitor current leads emf by π/2. Draw the phasor diagram for every theory answer on AC through L or C.
→The impedance of a series LCR circuit, Z = √(R² + (X_L − X_C)²), is derived from the phasor diagram. Draw V_R along the current, V_L up, V_C down, then add them.
→Meters read rms values. If the question says '220 V mains', that is e_rms. Multiply by √2 only when the peak value is asked for.
→Average power is e_rms·i_rms·cos φ, not e_rms·i_rms. In a purely inductive or capacitive circuit the power is zero even though a current flows: this is the wattless current.
→At resonance X_L = X_C, Z = R, current is maximum and power factor is 1. The voltage across L or C alone can then be much larger than the supply voltage (Q times).
→Remember why a choke coil is used instead of a resistor to reduce current in AC circuits: it has high inductance and low resistance, so it wastes very little power.

📐 Formulas(14)

Alternating EMF and Current

e = e_{0} sin ω t, i = i_{0} sin (ω t - ϕ), ω = 2 π f = \frac{2 π}{T}

Symbol	Meaning
$e_{0}$	Peak emf (V)
$i_{0}$	Peak current (A)
$ω$	Angular frequency (rad s⁻¹)
$f$	Frequency (Hz)
$ϕ$	Phase angle by which the current lags the emf (rad)

Mean (Average) Value over Half a Cycle

i_{a v} = \frac{2 i _{0}}{π} \approx 0.637 i_{0}, e_{a v} = \frac{2 e _{0}}{π}

RMS (Effective) Value★ Board fav

i_{r m s} = \frac{i _{0}}{2} \approx 0.707 i_{0}, e_{r m s} = \frac{e _{0}}{2}

AC Through a Pure Resistor

i = \frac{e _{0}}{R} sin ω t, i_{0} = \frac{e _{0}}{R}

Inductive Reactance

X_{L} = ω L = 2 π f L, i = i_{0} sin (ω t - \frac{π}{2})

Symbol	Meaning
$X_{L}$	Inductive reactance (Ω)
$L$	Self-inductance (H)

Capacitive Reactance

X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}, i = i_{0} sin (ω t + \frac{π}{2})

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

Impedance of a Series LCR Circuit★ Board fav

Z = R^{2} + (X_{L} - X_{C})^{2}, i_{0} = \frac{e _{0}}{Z}, i_{r m s} = \frac{e _{r m s}}{Z}

Symbol	Meaning
$Z$	Impedance (Ω)
$R$	Resistance (Ω)

Phase Angle in a Series LCR Circuit

tan ϕ = \frac{X _{L} - X _{C}}{R}

Average Power in an AC Circuit★ Board fav

P_{a v} = e_{r m s} i_{r m s} cos ϕ

Power Factor

cos ϕ = \frac{R}{Z}

Wattless Current

i_{w a ttl ess} = i_{r m s} sin ϕ

Resonant Frequency of a Series LCR Circuit★ Board fav

f_{r} = \frac{1}{2 π L C}, ω_{r} = \frac{1}{L C}

Quality Factor (Sharpness of Resonance)

Q = \frac{ω _{r}}{ω _{1} - ω _{2}} = \frac{ω _{r}}{2Δ ω} = \frac{ω _{r} L}{R} = \frac{1}{R} \frac{L}{C}

Symbol	Meaning
$Q$	Quality factor (no unit)
$ω_{r}$	Resonant angular frequency (rad s⁻¹)
$ω_{1}, ω_{2}$	Half-power angular frequencies; ω₁ − ω₂ = 2Δω is the bandwidth (rad s⁻¹)

LC Oscillations

q = q_{0} cos ω_{0} t, ω_{0} = \frac{1}{L C}, U = \frac{q _{0}^{2}}{2 C} = \frac{1}{2} L i_{0}^{2}

Symbol	Meaning
$q_{0}$	Initial (maximum) charge on the capacitor (C)
$U$	Total energy of the circuit (J)

✏️ Solved Examples

1Solved Exampleeasy3 steps

The domestic supply is 230 V, 50 Hz. Find the peak value of the voltage, its mean value over a half cycle, and its angular frequency.

The quoted 230 V is the rms value. Peak value:

e_{0} = 2 e_{r m s} = 1.414 \times 230 \approx 325 V

2Solved Exampleboard3 steps

A 10 μF capacitor is connected to a 230 V, 50 Hz supply. Find its reactance, the rms current and the average power consumed.

Capacitive reactance.

X_{C} = \frac{1}{2 π f C} = \frac{1}{2 π \times 50 \times 10 \times 1 0 ^{- 6}} \approx 318.3 Ω

3Solved Exampleboard5 steps

A series LCR circuit has R = 80 Ω, L = 0.1 H and C = 25 μF. It is connected to a 220 V (rms) source of angular frequency 1000 rad/s. Find the impedance, rms current, power factor and average power. Does the current lead or lag?

Reactances.

X_{L} = ω L = 1000 \times 0.1 = 100 Ω, X_{C} = \frac{1}{ω C} = \frac{1}{1000 \times 25 \times 1 0 ^{- 6}} = 40 Ω

4Solved ExampleHOTS5 steps

A series circuit with L = 0.5 H, C = 8 μF and R = 10 Ω is connected to a 50 V (rms) source of variable frequency. Find the resonant frequency, the Q-factor, and the rms voltage across the inductor at resonance.

Resonant angular frequency.

ω_{r} = \frac{1}{L C} = \frac{1}{0.5 \times 8 \times 1 0 ^{- 6}} = \frac{1}{2 \times 1 0 ^{- 3}} = 500 rad s^{- 1}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Treating the given mains voltage as the peak value
✓Quoted AC values (230 V, 220 V) are rms. Peak = √2 × rms.
2
Adding R, X_L and X_C directly: Z = R + X_L + X_C
✓They are out of phase. Z = √(R² + (X_L − X_C)²); X_L and X_C subtract because they are 180° apart.
3
Saying the current leads the emf in an inductor
✓Inductor: current lags by π/2. Capacitor: current leads by π/2. Remember ‘L lags’.
4
Writing average power as e_rms × i_rms
✓P_av = e_rms i_rms cos φ. The cos φ factor makes the power zero for pure L or C.
5
Using f instead of ω in X_L = ωL and X_C = 1/ωC
✓ω = 2πf. With f = 50 Hz, ω ≈ 314 rad/s, not 50.
6
Stating that the mean value of AC over a full cycle is 2i₀/π
✓2i₀/π is the mean over a half cycle. Over a full cycle the mean of a sinusoidal current is zero.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
A pure inductor of 0.2 H is connected to a 220 V, 50 Hz supply. Find its reactance and the rms current.
Q2
The peak value of an alternating current is 5 A. Find its rms value and its mean value over a half cycle.
Q3
Find the resonant frequency of a series circuit with L = 10 mH and C = 1 μF.
Q4
A series LCR circuit with R = 60 Ω has impedance 100 Ω on a 200 V (rms) supply. Find the power factor, the average power and the wattless component of current.
Q5
Calculate the Q-factor of a series LCR circuit with R = 20 Ω, L = 0.4 H and C = 10 μF.
Q6
What capacitance must be connected to a 25 mH inductor so that the LC circuit oscillates at 1 kHz?

📝 Notes

AC Circuits — Maharashtra HSC Overview

Chapter 13 of the Maharashtra Board Std XII Physics textbook follows on from the AC generator of Chapter 12. It asks how resistors, inductors and capacitors behave when the emf keeps reversing, and how much power such circuits actually use.

Peak, mean and rms values

Every AC numerical begins by identifying which value is given. Peak values ( $e_0$ , $i_0$ ) appear in the equations. Meters and supply ratings give rms values ( $e_0/\sqrt{2}$ ). The mean over a half cycle ( $2e_0/\pi$ ) is the third, less common one.

One current, three voltages

In a series circuit the current is common, so take it as the reference phasor. $V_R$ is in phase with it, $V_L$ is $90^\circ$ ahead and $V_C$ is $90^\circ$ behind. Add them as vectors to get $Z = \sqrt{R^2 + (X_L - X_C)^2}$ and $\tan\phi = (X_L - X_C)/R$ . The same diagram also gives the power factor $\cos\phi = R/Z$ .

Resonance

As the frequency rises, $X_L$ increases and $X_C$ decreases. They become equal at $f_r = 1/(2\pi\sqrt{LC})$ . At this frequency a series circuit offers only $R$ , the current is maximum and $\cos\phi = 1$ . A parallel LC circuit does the opposite: its impedance is maximum. The Q-factor tells you how sharp the series resonance peak is, and by how much the voltage across L or C is magnified.

Power and the choke coil

Power is consumed only in resistance. An ideal inductor or capacitor stores energy for a quarter cycle and returns it in the next, so the average power is zero. This is why a choke coil, a large inductance with small resistance, can cut down AC current in a fluorescent tube circuit with almost no energy wasted as heat. A series resistor doing the same job would waste energy as $i^2R$ heat.