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CBSE · Class 12 · Physics

Chapter 7: Alternating Current

7 concepts · 56 practice questions with answers · Free to read

AC voltage applied to a resistor and rms values

From daily life

The wall socket of a house gives a voltage that reverses its direction 100 times every second, yet an iron or a heater works just as well as on a battery. Link to earlier classes : Ohm's law and heating effect from Chapter 3, and the AC generator of Chapter 6.

Simple definition

An alternating voltage changes its size and direction periodically, usually as a sine wave. In a pure resistor the current rises and falls together with the voltage, that is, they are in phase. Since the average of an AC over a full cycle is zero, we describe it by its rms (root mean square) value, which gives the same heating as a DC of that value.

Rule 1: Voltage, current and rms values — v = v_m sin ωt, i = i_m sin ωt with i_m = v_m ÷ R. The rms values are V_rms = v_m ÷ √2 and I_rms = i_m ÷ √2 (about 0.707 of the peak).

  • Easy : The household mains of 220 V rms has a peak of 220 × 1.414 = 311 V.
  • Medium : For R = 50 Ω and V_rms = 220 V, I_rms = 4.4 A and the peak current is 4.4 × 1.414 = 6.2 A.
  • Exam : An AC meter always shows the rms value, not the peak.

Rule 2: Power in a resistor — The average power is P = V_rms I_rms = I_rms² R = ½ i_m² R.

  • Easy : For the 50 Ω resistor above, P = 220 × 4.4 = 968 W.
  • Medium : A peak current of 5 A through 20 Ω gives P = ½ × 25 × 20 = 250 W.
  • Exam : The equivalent DC for a 5 A peak AC is 5 ÷ √2 = 3.5 A.
Remember

rms = peak ÷ √2; in a resistor, v and i are in phase; P = I_rms²R.

Common mistake

Using the peak current in P = I²R. Use the rms value, or the factor ½ with the peak.

Shortcut

Memory clue : "rms is 0.707 of the peak: peak over root two."

Spot the difference

ValueMeaning
Peak (v_m)the largest value
rmsgives the same heating as DC
Average over a cyclezero

Practice questions

Easy
  1. 1Write the relation between rms and peak values.
  2. 2Find the peak voltage of 220 V rms mains.
  3. 3What is the average value of AC over a full cycle?
Medium
  1. 4The peak voltage is 311 V. Find the rms voltage.
  2. 5A 50 Ω resistor is connected to 220 V rms. Find the rms and peak currents.
  3. 6Find the power consumed by that resistor.
Challenging
  1. 7A peak current of 5 A flows in a 20 Ω resistor. Find the average power.
  2. 8Which DC current gives the same heating as that AC?
Show answers (try them yourself first)
  1. 1V_rms = V_peak ÷ √2.
  2. 2311 V.
  3. 3Zero.
  4. 4V_rms = 311 ÷ 1.414 ≈ 220 V.
  5. 5I_rms = 4.4 A; peak = 6.2 A.
  6. 6P = 220 × 4.4 = 968 W.
  7. 7P = ½ i_m² R = 250 W.
  8. 85 ÷ √2 ≈ 3.5 A.

AC voltage applied to an inductor

From daily life

A choke coil in a tube light limits the current without wasting power as heat. Link to earlier classes : self-inductance from Chapter 6.

Simple definition

An inductor opposes a changing current through self-induced emf. In an AC circuit the opposition is called inductive reactance X_L. The current in a pure inductor lags the voltage by 90° (π ÷ 2).

Rule 1: Inductive reactance — X_L = ω L = 2π f L (unit ohm); the rms current is I = V ÷ X_L. X_L grows with frequency and is zero for DC.

  • Easy : L = 0.50 H at 50 Hz ⇒ X_L = 314 × 0.5 = 157 Ω.
  • Medium : On 220 V, I_rms = 220 ÷ 157 = 1.40 A; at 100 Hz, X_L = 314 Ω and the current halves.
  • Exam : For X_L = 100 Ω at 50 Hz, L = 100 ÷ 314 = 0.318 H.

Rule 2: Power in a pure inductor — The average power over a cycle is zero. Energy is stored in the magnetic field during one quarter of the cycle and returned in the next.

  • Easy : A pure inductor consumes no power, though a current flows.
  • Medium : It passes low frequencies easily and opposes high frequencies.
  • Exam : For DC, ω = 0 so X_L = 0: the inductor behaves like a plain wire.
Remember

X_L = ωL; current lags by 90°; average power zero.

Common mistake

Thinking that reactance X_L is a true resistance that wastes energy as heat. It does not.

Shortcut

Memory clue : "ELI the ICE man: in L (inductor), voltage E comes first, current I lags."

Spot the difference

SourceEffect of higher f on X_L
Inductorincreases
Resistornone

Practice questions

Easy
  1. 1Write the formula for inductive reactance.
  2. 2What is the phase relation between current and voltage in a pure inductor?
  3. 3What is X_L for DC?
Medium
  1. 4Find X_L of a 0.50 H inductor at 50 Hz.
  2. 5Find the rms current on a 220 V, 50 Hz supply.
  3. 6Find X_L at 100 Hz.
Challenging
  1. 7How much power does a pure inductor consume?
  2. 8Find the inductance with X_L = 100 Ω at 50 Hz.
Show answers (try them yourself first)
  1. 1X_L = ωL = 2πfL.
  2. 2The current lags the voltage by 90°.
  3. 3Zero.
  4. 4X_L ≈ 157 Ω.
  5. 5I = 220 ÷ 157 ≈ 1.40 A.
  6. 6X_L ≈ 314 Ω.
  7. 7Zero (on average).
  8. 8L = 100 ÷ 314 ≈ 0.318 H.

AC voltage applied to a capacitor

From daily life

A capacitor in a fan or in a radio blocks direct current but lets alternating signals pass. Link to earlier classes : capacitors from Chapter 2.

Simple definition

A capacitor charges and discharges as the AC reverses. Its opposition to the current is the capacitive reactance X_C. The current in a pure capacitor leads the voltage by 90° (π ÷ 2).

Rule 1: Capacitive reactance — X_C = 1 ÷ (ω C) = 1 ÷ (2π f C) (unit ohm); the rms current is I = V ÷ X_C. X_C falls as frequency rises and is infinite for DC.

  • Easy : C = 20 μF at 50 Hz ⇒ X_C = 1 ÷ (314 × 20 × 10⁻⁶) = 159 Ω.
  • Medium : On 220 V, I_rms = 220 ÷ 159 = 1.38 A; at 100 Hz, X_C = 79.6 Ω.
  • Exam : For X_C = 100 Ω at 50 Hz, C = 1 ÷ (314 × 100) = 31.8 μF.

Rule 2: Power in a pure capacitor — The average power is zero. The capacitor stores energy in its electric field and gives it back.

  • Easy : A capacitor in a DC circuit blocks the current once it is charged.
  • Medium : It passes high frequencies easily, the opposite of an inductor.
  • Exam : The current is largest when the voltage across the capacitor is zero.
Remember

X_C = 1/ωC; current leads by 90°; average power zero.

Common mistake

Writing X_C = ωC. The reactance of a capacitor is the inverse.

Shortcut

Memory clue : "ELI the ICE man: in C (capacitor), current I leads voltage E."

Spot the difference

ComponentReactancePhase of currentBlocks
InductorωLlags by 90°high frequency
Capacitor1/ωCleads by 90°DC (low frequency)

Practice questions

Easy
  1. 1Write the formula for capacitive reactance.
  2. 2What is the phase relation between current and voltage in a pure capacitor?
  3. 3How does a capacitor behave for DC?
Medium
  1. 4Find X_C for 20 μF at 50 Hz.
  2. 5Find the rms current on a 220 V, 50 Hz supply.
  3. 6Find X_C at 100 Hz.
Challenging
  1. 7How much power does a pure capacitor consume?
  2. 8Find the capacitance for which X_C = 100 Ω at 50 Hz.
Show answers (try them yourself first)
  1. 1X_C = 1 ÷ (ωC).
  2. 2The current leads the voltage by 90°.
  3. 3It blocks the current (infinite reactance).
  4. 4X_C ≈ 159 Ω.
  5. 5I ≈ 1.38 A.
  6. 6X_C ≈ 79.6 Ω.
  7. 7Zero (on average).
  8. 8C ≈ 31.8 μF.

The series LCR circuit and impedance

From daily life

A radio set picks one station out of many using a circuit with a coil and a capacitor. Link to earlier classes : the previous three sections and Pythagoras' theorem.

Simple definition

In a series LCR circuit the same current flows through R, L and C. The voltages across them are not in phase, so they add as vectors (phasors). The total opposition is the impedance Z.

Rule 1: Impedance and phase — Z = √[R² + (X_L − X_C)²], I = V ÷ Z and tan φ = (X_L − X_C) ÷ R. The current lags the voltage if X_L > X_C (the circuit is inductive) and leads it if X_C > X_L (capacitive).

  • Easy : R = 30 Ω, X_L = 80 Ω, X_C = 40 Ω ⇒ Z = √(900 + 1600) = 50 Ω.
  • Medium : On 220 V, I = 4.4 A and tan φ = 40 ÷ 30, so φ = 53° with the current lagging.
  • Exam : The voltages are V_R = 4.4 × 30 = 132 V, V_L = 4.4 × 80 = 352 V and V_C = 4.4 × 40 = 176 V. Here V_L is larger than the supply, but V = √[V_R² + (V_L − V_C)²] = √(132² + 176²) = 220 V.

Rule 2: Phasor diagram — V_R is along the current, V_L is 90° ahead and V_C is 90° behind. The supply voltage is the vector sum.

  • Easy : V_L and V_C point in opposite directions, so they partly cancel.
  • Medium : If X_L = X_C the net reactance is zero and Z = R (resonance).
  • Exam : A voltage across L or C can be greater than the supply voltage without any contradiction, as they are 180° out of phase.
Phasor diagram of a series LCR circuit when XL is greater than XCV_RV_LV_CVφV_L − V_C= 176 VV_R = 132 VV = 220 V

In the figure : Phasor diagram of a series LCR circuit with X_L > X_C: V_L above, V_C below and the resultant V at an angle φ to V_R.

Remember

Z = √[R² + (X_L − X_C)²]; tan φ = (X_L − X_C)/R; add voltages as vectors.

Common mistake

Adding R, X_L and X_C directly as numbers.

Shortcut

Memory clue : "Z is the hypotenuse: R along, (X_L minus X_C) up."

Spot the difference

ConditionCircuit isCurrent
X_L > X_Cinductivelags voltage
X_C > X_Lcapacitiveleads voltage
X_L = X_Cresistivein phase

Practice questions

Easy
  1. 1Write the formula for the impedance of a series LCR circuit.
  2. 2What is the unit of impedance?
  3. 3If X_L > X_C, does the current lead or lag the voltage?
Medium
  1. 4Find Z for R = 30 Ω, X_L = 80 Ω and X_C = 40 Ω.
  2. 5Find the current on a 220 V supply.
  3. 6Find the phase angle between the voltage and the current.
Challenging
  1. 7Find the rms voltages across R, L and C in the circuit above.
  2. 8Verify that these voltages combine to give the 220 V supply.
Show answers (try them yourself first)
  1. 1Z = √[R² + (X_L − X_C)²].
  2. 2Ohm.
  3. 3It lags.
  4. 4Z = 50 Ω.
  5. 5I = 220 ÷ 50 = 4.4 A.
  6. 6φ = tan⁻¹(40 ÷ 30) ≈ 53°; the current lags.
  7. 7V_R = 132 V, V_L = 352 V, V_C = 176 V.
  8. 8√[132² + (352 − 176)²] = √(132² + 176²) = 220 V.

Resonance in the LCR circuit

From daily life

Turning the tuning knob of a radio changes a capacitor so that the circuit resonates with one station. Link to earlier classes : resonance of oscillations from Class 11.

Simple definition

At resonance the inductive and capacitive reactances are equal. Then the impedance is least (equal to R), the current is largest and the voltage and current are in phase.

Rule 1: Resonant frequency — ω₀ = 1 ÷ √(L C), so f₀ = 1 ÷ (2π √(L C)). At resonance Z = R and I = V ÷ R.

  • Easy : L = 0.10 H and C = 10 μF ⇒ ω₀ = 1 ÷ √(10⁻⁶) = 1000 rad s⁻¹ and f₀ = 159 Hz.
  • Medium : With R = 10 Ω and V = 100 V, the current at resonance is 10 A.
  • Exam : To tune a radio to a higher frequency, reduce C or L.

Rule 2: Quality factor — Q = ω₀ L ÷ R = 1 ÷ (ω₀ C R). A higher Q means a sharper resonance. The bandwidth is 2Δω = R ÷ L, and Q = ω₀ ÷ (2Δω).

  • Easy : Q = 1000 × 0.10 ÷ 10 = 10.
  • Medium : The bandwidth is R ÷ L = 10 ÷ 0.10 = 100 rad s⁻¹, and ω₀ ÷ 100 = 10, as before.
  • Exam : At resonance V_L = I X_L = 10 × 100 = 1000 V, which is Q times the supply voltage of 100 V.
Current against frequency in a series LCR circuit: resonance peakfIf₀small R (sharp)large R (flat)

In the figure : The current against frequency: a peak at f₀ whose sharpness depends on R.

Remember

ω₀ = 1/√LC; Z = R and I is largest at resonance; Q = ω₀L/R.

Common mistake

Thinking that resonance occurs when R is zero. It occurs when X_L = X_C.

Shortcut

Memory clue : "Resonance: the two reactances cancel and only R is left."

Spot the difference

QuantityAt resonance
Impedanceminimum (= R)
Currentmaximum
Phase anglezero

Practice questions

Easy
  1. 1State the condition of resonance in a series LCR circuit.
  2. 2What is the impedance at resonance?
  3. 3What is the phase angle at resonance?
Medium
  1. 4Find ω₀ and f₀ for L = 0.10 H and C = 10 μF.
  2. 5Find the current at resonance for R = 10 Ω and V = 100 V rms.
  3. 6Find the quality factor.
Challenging
  1. 7Find the voltage across the inductor at resonance.
  2. 8Find the bandwidth 2Δω.
Show answers (try them yourself first)
  1. 1X_L = X_C, that is ω₀ = 1 ÷ √(LC).
  2. 2Z = R, the minimum.
  3. 3Zero.
  4. 4ω₀ = 1000 rad s⁻¹; f₀ ≈ 159 Hz.
  5. 5I = V ÷ R = 10 A.
  6. 6Q = ω₀L ÷ R = 10.
  7. 7V_L = I X_L = 10 × 100 = 1000 V.
  8. 82Δω = R ÷ L = 100 rad s⁻¹.

Power in an AC circuit and the power factor

From daily life

A factory pays extra if its machines have a low power factor, since the supply must carry a larger current for the same useful power. Link to earlier classes : power from Chapter 3 and phase from the earlier sections.

Simple definition

The useful (average) power in an AC circuit depends on the phase angle φ between voltage and current. The factor cos φ is the power factor; it equals R ÷ Z.

Rule 1: Average power — P = V_rms I_rms cos φ, where cos φ = R ÷ Z. Also P = I_rms² R.

  • Easy : For R = 30 Ω, Z = 50 Ω, V = 220 V and I = 4.4 A, cos φ = 0.6 and P = 220 × 4.4 × 0.6 = 581 W.
  • Medium : Check: I²R = 4.4² × 30 = 581 W.
  • Exam : A load that draws 2 A at 220 V with a power factor of 0.8 uses P = 220 × 2 × 0.8 = 352 W.

Rule 2: Special cases — For a pure R, φ = 0 and cos φ = 1. For a pure L or C, φ = 90° and cos φ = 0, so the power is zero (the current is "wattless"). At resonance cos φ = 1.

  • Easy : A pure inductor draws a current but uses no power.
  • Medium : A low power factor needs a larger current for the same power, so more is lost in the supply lines.
  • Exam : Power factor is corrected by adding a capacitor to an inductive load.
Remember

P = VI cos φ; cos φ = R/Z; wattless current for L or C only.

Common mistake

Using P = VI for a circuit that contains L or C.

Shortcut

Memory clue : "Real power = V times I times cosine phi."

Spot the difference

CircuitφPower factor
Pure R0°1
Pure L or C90°0
LCR at resonance0°1

Practice questions

Easy
  1. 1Write the formula for the average power in an AC circuit.
  2. 2What is the power factor of a pure inductor?
  3. 3What is the power factor at resonance?
Medium
  1. 4For R = 30 Ω, Z = 50 Ω at 220 V, find the power factor.
  2. 5Find the power consumed in that circuit.
  3. 6Check the power using I²R.
Challenging
  1. 7A load draws 2 A at 220 V with a power factor of 0.8. Find the power used.
  2. 8Why is a current in a pure inductor called wattless?
Show answers (try them yourself first)
  1. 1P = V_rms I_rms cos φ.
  2. 2Zero.
  3. 3One.
  4. 4cos φ = R ÷ Z = 0.6.
  5. 5P = 220 × 4.4 × 0.6 ≈ 581 W.
  6. 6I²R = 19.36 × 30 ≈ 581 W.
  7. 7P = 220 × 2 × 0.8 = 352 W.
  8. 8The phase angle is 90°, so the average power is zero.

Transformers

From daily life

The grey box on a pole near your home steps the high voltage of the line down to 220 V; a phone charger steps it down further. Link to earlier classes : mutual induction from Chapter 6.

Simple definition

A transformer changes an AC voltage using mutual induction between two coils wound on a common soft-iron core. It does not work on DC. A step-up transformer has more turns in the secondary (N_s > N_p), a step-down transformer has fewer.

Rule 1: Ideal transformer — V_s ÷ V_p = N_s ÷ N_p = I_p ÷ I_s (p = primary, s = secondary). The output power equals the input power in an ideal case, so a rise in voltage means a fall in current.

  • Easy : N_p = 1000, N_s = 50 and V_p = 220 V ⇒ V_s = 11 V.
  • Medium : If the load current is 2 A, I_p = 2 × 11 ÷ 220 = 0.10 A.
  • Exam : A turns ratio of 10 : 1 (step-up) turns 220 V into 2200 V.

Rule 2: Efficiency and losses — Efficiency η = output power ÷ input power. Losses are the heating of the windings (copper loss), eddy currents and hysteresis in the core (iron loss), leakage of flux and humming caused by vibration.

  • Easy : An input of 500 W and an output of 440 W gives η = 0.88, or 88 %.
  • Medium : The loss is 60 W.
  • Exam : Laminating the core cuts eddy currents, and using soft iron cuts hysteresis loss.
Remember

Vs/Vp = Ns/Np = Ip/Is; power in = power out (ideal); works only with AC.

Common mistake

Thinking that a step-up transformer produces extra power. The current falls as the voltage rises.

Shortcut

Memory clue : "Voltage follows turns; current goes the opposite way."

Spot the difference

TypeTurnsVoltageCurrent
Step-upNs > Nprisesfalls
Step-downNs < Npfallsrises

Practice questions

Easy
  1. 1On which principle does a transformer work?
  2. 2How do the turns compare in a step-up transformer?
  3. 3Does a transformer work on DC?
Medium
  1. 4A transformer has 1000 primary and 50 secondary turns and 220 V input. Find the output voltage.
  2. 5If the secondary delivers 2 A, find the primary current (ideal case).
  3. 6Find the output of a step-up transformer with a turns ratio of 10 and 220 V input.
Challenging
  1. 7A transformer takes 500 W and delivers 440 W. Find its efficiency and the loss.
  2. 8Name two causes of energy loss in a transformer and how they are reduced.
Show answers (try them yourself first)
  1. 1Mutual induction.
  2. 2The secondary has more turns than the primary.
  3. 3No.
  4. 4V_s = 220 × 50 ÷ 1000 = 11 V.
  5. 5I_p = 2 × 11 ÷ 220 = 0.10 A.
  6. 62200 V.
  7. 7η = 440 ÷ 500 = 88 %; loss = 60 W.
  8. 8Eddy currents (laminated core) and hysteresis (soft iron core); also copper loss (thick wires).