Class 12 Physics · Chapter 7 NotesAlternating Current

Revise Class 12 Physics Alternating Current with clear notes on rms values, reactance, impedance, LCR resonance, power factor and transformers.

6 topics5 sample MCQs5 practice questions
Chapter contents

Chapter summary

Alternating current is the form in which electrical energy reaches our homes, schools and industries. Unlike direct current, an alternating voltage reverses its direction periodically and varies sinusoidally with time. This chapter explains how such a voltage drives current through resistors, inductors and capacitors, and how these elements behave differently because of their reactance. You will learn the meaning of rms values, why household 220 V is an rms value, and how phasors help us add voltages that are out of phase. The chapter then combines R, L and C into a series LCR circuit, introduces impedance and resonance, and shows how the tuning circuit of a radio works. Finally, it explains the power factor, the idea of wattless current, and how transformers step voltage up or down for efficient long-distance transmission of electrical energy.

What you'll learn

1Define alternating voltage and alternating current and write their sinusoidal expressions.
2Distinguish between peak value and rms value, and calculate one from the other.
3Explain the phase relationship between voltage and current for a pure resistor, a pure inductor and a pure capacitor.
4Define inductive reactance, capacitive reactance and impedance, and use them to find current amplitude.
5Analyse a series LCR circuit using phasors and determine the phase angle and power factor.
6Describe resonance in a series LCR circuit and calculate the resonant frequency.
7Explain the working of a transformer and use the turns ratio to relate voltages and currents.
8List the main energy losses in a real transformer and how they are reduced.

Chapter at a glance

01AC Generator and Alternating EMF
02AC Voltage, Current, and Power
03AC Circuit Elements and Impedance
04Transformers and Power Transmission
05Representation of AC Current and Voltage by Rotating Vectors — Phasors
06Resonance

Detailed chapter notes

01

AC Voltage Applied to a Resistor

When a sinusoidal voltage v = vₘ sin ωt is applied across a resistor R, Kirchhoff's loop rule gives vₘ sin ωt = iR, so the current is i = iₘ sin ωt with iₘ = vₘ/R. This is Ohm's law and it works for ac just as it does for dc. The important point is that voltage and current reach zero, minimum and maximum at exactly the same instants, so they are in phase. Even though the average current over a full cycle is zero, the average power is not zero, because Joule heating depends on i², which is always positive. The instantaneous power is p = iₘ²R sin²ωt, and its average over a cycle works out to (1/2)iₘ²R.

  • v = vₘ sin ωt, i = iₘ sin ωt, iₘ = vₘ/R
  • Voltage and current are in phase for a pure resistor
  • Average power P = (1/2) iₘ²R
02

RMS Current, RMS Voltage and Average Power

To write ac power in the same form as dc power, we define the root mean square or effective current I = iₘ/√2 = 0.707 iₘ, and similarly the rms voltage V = vₘ/√2 = 0.707 vₘ. Using these, the average power becomes P = I²R = IV, and the relation between current and voltage becomes V = IR, exactly like the dc case. This is why rms values are used to specify ac quantities. For example, the household supply of 220 V is an rms value; its peak value is vₘ = √2 × 220 V ≈ 311 V. The rms current is the equivalent dc current that would produce the same average power loss.

  • I = iₘ/√2 ≈ 0.707 iₘ
  • V = vₘ/√2 ≈ 0.707 vₘ
  • P = I²R = IV and V = IR
  • 220 V mains is an rms value; peak is about 311 V
03

Phasors: Representing AC Voltage and Current by Rotating Vectors

A phasor is a vector that rotates about the origin with angular speed ω. Its magnitude represents the amplitude (peak value) of the alternating quantity, and its vertical component at any instant gives the instantaneous value. Phasor diagrams make it easy to see the phase relationship between voltage and current in a circuit. For a pure resistor, the voltage and current phasors point in the same direction, showing zero phase difference. Phasors are not vectors in the true sense because voltage and current are scalar quantities; they are used only because harmonically varying scalars combine in the same way as the projections of rotating vectors.

  • Phasor magnitude = amplitude of the quantity
  • Vertical component = instantaneous value
  • For a resistor, V and I phasors are in the same direction
04

AC Voltage Applied to an Inductor

For a purely inductive circuit, Kirchhoff's loop rule gives v − L(di/dt) = 0. Integrating, the current is i = iₘ sin(ωt − π/2), so the current lags the voltage by π/2, or one-quarter of a cycle. The quantity ωL plays the role of resistance and is called inductive reactance, X_L = ωL, measured in ohm. The current amplitude is iₘ = vₘ/X_L. Inductive reactance increases with frequency and with inductance, so a high-frequency ac is opposed more strongly by an inductor. The instantaneous power is p = −(iₘvₘ/2) sin 2ωt, whose average over a complete cycle is zero, so an ideal inductor consumes no net power.

  • X_L = ωL, unit ohm
  • iₘ = vₘ/X_L
  • Current lags voltage by π/2
  • Average power over a cycle is zero
05

AC Voltage Applied to a Capacitor

When an ac source is connected to a capacitor, the capacitor charges and discharges alternately, so current flows continuously, unlike in a dc circuit where the current stops once the capacitor is fully charged. The current is i = iₘ sin(ωt + π/2), so the current leads the voltage by π/2. The quantity 1/ωC is called capacitive reactance, X_C = 1/ωC, and the current amplitude is iₘ = vₘ/X_C. Capacitive reactance decreases as frequency or capacitance increases. As with the inductor, the average power supplied to a pure capacitor over a complete cycle is zero, because the average of sin 2ωt over a cycle is zero.

  • X_C = 1/ωC, unit ohm
  • iₘ = vₘ/X_C
  • Current leads voltage by π/2
  • Average power over a cycle is zero
06

AC Voltage Applied to a Series LCR Circuit

In a series LCR circuit the same current flows through all three elements. Using phasors, V_R is in phase with I, V_L leads I by π/2 and V_C lags I by π/2. Combining these, the source voltage amplitude satisfies vₘ² = v_Rm² + (v_Cm − v_Lm)², which gives iₘ = vₘ/Z, where Z = √(R² + (X_C − X_L)²) is the impedance of the circuit. The phase angle is given by tan φ = (X_C − X_L)/R. If X_C > X_L the circuit is predominantly capacitive and current leads the voltage; if X_C < X_L it is predominantly inductive and current lags the voltage. The impedance diagram is a right triangle with Z as the hypotenuse.

  • Z = √(R² + (X_C − X_L)²)
  • iₘ = vₘ/Z
  • tan φ = (X_C − X_L)/R
  • Capacitive if X_C > X_L; inductive if X_C < X_L
07

Resonance in a Series LCR Circuit

Resonance occurs when the inductive and capacitive reactances become equal, X_C = X_L, so the impedance is minimum and equal to R. The frequency at which this happens is the resonant frequency ω₀ = 1/√(LC). At resonance the current amplitude is maximum, iₘ = vₘ/R, and the voltages across L and C cancel each other because they are out of phase. Resonance is possible only when both L and C are present; an RL or RC circuit cannot show resonance. This principle is used in the tuning circuit of a radio or TV, where the capacitance is varied so that the circuit resonates at the frequency of the desired station.

  • Resonant frequency ω₀ = 1/√(LC)
  • At resonance Z = R and iₘ = vₘ/R
  • Voltages across L and C cancel
  • Used in radio and TV tuning circuits
08

Power in an AC Circuit and the Power Factor

For a series LCR circuit driven by v = vₘ sin ωt, the current is i = iₘ sin(ωt + φ). The instantaneous power is p = (vₘiₘ/2)[cos φ − cos(2ωt + φ)], and the average power over a cycle is P = (vₘiₘ/2) cos φ = V I cos φ. The quantity cos φ is called the power factor. For a pure resistor φ = 0, so cos φ = 1 and power dissipation is maximum. For a purely inductive or capacitive circuit φ = π/2, so cos φ = 0 and no net power is dissipated; such a current is called wattless current. In an LCR circuit, power is dissipated only in the resistor. A low power factor in transmission means a larger current for the same power, increasing I²R losses.

  • P = V I cos φ
  • cos φ is the power factor
  • cos φ = 1 for pure R; cos φ = 0 for pure L or C
  • Wattless currentno net power dissipation
09

Transformers and Power Transmission

A transformer changes an alternating voltage from one value to another using mutual induction. It has a primary coil of N_p turns and a secondary coil of N_s turns wound on a soft-iron core. For an ideal transformer, V_s/V_p = N_s/N_p and I_s/I_p = N_p/N_s. If N_s > N_p the voltage is stepped up and the current is reduced; if N_s < N_p the voltage is stepped down and the current is increased. In practice, energy losses occur due to flux leakage, resistance of windings, eddy currents and hysteresis, and these are reduced by careful winding, thick wire, laminated cores and low-hysteresis material. For transmission, voltage is stepped up to reduce current and hence I²R loss, then stepped down near consumers.

  • V_s/V_p = N_s/N_p and I_s/I_p = N_p/N_s
  • Step-up transformerN_s > N_p
  • Step-down transformerN_s < N_p
  • Lossesflux leakage, winding resistance, eddy currents, hysteresis
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Quick revision: key points

  • Alternating voltage and current vary sinusoidally: v = vₘ sin ωt, i = iₘ sin ωt.
  • RMS values: I = iₘ/√2 ≈ 0.707 iₘ and V = vₘ/√2 ≈ 0.707 vₘ; P = I²R = IV.
  • For a pure resistor, voltage and current are in phase; average power is (1/2)iₘ²R.
  • For a pure inductor, current lags voltage by π/2; X_L = ωL; average power is zero.
  • For a pure capacitor, current leads voltage by π/2; X_C = 1/ωC; average power is zero.
  • Series LCR: Z = √(R² + (X_C − X_L)²), tan φ = (X_C − X_L)/R, iₘ = vₘ/Z.
  • Resonance occurs at ω₀ = 1/√(LC), where Z = R and current is maximum.
  • Average power P = V I cos φ; cos φ is the power factor; pure L or C gives wattless current.
  • Transformer relations: V_s/V_p = N_s/N_p and I_s/I_p = N_p/N_s.
  • Real transformer losses: flux leakage, winding resistance, eddy currents and hysteresis.

Test yourself

Try each question first, then reveal the answer.

Question 01

In an AC generator, the alternating EMF is produced due to the rotation of a coil in a magnetic field. What is the frequency of the alternating EMF if the coil completes 50 rotations per second?

  • A25 Hz
  • B50 Hz
  • C100 Hz
  • D200 Hz
Show answer
Answer: (B) 50 Hz

Frequency is equal to the number of complete cycles per second. Since one complete rotation produces one complete cycle of EMF, 50 rotations per second means 50 Hz frequency.

Question 02

What is the peak voltage (V₀) if the RMS voltage of an AC supply is 220 V?

  • A155.6 V
  • B311.1 V
  • C440 V
  • D110 V
Show answer
Answer: (B) 311.1 V

The relationship between RMS voltage and peak voltage is V_rms = V₀/√2. Therefore, V₀ = V_rms × √2 = 220 × 1.414 ≈ 311.1 V.

Question 03

In an AC circuit containing only a pure resistance R, what is the phase difference between voltage and current?

  • A0°
  • B90°
  • C45°
  • D180°
Show answer
Answer: (A) 0°

In a pure resistive circuit, voltage and current are in phase with each other, so the phase difference is 0°. This is because resistance does not store energy and follows Ohm's law directly.

Question 04

A transformer has 100 turns in the primary coil and 500 turns in the secondary coil. If the primary voltage is 220 V, what is the secondary voltage?

  • A44 V
  • B1100 V
  • C220 V
  • D550 V
Show answer
Answer: (B) 1100 V

Using the transformer equation Vs/Vp = Ns/Np, we get Vs = 220 × (500/100) = 1100 V. This is a step-up transformer.

Question 05

In a phasor diagram for an AC circuit, what does the length of a phasor represent?

  • AThe rms value of the quantity
  • BThe peak value of the quantity
  • CThe instantaneous value of the quantity
  • DThe average value of the quantity
Show answer
Answer: (B) The peak value of the quantity

The magnitude of a phasor represents the amplitude or peak value of the oscillating quantity (voltage or current).

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Sample questions and answers

Sample question3 marks

Q1. What is the principle of an AC generator? Explain the role of slip rings in it.

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Model answer

An AC generator works on the principle of electromagnetic induction. When a coil is rotated in a uniform magnetic field, the magnetic flux linked with the coil changes, inducing an alternating emf. Slip rings are used to maintain a continuous connection between the rotating coil and the external circuit, allowing the generated alternating current to be drawn out.

Sample question3 marks

Q2. Define root mean square (rms) value of an alternating current. How is it related to the peak value?

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Model answer

The rms value of an alternating current is the equivalent dc current that would produce the same average power loss in a resistor. It is defined as I = i_m / √2 = 0.707 i_m, where i_m is the peak current.

Sample question3 marks

Q3. Define inductive reactance and capacitive reactance. Write their SI units and state how they vary with frequency.

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Model answer

Inductive reactance (XL) is the opposition offered by an inductor to ac, given by XL = ωL = 2πfL. Its SI unit is ohm (Ω). It is directly proportional to frequency. Capacitive reactance (XC) is the opposition offered by a capacitor to ac, given by XC = 1/ωC = 1/(2πfC). Its SI unit is ohm (Ω). It is inversely proportional to frequency.

Sample question3 marks

Q4. State the principle of a transformer. Explain why the core of a transformer is laminated.

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Model answer

A transformer works on the principle of mutual induction. The core is laminated to minimise eddy current losses. Laminations increase the resistance of the core, reducing the magnitude of eddy currents and thus the heating effect.

Sample question3 marks

Q5. What is a phasor? How does it represent a sinusoidally varying alternating voltage or current?

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Model answer

A phasor is a vector that rotates about the origin with angular speed omega. Its magnitude represents the amplitude (peak value) of the alternating quantity, and its vertical component at any instant gives the instantaneous value of the voltage or current. For example, a voltage phasor of length v_m rotating with angular speed omega has vertical projection v_m sin(omega t), representing v = v_m sin(omega t).

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Frequently asked questions

What is meant by rms value of alternating current?

The rms or effective current is the value of dc current that would produce the same average power loss in a resistor as the alternating current. It is given by I = iₘ/√2 ≈ 0.707 iₘ, where iₘ is the peak current. Similarly, the rms voltage is V = vₘ/√2.

Why is household supply of 220 V called an rms value?

The 220 V supplied to homes is the rms value of the alternating voltage. The actual voltage oscillates sinusoidally, and its peak value is vₘ = √2 × 220 V ≈ 311 V. Instruments and ratings normally quote rms values because they directly relate to average power.

What is the difference between resistance and reactance?

Resistance R opposes current and dissipates electrical energy as heat in both dc and ac circuits. Reactance, which may be inductive (X_L = ωL) or capacitive (X_C = 1/ωC), also limits current in an ac circuit but does not dissipate net power over a cycle. Reactance depends on frequency, while resistance does not.

Why is the average power consumed by a pure inductor or capacitor zero?

In a pure inductor the current lags the voltage by π/2, and in a pure capacitor it leads by π/2. The instantaneous power varies as sin 2ωt, whose average over a complete cycle is zero. Energy is stored and returned each cycle, so no net power is dissipated.

What is resonance in a series LCR circuit?

Resonance occurs when the inductive and capacitive reactances become equal, X_L = X_C, at the resonant frequency ω₀ = 1/√(LC). At this frequency the impedance is minimum and equal to R, the current amplitude is maximum, and the voltages across L and C cancel each other.

How does a transformer change voltage and current?

For an ideal transformer, V_s/V_p = N_s/N_p and I_s/I_p = N_p/N_s, where N_p and N_s are the turns in the primary and secondary coils. A step-up transformer has more secondary turns and raises voltage while reducing current; a step-down transformer does the opposite.

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