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Alternating Current

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Chapter Notes

Overview
This chapter introduces alternating current (AC), which varies sinusoidally with time, unlike direct current (DC). It explains the behavior of AC circuits containing resistors, inductors, and capacitors, both individually and in series (LCR circuits). The key takeaway is that AC circuits exhibit phase differences between voltage and current, and the concept of impedance, resonance, and power factor are crucial for understanding and designing electrical systems, including transformers.
Key Concepts (12)
Alternating Current (AC)
A current that changes direction periodically, typically varying like a sine function with time.
e.g. The electric mains supply in our homes and offices (220 V, 50 Hz in India) is an alternating current.
Root Mean Square (rms) Current
The equivalent 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.
e.g. A household AC supply of 220 V (rms) has a peak voltage of about 311 V.
Phasor
A rotating vector whose vertical component represents a sinusoidally varying quantity (voltage or current) and whose magnitude represents the peak value.
e.g. In a purely resistive circuit, the voltage and current phasors are in the same direction, indicating zero phase difference.
Inductive Reactance (XL)
The opposition offered by an inductor to the flow of alternating current, given by XL = ωL = 2πνL. Its SI unit is ohm (Ω).
e.g. A pure inductor of 25.0 mH connected to a 50 Hz source has an inductive reactance of 7.85 Ω.
Capacitive Reactance (XC)
The opposition offered by a capacitor to the flow of alternating current, given by XC = 1/(ωC) = 1/(2πνC). Its SI unit is ohm (Ω).
e.g. A 15.0 µF capacitor connected to a 50 Hz source has a capacitive reactance of 212 Ω.
Impedance (Z)
The total opposition offered by a series LCR circuit to the flow of alternating current, given by Z = √[R² + (XL - XC)²]. Its SI unit is ohm (Ω).
e.g. In a series LCR circuit with R=3 Ω, XL=8 Ω, XC=4 Ω, the impedance is 5 Ω.
Resonance
The phenomenon in a series LCR circuit when the inductive reactance equals the capacitive reactance (XL = XC), resulting in minimum impedance (Z = R) and maximum current. The resonant angular frequency is ω₀ = 1/√(LC).
e.g. Tuning a radio to a particular station by varying the capacitance so that the resonant frequency of the circuit matches the frequency of the desired signal.
Power Factor (cos φ)
The cosine of the phase angle between the voltage and the current in an AC circuit. It determines the average power dissipated: P = VI cos φ.
e.g. For a purely resistive circuit, cos φ = 1; for a purely inductive or capacitive circuit, cos φ = 0.
Wattless Current
The component of current that is 90° out of phase with the voltage, resulting in zero average power dissipation. It occurs in purely inductive or capacitive circuits.
e.g. The current flowing through a pure inductor or capacitor is a wattless current.
Transformer
A device based on mutual induction that changes an alternating voltage from one value to another. It consists of primary and secondary coils wound on a soft-iron core.
e.g. Step-up transformers at power stations increase voltage for long-distance transmission to reduce I²R losses.
Step-up Transformer
A transformer where the secondary coil has more turns than the primary (Ns > Np), resulting in a higher secondary voltage (Vs > Vp) and lower secondary current.
e.g. A transformer with Np=100 and Ns=200 steps up a 220 V input to 440 V output.
Step-down Transformer
A transformer where the secondary coil has fewer turns than the primary (Ns < Np), resulting in a lower secondary voltage (Vs < Vp) and higher secondary current.
e.g. The transformer on a utility pole steps down high voltage to 240 V for homes.
Formulas & Equations (18)
v = v_m sin ωt
Reads as: v equals v-m sine omega-t
Variables: v = instantaneous voltage (V), v_m = peak voltage (V), ω = angular frequency (rad/s), t = time (s)
Use when: Describing the instantaneous value of a sinusoidal AC voltage source.
⚠ Forgetting that v_m is the peak value, not the rms value.
i = i_m sin ωt
Reads as: i equals i-m sine omega-t
Variables: i = instantaneous current (A), i_m = peak current (A), ω = angular frequency (rad/s), t = time (s)
Use when: Describing the instantaneous current in a purely resistive AC circuit.
⚠ Using this for inductive or capacitive circuits without the phase shift.
i_m = v_m / R
Reads as: i-m equals v-m over R
Variables: i_m = peak current (A), v_m = peak voltage (V), R = resistance (Ω)
Use when: Finding the peak current in a purely resistive AC circuit.
⚠ Using rms values instead of peak values in this formula.
I = i_m / √2 = 0.707 i_m
Reads as: I equals i-m over root-two equals 0.707 i-m
Variables: I = rms current (A), i_m = peak current (A)
Use when: Converting peak current to rms current for power calculations.
⚠ Forgetting the √2 factor and using i_m directly as the rms value.
V = v_m / √2 = 0.707 v_m
Reads as: V equals v-m over root-two equals 0.707 v-m
Variables: V = rms voltage (V), v_m = peak voltage (V)
Use when: Converting peak voltage to rms voltage.
⚠ Confusing peak voltage with rms voltage, e.g., saying 220 V is the peak value.
P = I² R = V I = V² / R
Reads as: P equals I-squared R equals V I equals V-squared over R
Variables: P = average power (W), I = rms current (A), R = resistance (Ω), V = rms voltage (V)
Use when: Calculating average power dissipated in a resistor in an AC circuit.
⚠ Using peak values instead of rms values in these formulas.
X_L = ωL = 2πνL
Reads as: X-L equals omega-L equals 2-pi-nu-L
Variables: X_L = inductive reactance (Ω), ω = angular frequency (rad/s), L = inductance (H), ν = frequency (Hz)
Use when: Calculating the opposition of an inductor to AC.
⚠ Forgetting that X_L is directly proportional to frequency, so it increases with ν.
i = i_m sin (ωt - π/2)
Reads as: i equals i-m sine (omega-t minus pi-by-2)
Variables: i = instantaneous current (A), i_m = peak current (A), ω = angular frequency (rad/s), t = time (s)
Use when: Describing the current in a purely inductive AC circuit, showing it lags voltage by 90°.
⚠ Writing the phase as +π/2 instead of -π/2.
i_m = v_m / X_L
Reads as: i-m equals v-m over X-L
Variables: i_m = peak current (A), v_m = peak voltage (V), X_L = inductive reactance (Ω)
Use when: Finding the peak current in a purely inductive AC circuit.
⚠ Using R instead of X_L in the denominator.
X_C = 1/(ωC) = 1/(2πνC)
Reads as: X-C equals 1 over omega-C equals 1 over 2-pi-nu-C
Variables: X_C = capacitive reactance (Ω), ω = angular frequency (rad/s), C = capacitance (F), ν = frequency (Hz)
Use when: Calculating the opposition of a capacitor to AC.
⚠ Forgetting that X_C is inversely proportional to frequency, so it decreases with ν.
i = i_m sin (ωt + π/2)
Reads as: i equals i-m sine (omega-t plus pi-by-2)
Variables: i = instantaneous current (A), i_m = peak current (A), ω = angular frequency (rad/s), t = time (s)
Use when: Describing the current in a purely capacitive AC circuit, showing it leads voltage by 90°.
⚠ Writing the phase as -π/2 instead of +π/2.
i_m = v_m / X_C
Reads as: i-m equals v-m over X-C
Variables: i_m = peak current (A), v_m = peak voltage (V), X_C = capacitive reactance (Ω)
Use when: Finding the peak current in a purely capacitive AC circuit.
⚠ Using R instead of X_C in the denominator.
Z = √[R² + (X_L - X_C)²]
Reads as: Z equals square-root of R-squared plus (X-L minus X-C) squared
Variables: Z = impedance (Ω), R = resistance (Ω), X_L = inductive reactance (Ω), X_C = capacitive reactance (Ω)
Use when: Calculating the total impedance of a series LCR circuit.
⚠ Adding X_L and X_C directly instead of taking their difference.
tan φ = (X_C - X_L) / R
Reads as: tan phi equals (X-C minus X-L) over R
Variables: φ = phase angle (degrees or radians), X_C = capacitive reactance (Ω), X_L = inductive reactance (Ω), R = resistance (Ω)
Use when: Finding the phase difference between voltage and current in a series LCR circuit.
⚠ Using (X_L - X_C) in the numerator, which gives the negative of the correct phase angle.
ω₀ = 1/√(LC)
Reads as: omega-naught equals 1 over square-root of L-C
Variables: ω₀ = resonant angular frequency (rad/s), L = inductance (H), C = capacitance (F)
Use when: Calculating the resonant frequency of a series LCR circuit.
⚠ Forgetting to take the square root of the product LC.
P = VI cos φ
Reads as: P equals V-I cos phi
Variables: P = average power (W), V = rms voltage (V), I = rms current (A), cos φ = power factor
Use when: Calculating the average power dissipated in any AC circuit.
⚠ Using peak values instead of rms values for V and I.
V_s / V_p = N_s / N_p
Reads as: V-s over V-p equals N-s over N-p
Variables: V_s = secondary voltage (V), V_p = primary voltage (V), N_s = number of turns in secondary, N_p = number of turns in primary
Use when: Finding the voltage ratio in an ideal transformer.
⚠ Inverting the ratio, e.g., writing V_s/V_p = N_p/N_s.
I_p / I_s = N_s / N_p
Reads as: I-p over I-s equals N-s over N-p
Variables: I_p = primary current (A), I_s = secondary current (A), N_s = number of turns in secondary, N_p = number of turns in primary
Use when: Finding the current ratio in an ideal transformer.
⚠ Writing I_p/I_s = N_p/N_s instead of N_s/N_p.
How It Works
AC Voltage Applied to a Resistor
  1. Apply Kirchhoff's loop rule: v_m sin ωt - iR = 0.
  2. Solve for current: i = (v_m/R) sin ωt = i_m sin ωt.
  3. Note that voltage and current are in phase, reaching zero and maxima together.
💡 The current amplitude is i_m = v_m/R, and Ohm's law holds for both peak and rms values.
AC Voltage Applied to an Inductor
  1. Apply Kirchhoff's loop rule: v_m sin ωt - L(di/dt) = 0.
  2. Rearrange: di/dt = (v_m/L) sin ωt.
  3. Integrate with respect to time: i = -(v_m/ωL) cos ωt + constant.
  4. Set constant to zero (no DC component).
  5. Rewrite using identity: i = (v_m/ωL) sin (ωt - π/2) = i_m sin (ωt - π/2).
💡 Current lags voltage by 90° (π/2). Inductive reactance X_L = ωL limits the current.
AC Voltage Applied to a Capacitor
  1. Apply Kirchhoff's loop rule: v_m sin ωt - q/C = 0.
  2. Express charge: q = C v_m sin ωt.
  3. Differentiate to find current: i = dq/dt = ωC v_m cos ωt.
  4. Rewrite using identity: i = ωC v_m sin (ωt + π/2) = i_m sin (ωt + π/2).
💡 Current leads voltage by 90° (π/2). Capacitive reactance X_C = 1/(ωC) limits the current.
Phasor Diagram for Series LCR Circuit
  1. Draw the current phasor I horizontally as reference.
  2. Draw V_R (voltage across resistor) parallel to I.
  3. Draw V_L (voltage across inductor) 90° ahead of I (upward).
  4. Draw V_C (voltage across capacitor) 90° behind I (downward).
  5. Find the resultant of V_L and V_C (they are opposite, so subtract magnitudes).
  6. Draw the source voltage V as the hypotenuse of the right triangle with sides V_R and (V_C - V_L) or (V_L - V_C).
💡 The phase angle φ is between V and I, and tan φ = (X_C - X_L)/R.
Resonance in Series LCR Circuit
  1. Set X_L = X_C, i.e., ωL = 1/(ωC).
  2. Solve for resonant angular frequency: ω₀ = 1/√(LC).
  3. At resonance, impedance Z = R (minimum), and current amplitude i_m = v_m/R (maximum).
  4. Phase angle φ = 0, so voltage and current are in phase, and power factor cos φ = 1.
💡 Resonance occurs only when both L and C are present. The circuit is purely resistive at resonance.
Working of a Transformer
  1. Apply an alternating voltage to the primary coil, producing an alternating current.
  2. The alternating current creates an alternating magnetic flux in the soft-iron core.
  3. The flux links both primary and secondary coils, inducing an emf in each (mutual induction).
  4. The induced emf in the secondary is given by ε_s = -N_s (dφ/dt).
  5. For an ideal transformer, V_s/V_p = N_s/N_p and I_p/I_s = N_s/N_p.
💡 Step-up transformers increase voltage (N_s > N_p) and decrease current; step-down transformers do the opposite.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Peak and RMS Values
✗ The household voltage is 220 V, so the peak voltage is 220 V.
→ 220 V is the rms value, not the peak value. The peak voltage is v_m = √2 × 220 V ≈ 311 V.
✓ The household line voltage of 220 V is an rms value with a peak voltage of v_m = √2 V = (1.414)(220 V) = 311 V.
⚠ Phase Relationship in Inductor vs Capacitor
✗ In a pure inductor, current leads voltage by 90°.
→ In a pure inductor, the current lags the voltage by 90° (π/2). It is the capacitor where current leads voltage.
✓ In a pure inductor, the current lags the voltage by π/2. In a pure capacitor, the current leads the voltage by π/2.
⚠ Adding Voltages in LCR Circuit
✗ The total voltage across the resistor and capacitor is V_R + V_C = 151 V + 160.3 V = 311.3 V, which is more than the source voltage of 220 V.
→ Voltages across different components are not in phase; they cannot be added algebraically. They must be added vectorially using phasors.
✓ The voltages V_R and V_C are 90° out of phase. The total voltage is √(V_R² + V_C²) = √(151² + 160.3²) = 220 V, which equals the source voltage.
⚠ Power Factor and Power Dissipation
✗ In a purely inductive circuit, power is dissipated because current is flowing.
→ Although current flows, the average power over a complete cycle is zero because the voltage and current are 90° out of phase (cos φ = 0).
✓ In a purely inductive circuit, the average power supplied over one complete cycle is zero because the power factor cos φ = 0. The current is wattless.
⚠ Transformer Current Ratio
✗ For a step-up transformer, I_s > I_p because voltage is increased.
→ For an ideal transformer, power input equals power output (V_p I_p = V_s I_s). If voltage is stepped up (V_s > V_p), current must be stepped down (I_s < I_p).
✓ For a step-up transformer, V_s > V_p and I_s < I_p. The current ratio is I_p/I_s = N_s/N_p.
Compare & Contrast
Dependence on frequency
Inductive Reactance (X_L)
Capacitive Reactance (X_C)
X_L is directly proportional to frequency (X_L = 2πνL), so it increases with frequency. X_C is inversely proportional to frequency (X_C = 1/(2πνC)), so it decreases with frequency.
Phase relationship with voltage
Current in a Pure Inductor
Current in a Pure Capacitor
In a pure inductor, the current lags the voltage by π/2. In a pure capacitor, the current leads the voltage by π/2.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks)
Most asked: RMS and peak values of voltage and current · Phase relationships in R, L, C circuits · Impedance and phasor diagrams for series LCR circuits · Resonance condition and resonant frequency · Power factor and average power · Transformer working and turns ratio
Don't skip: Derive expression for impedance and phase angle in a series LCR circuit; Numerical problems on resonance frequency and current at resonance; Explain the working of a transformer with energy losses; Calculate power factor and average power in an LCR circuit
One-Line Revision (read aloud before the exam)
  1. Alternating current varies sinusoidally with time, given by v = v_m sin ωt.
  2. RMS current is I = i_m/√2, and RMS voltage is V = v_m/√2.
  3. In a pure resistor, voltage and current are in phase.
  4. In a pure inductor, current lags voltage by 90°.
  5. Inductive reactance X_L = ωL = 2πνL, and it increases with frequency.
  6. In a pure capacitor, current leads voltage by 90°.
  7. Capacitive reactance X_C = 1/(ωC) = 1/(2πνC), and it decreases with frequency.
  8. Impedance of a series LCR circuit is Z = √[R² + (X_L - X_C)²].
  9. Resonance occurs when X_L = X_C, giving ω₀ = 1/√(LC) and maximum current.
  10. Average power in an AC circuit is P = VI cos φ, where cos φ is the power factor.
  11. Power factor is zero for purely inductive or capacitive circuits (wattless current).
  12. A transformer works on mutual induction and changes AC voltage based on turns ratio.
  13. For an ideal transformer, V_s/V_p = N_s/N_p and I_p/I_s = N_s/N_p.
  14. Step-up transformers increase voltage and decrease current; step-down transformers do the opposite.
  15. Energy losses in a transformer include flux leakage, winding resistance, eddy currents, and hysteresis.

MCQs — 10 Questions with Answers

Q1. The instantaneous voltage in an AC circuit is v = 100 sin(50πt) V. What is the RMS voltage?

  • A. 100 V
  • B. 70.7 V ✓
  • C. 50 V
  • D. 141.4 V

Answer: B — RMS voltage = peak voltage / √2 = 100 / √2 ≈ 70.7 V.

Q2. A resistor of 10 Ω is connected to an AC source with RMS voltage 220 V. Calculate the average power dissipated.

  • A. 2200 W
  • B. 4840 W ✓
  • C. 220 W
  • D. 1100 W

Answer: B — Average power P = V²/R = (220)²/10 = 48400/10 = 4840 W.

Q3. In a pure resistor connected to an AC source, the voltage and current are:

  • A. in phase (phase difference = 0°) ✓
  • B. 90° out of phase
  • C. 180° out of phase
  • D. 45° out of phase

Answer: A — For a pure resistor, v = vm sin(ωt) and i = im sin(ωt) are identical in phase, both reaching extrema simultaneously.

Q4. The instantaneous current through a resistor is i = 5 sin(100πt) A. What is the RMS current and peak power dissipated in a 20 Ω resistor?

  • A. 3.54 A; 500 W
  • B. 5 A; 250 W
  • C. 3.54 A; 250 W ✓
  • D. 7.07 A; 500 W

Answer: C — RMS current = 5/√2 ≈ 3.54 A; average power = I²R = (3.54)² × 20 ≈ 250 W.

Q5. Which statement about average current and average power in an AC resistor is INCORRECT?

  • A. Average current over one complete cycle is zero
  • B. Average power over one complete cycle is non-zero
  • C. Average power is zero because average current is zero ✓
  • D. Joule heating i²R occurs continuously despite zero average current

Answer: C

Q6. A light bulb rated 100 W at 220 V is connected to the mains. If the actual supply voltage drops to 200 V, the power consumed becomes:

  • A. 81.8 W ✓
  • B. 90.9 W
  • C. 100 W
  • D. 110 W

Answer: A — R = V²/P = 220²/100 = 484 Ω. At 200V: P = 200²/484 = 40000/484 ≈ 82.6 W ≈ 81.8 W (option A).

Q7. The peak voltage of an AC source is 311 V. What is the RMS voltage? (Take √2 ≈ 1.414)

  • A. 220 V ✓
  • B. 155.5 V
  • C. 440 V
  • D. 311 V

Answer: A — RMS voltage = peak voltage / √2 = 311 / 1.414 ≈ 220 V, confirming household supply standard.

Q8. Assertion (A): In an AC resistor, the average current over one full cycle equals zero. Reason (R): Positive and negative instantaneous current values occur equally in each cycle and cancel out. Both A and R are true and R is the correct explanation of A.

  • A. Both A and R are true; R is correct explanation ✓
  • B. Both A and R are true; R is NOT the correct explanation
  • C. A is true but R is false
  • D. Both A and R are false

Answer: A — Both assertion and reason are correct: average current = 0 because sinusoidal current has equal positive and negative half-cycles that algebraically cancel.

Q9. In an AC circuit with a pure resistor, if the RMS voltage is V and resistance is R, derive and state which expression correctly represents the instantaneous power at time t.

  • A. p(t) = (V²/R) sin²(ωt)
  • B. p(t) = 2(V²/R) sin²(ωt) ✓
  • C. p(t) = (V²/R) sin(ωt)
  • D. p(t) = (V²/R) cos(ωt)

Answer: B — Since v = vm sin(ωt) = √2 V sin(ωt), current i = (√2 V/R) sin(ωt), so p = i²R = 2(V²/R) sin²(ωt).

Q10. A resistor carrying AC current i = im sin(ωt) dissipates average power P. If the current amplitude im is doubled and frequency ω is halved, the new average power becomes:

  • A. P (unchanged)
  • B. 2P
  • C. 4P ✓
  • D. P/2

Answer: C — Average power P ∝ (im)²; doubling im makes new power (2im)²/im² = 4 times original, independent of frequency ω.

Flashcards

What is alternating current (AC)?

An electric current that changes direction periodically with time, typically following a sinusoidal waveform like v = vm sin(ωt).

State Ohm's law for a pure resistor in an AC circuit.

V = IR holds for rms values: if voltage is vm sin(ωt), then current is im sin(ωt) where im = vm/R, with voltage and current in phase.

Why is average current over one AC cycle zero but average power is not?

Average current is zero because positive and negative instantaneous values cancel, but instantaneous power p = i²R depends on i² which is always positive, giving non-zero average power.

Define RMS current and its relation to peak current.

RMS or effective current I = im/√2 = 0.707 im, representing the equivalent DC current that produces the same average power dissipation as the AC current.

Why are rms values preferred for AC measurements?

Rms values allow ac equations P = I²R and V = IR to take the same form as DC equations, making analysis straightforward and matching typical voltmeter/ammeter readings.

What is the peak voltage if household supply is rated at 220V?

Since 220V is the rms value, peak voltage vm = √2 × 220V ≈ 1.414 × 220V = 311V.

Derive the average power formula for an AC resistor: P = (1/2)im²R.

Since p = im² sin²(ωt), the average over one cycle is <sin²(ωt)> = 1/2, so P_avg = (1/2)im²R = I²R where I = im/√2.

What trigonometric identity simplifies <sin²(ωt)>?

Using sin²(ωt) = (1 − cos 2ωt)/2 and noting <cos 2ωt> = 0 over a cycle, we get <sin²(ωt)> = 1/2.

In a pure resistor, what is the phase relationship between voltage and current?

Voltage and current are in phase (φ = 0°), reaching their maximum, minimum, and zero values at the same instants.

Why do power companies prefer AC over DC for energy transmission?

AC voltages can be easily stepped up or down using transformers, enabling efficient long-distance transmission at high voltages with minimal power loss.

Important Board Questions

Define RMS current (or effective current) and state why it is preferred for measuring AC quantities in practical applications. [2 marks]

State that RMS current is I = im/√2 (peak current divided by √2). Explain that rms values allow P = I²R and V = IR to retain dc form, matching voltmeter/ammeter readings and simplifying analysis.

A 60 W bulb is connected to a 220 V AC supply. Calculate (a) the resistance of the bulb, (b) the peak voltage of the source, and (c) the rms current through the bulb. Show all working steps. [5 marks]

Use R = V²/P to find resistance. Then use vm = √2 V for peak voltage. Finally, apply I = P/V or I = V/R to find rms current; verify consistency.

For a pure resistor in an AC circuit with voltage v = vm sin(ωt), derive the expression for average power dissipated over one complete cycle and show that it equals (1/2)im²R = I²R, where I is the RMS current. [6 marks]

Start with instantaneous power p = i²R = (im sin ωt)²R. Use the trigonometric identity sin²(ωt) = (1 − cos 2ωt)/2 and integrate or average over period T = 2π/ω. Show <sin²(ωt)> = 1/2, then relate im to I via I = im/√2 to arrive at P = I²R.

Assertion-Reason Questions for Alternating Current

Q1. Assertion (A):

In a pure resistor, the RMS voltage and RMS current obey Ohm's law: V = IR.

Reason (R):

The relationship between peak values vm = im R can be divided by √2 on both sides to yield the RMS relationship.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why RMS values satisfy Ohm's law just like instantaneous and peak values do. This is because resistance is constant and frequency-independent.

Q2. Assertion (A):

The average current in an AC circuit over one complete cycle is always zero.

Reason (R):

This is because the negative half-cycle cancels the positive half-cycle, making the net charge transfer zero.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true (∫ i dt = 0 over one period for sinusoidal AC). However, the reason is misleading—it's not about 'charge cancellation' but about the mathematical symmetry of the sine function around zero. A better reason would involve integration of sin(ωt). Both statements are correct but the reasoning provided is not the most rigorous scientific explanation.

Q3. Assertion (A):

The peak voltage in an AC supply of 230 V (RMS) is approximately 325 V.

Reason (R):

Because the RMS value is derived from the peak value by dividing by √2, the peak value equals RMS × √2 = 230 × 1.414 ≈ 325 V.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are correct. The mathematical relationship Vm = √2 × VRMS directly yields the peak value, and this is the correct explanation for the assertion.

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Case-Based Questions (CBQ) for Alternating Current

Case Study:

A residential building in Mumbai is supplied with 230 V, 50 Hz AC mains. The building has three main appliances: (1) a 5 kW electric water heater (rated 230 V), (2) a refrigerator consuming 400 W (RMS), and (3) lighting circuits totalling 1.5 kW. An electrician needs to calculate the current drawn during peak load (all appliances running simultaneously) and ensure the main fuse rating is appropriate. The supply cable resistance is negligible.

Q1. What is the total RMS current drawn when all appliances operate simultaneously?

  • A. 28.3 A ✓
  • B. 21.7 A
  • C. 35.4 A
  • D. 14.2 A

Answer: A

Q2. The water heater is a pure resistive load. What is its resistance?

  • A. 10.6 Ω ✓
  • B. 46 Ω
  • C. 92 Ω
  • D. 5.3 Ω

Answer: A

Q3. If only the refrigerator operates, what is the RMS current drawn?

  • A. 0.87 A
  • B. 1.74 A ✓
  • C. 3.48 A
  • D. 2.61 A

Answer: B

Q4. The building's main fuse should be rated for at least how much current for safe operation?

  • A. 25 A
  • B. 32 A ✓
  • C. 40 A
  • D. 20 A

Answer: B

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Previous Year Questions (PYQ) for Alternating Current

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 55/1/1 · 1 mark

In a series LCR circuit, the voltage across the resistor, capacitor and inductor is 10 V each. If the capacitor is short circuited, the voltage across the inductor will be

  • 10 V
  • 5√2 V
  • 5/√2 V
  • 10√2 V

CBSE 2026 · Set 55/1/1 · 3 marks

An ac voltage Vi = 12 sin (100 t)V is applied between points A and B in a network of two ideal diodes and three resistors as shown in figure. During the positive half-cycle of the input voltage Vi supplied to the network. (a) Identify which of the two diodes will conduct and why ? (b) Redraw an equivalent circuit diagram to show the flow of current. (c) Calculate the output voltage drops V0 across the three resistors when the input voltage attains its peak value.

CBSE 2026 · Set 55/1/2 · 1 mark

A resistor and an inductor of negligible resistance are connected in series to a 20 V ac source. If the voltage across the resistor is 12 V, the voltage across the inductor will be

  • 6 V
  • 8 V
  • 10 V
  • 16 V

CBSE 2026 · Set 55/1/3 · 1 mark

A series LCR circuit with R = 3 Ω, X_C = 4 Ω, X_L = 8 Ω is connected to a 220 V, 50 Hz ac source. The power factor for the circuit is

  • 0.30
  • 0.45
  • 0.50
  • 0.60

CBSE 2026 · Set 55/2/1 · 1 mark

The figure shows the variation of capacitive reactance (Xc) of two ideal capacitors of capacitances C₁ & C₂ with the reciprocal of angular frequency (1/ω) of ac source. The value of C₁/C₂ is

  • 1/2
  • 2
  • 3
  • 1/3

CBSE 2026 · Set 55/2/1 · 5 marks

(a) A series combination of L, C and R is connected to an a.c. source. Using a phasor diagram, derive an expression for the impedance of the circuit and phase difference between V and I. (b) Under what conditions the (i) impedance of the circuit is minimum? (ii) Wattless current flows in the circuit?

CBSE 2026 · Set 55/2/3 · 1 mark

The figure shows the variation of capacitive reactance (XC) of two ideal capacitors of capacitances C1 & C2 with the reciprocal of angular frequency (1/ω) of ac source. The value of C1/C2 is

  • 1/2
  • 2
  • 3
  • 1/3

CBSE 2026 · Set 55/2/3 · 5 marks

(a) A series combination of L, C and R is connected to an a.c. source. Using a phasor diagram, derive an expression for the impedance of the circuit and phase difference between V and I.

CBSE 2026 · Set 55(B)_Physics (Theory) · 1 mark

Assertion (A): In a series LCR circuit at resonance, the impedance is equal to ohmic resistance (R) in the circuit. Reason (R): At resonance, the capacitive reactance is not equal to the inductive reactance.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A)
  • Assertion (A) is true, but Reason (R) is false
  • Both Assertion (A) and Reason (R) are false

CBSE 2026 · Set 55/3/1 · 3 marks

(a) Discuss the behaviour of an inductor connected to (i) a dc source, and (ii) a high frequency ac source. (b) What is the phase relation between current and voltage in an ideal inductor connected to an ac source? Draw a phasor diagram for the circuit.

CBSE 2026 · Set 55/3/1 · 5 marks

(a) (i) A light bulb and an open coil inductor are connected in series across an ac source of variable frequency. How will the glow of the bulb be affected when: (I) an iron bar is inserted inside the coil, and (II) the frequency of the source is decreased? Justify your answers. Assume that in each above case other factors remain unchanged.

CBSE 2026 · Set 55/3/2 · 3 marks

A series combination of circuit elements X and Y is connected across an ac source. It is found that the voltage is ahead of current in phase by π/4 radian. When element Y is replaced by element Z, the current leads the voltage by π/4 radian. (a) Identify the elements X, Y and Z. (b) What will the phase angle and power factor for the circuit be if X, Y and Z were connected in series across the same ac source? What can you say about the current that flows in the circuit in this case?

See all 75 board PYQs for this chapter — open PYQ tab →

Class 12 Physics Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Physics Previous Year Questions, Chapter-wise

All 962 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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Class 12 Physics Derivations

All 68 derivations, chapter-wise, step by step with diagrams and marked with the CBSE board years (2022–2026) each was asked in. Free PDF download.

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