Important Questions Class 12 Physics Chapter 7: Alternating Current

Alternating current is an electric current that changes its magnitude and direction periodically with time. An AC source usually produces sinusoidal voltage, and household supply values are specified as rms values.

Alternating current explains why electrical energy can travel long distances efficiently through transformers. Important Questions Class 12 Physics Chapter 7 help students practise rms values, phasors, reactance, LCR circuits, resonance, power factor, wattless current, and transformers. The CBSE 2026 chapter contains numerical questions, phasor-based reasoning, phase relations, and circuit applications from NCERT Alternating Current.

Key Takeaways

• RMS Current: The effective current equals I = im/√2 for sinusoidal AC.
• Pure Inductor: Current lags voltage by π/2 in an ideal inductive AC circuit.
• Pure Capacitor: Current leads voltage by π/2 in an ideal capacitive AC circuit.
• Series LCR Resonance: Resonance occurs when XL = XC and ω0 = 1/√LC.

Important Questions Class 12 Physics Chapter 7 Structure 2026

Important Questions Class 12 Physics Chapter 7 with Answers

AC circuits use voltage and current that vary with time, usually as sine functions.
Students must track rms values, phase difference, impedance, and average power.
These alternating current class 12 important questions follow the NCERT 2026 chapter sequence.

1. What does Important Questions Class 12 Physics Chapter 7 mainly test?

Important Questions Class 12 Physics Chapter 7 mainly test AC voltage, rms values, reactance, LCR circuits, resonance, power factor, and transformers. The chapter combines numerical formulas with phasor reasoning.

1. AC Basics: Use v = vm sin ωt and i = im sin ωt.
2. Reactance: Use XL = ωL and XC = 1/ωC.
3. LCR Circuit: Use Z = √[R² + (XL − XC)²].
4. Transformer: Use Vs/Vp = Ns/Np.
5. Final Result: The chapter tests sinusoidal circuits and transformer action.

2. What is alternating current in Class 12 Physics?

Alternating current is current that changes direction and magnitude periodically. A sinusoidal AC current follows i = im sin ωt.

1. Current Type: Alternating current.
2. General Form: i = im sin ωt.
3. Frequency: Source frequency decides the number of cycles per second.
4. Final Result: AC reverses direction every half cycle.

3. Why is AC preferred over DC for power transmission?

AC is preferred because transformers can easily change AC voltage levels. High voltage transmission reduces current and line loss.

1. Transformer Use: AC voltage can step up or step down.
2. Line Loss: Loss varies as I²R.
3. High Voltage Effect: Current becomes smaller for the same power.
4. Final Result: AC reduces long-distance transmission loss.

AC Voltage Applied to Resistor Class 12 Questions

A pure resistor gives the simplest AC circuit because voltage and current stay in phase.
Ohm’s law works for rms values in the same form as DC circuits.
These AC voltage applied to resistor Class 12 questions cover power, rms values, and phase relation.

4. What happens when AC voltage is applied to a pure resistor?

Current through a pure resistor stays in phase with voltage. Both reach zero, maximum, and minimum together.

1. Applied Voltage: v = vm sin ωt.
2. Ohm’s Law: v = iR.
3. Current: i = im sin ωt.
4. Current Amplitude: im = vm/R.
5. Final Result: Voltage and current are in phase in a resistor.

5. What is rms current in AC?

RMS current is the effective current that produces the same heating as DC. For sinusoidal AC, I = im/√2.

1. Peak Current: im.
2. RMS Formula: I = im/√2.
3. Decimal Form: I = 0.707 im.
4. Final Result: RMS current equals im/√2.

6. What is rms voltage in AC?

RMS voltage is the effective AC voltage used for power calculations. For sinusoidal AC, V = vm/√2.

1. Peak Voltage: vm.
2. RMS Formula: V = vm/√2.
3. Decimal Form: V = 0.707 vm.
4. Final Result: RMS voltage equals vm/√2.

7. What is the peak voltage of 220 V AC supply?

The peak voltage is 311 V. The 220 V supply value is an rms value.

1. Given Data:
   V = 220 V
2. Formula Used:
   vm = √2 V
3. Calculation:
   vm = 1.414 × 220
   vm = 311 V
4. Final Result: Peak voltage = 311 V.

8. A 100 W bulb works on 220 V AC. Find its resistance and rms current.

The resistance is 484 Ω, and rms current is 0.454 A. Use P = V²/R and P = VI.

1. Given Data:
   P = 100 W
   V = 220 V
2. Resistance:
   R = V²/P
   R = 220²/100
   R = 484 Ω
3. Current:
   I = P/V
   I = 100/220
   I = 0.454 A
4. Final Result: R = 484 Ω and I = 0.454 A.

9. Why is average current zero but average power not zero in a resistor?

Average current is zero because positive and negative current halves cancel. Average power remains positive because p = i²R.

1. Current Average: One full cycle gives zero.
2. Power Formula: p = i²R.
3. Sign: i² stays positive for both half cycles.
4. Final Result: A resistor dissipates power in AC.

RMS Current and RMS Voltage Class 12 Questions

RMS values make AC calculations look like DC calculations.
Meters and household supply ratings usually show rms values.
These rms current and rms voltage Class 12 questions cover effective values and heating effect.

10. Why do we use rms values in AC circuits?

We use rms values because they give the same heating effect as DC values. AC power calculations become simple.

1. Heating Effect: Depends on i²R.
2. RMS Current: Produces the same average heating as DC current.
3. Power Formula: P = I²R.
4. Final Result: RMS values express effective AC values.

11. The peak voltage of an AC source is 300 V. Find rms voltage.

The rms voltage is 212 V. Use V = vm/√2.

1. Given Data:
   vm = 300 V
2. Formula Used:
   V = vm/√2
3. Calculation:
   V = 300/1.414
   V = 212 V
4. Final Result: RMS voltage = 212 V.

12. The rms current in an AC circuit is 10 A. Find peak current.

The peak current is 14.14 A. Use im = √2 I.

1. Given Data:
   I = 10 A
2. Formula Used:
   im = √2 I
3. Calculation:
   im = 1.414 × 10
   im = 14.14 A
4. Final Result: Peak current = 14.14 A.

13. What is average power in a pure resistor?

Average power in a pure resistor is P = I²R. It also equals P = VI.

1. RMS Current: I.
2. Resistance: R.
3. Formula Used:
   P = I²R = V²/R = VI
4. Final Result: A pure resistor consumes real power.

Phasor Diagram Class 12 Physics Questions

A phasor represents a sinusoidally varying quantity through a rotating vector.
Phasors help compare phase differences across R, L, C, and LCR circuits.
These phasor diagram Class 12 Physics questions focus on current-voltage phase relations.

14. What is a phasor in AC circuits?

A phasor is a rotating vector used to represent sinusoidal voltage or current. Its projection gives instantaneous value.

1. Magnitude: Represents peak value.
2. Rotation: Angular speed equals ω.
3. Projection: Gives sinusoidal variation.
4. Final Result: Phasors show amplitude and phase together.

15. Are AC voltage and current real vectors?

AC voltage and current are not real vectors. Phasors only represent their amplitude and phase mathematically.

1. Voltage: Scalar quantity.
2. Current: Scalar quantity.
3. Phasor Use: Helps add sinusoidal quantities.
4. Final Result: Phasors are mathematical aids.

16. What is the phase difference between voltage and current in a resistor?

The phase difference is 0° in a pure resistor. Voltage and current remain in phase.

1. Voltage: v = vm sin ωt.
2. Current: i = im sin ωt.
3. Phase Angle: φ = 0.
4. Final Result: Voltage and current are in phase.

AC Voltage Applied to Inductor Class 12 Questions

A pure inductor opposes changes in current through self-induced emf.
Its reactance increases with frequency, so high-frequency AC faces more opposition.
These AC voltage applied to inductor Class 12 questions explain inductive reactance and phase lag.

17. What happens when AC voltage is applied to a pure inductor?

Current lags voltage by π/2 in a pure inductor. The inductor stores and returns magnetic energy.

1. Applied Voltage: v = vm sin ωt.
2. Current Form: i = im sin(ωt − π/2).
3. Phase Relation: Current lags voltage by 90°.
4. Final Result: Inductor current lags voltage by π/2.

18. What is inductive reactance?

Inductive reactance is the opposition offered by an inductor to AC. Its formula is XL = ωL.

1. Formula Used:
   XL = ωL
2. Also:
   XL = 2πfL
3. SI Unit: Ohm.
4. Final Result: Inductive reactance increases with frequency.

19. Find inductive reactance of a 25 mH inductor at 50 Hz.

The inductive reactance is 7.85 Ω. Use XL = 2πfL.

1. Given Data:
   L = 25 mH = 25 × 10^-3 H
   f = 50 Hz
2. Formula Used:
   XL = 2πfL
3. Calculation:
   XL = 2 × 3.14 × 50 × 25 × 10^-3
   XL = 7.85 Ω
4. Final Result: XL = 7.85 Ω.

20. A 25 mH inductor connects to 220 V, 50 Hz AC. Find rms current.

The rms current is 28 A. Use I = V/XL.

1. Given Data:
   V = 220 V
   XL = 7.85 Ω
2. Formula Used:
   I = V/XL
3. Calculation:
   I = 220/7.85
   I = 28 A
4. Final Result: I = 28 A.

21. Why is average power zero in a pure inductor?

Average power is zero because the inductor returns stored energy to the source. Its voltage and current differ by π/2.

1. Phase Difference: φ = 90°.
2. Power Factor: cos 90° = 0.
3. Average Power: P = VI cos φ.
4. Final Result: A pure inductor consumes no average power.

Inductive Reactance Class 12 Questions

Inductive reactance controls AC current without dissipating real power in an ideal inductor.
A larger inductance or frequency gives a larger XL.
These inductive reactance Class 12 questions connect formula, unit, and frequency effect.

22. How does inductive reactance change with frequency?

Inductive reactance increases directly with frequency. The formula XL = 2πfL shows this relation.

1. Formula: XL = 2πfL.
2. If Frequency Doubles: XL doubles.
3. Current Effect: I = V/XL decreases.
4. Final Result: Higher frequency gives higher inductive reactance.

23. Why does an iron rod reduce bulb brightness in an inductor circuit?

An iron rod increases inductance and inductive reactance. More voltage drops across the inductor.

1. Iron Rod Effect: Magnetic field inside coil increases.
2. Inductance: L increases.
3. Reactance: XL = ωL increases.
4. Final Result: Bulb brightness decreases.

24. Why does DC face no inductive reactance in steady state?

DC has zero frequency in steady state. Since XL = 2πfL, inductive reactance becomes zero.

1. DC Frequency: f = 0.
2. Formula: XL = 2πfL.
3. Result: XL = 0.
4. Final Result: Ideal inductor acts like a wire for steady DC.

AC Voltage Applied to Capacitor Class 12 Questions

A capacitor blocks steady DC after charging, but it allows AC through repeated charging.
Its reactance decreases when frequency or capacitance increases.
These AC voltage applied to capacitor Class 12 questions cover capacitive reactance and phase lead.

25. What happens when AC voltage is applied to a capacitor?

Current leads voltage by π/2 in a pure capacitor. The capacitor charges and discharges every half cycle.

1. Applied Voltage: v = vm sin ωt.
2. Current Form: i = im sin(ωt + π/2).
3. Phase Relation: Current leads voltage by 90°.
4. Final Result: Capacitor current leads voltage by π/2.

26. What is capacitive reactance?

Capacitive reactance is opposition offered by a capacitor to AC. Its formula is XC = 1/ωC.

1. Formula Used:
   XC = 1/ωC
2. Also:
   XC = 1/(2πfC)
3. SI Unit: Ohm.
4. Final Result: Capacitive reactance decreases with frequency.

27. Find capacitive reactance of 15 μF capacitor at 50 Hz.

The capacitive reactance is 212 Ω. Use XC = 1/(2πfC).

1. Given Data:
   C = 15 μF = 15 × 10^-6 F
   f = 50 Hz
2. Formula Used:
   XC = 1/(2πfC)
3. Calculation:
   XC = 1/(2 × 3.14 × 50 × 15 × 10^-6)
   XC = 212 Ω
4. Final Result: XC = 212 Ω.

28. A 15 μF capacitor connects to 220 V, 50 Hz AC. Find rms current.

The rms current is 1.04 A. Use I = V/XC.

1. Given Data:
   V = 220 V
   XC = 212 Ω
2. Formula Used:
   I = V/XC
3. Calculation:
   I = 220/212
   I = 1.04 A
4. Final Result: I = 1.04 A.

29. What happens to capacitive reactance when frequency doubles?

Capacitive reactance becomes half. The current becomes double for the same rms voltage.

1. Formula: XC = 1/(2πfC).
2. Frequency Change: f becomes 2f.
3. Reactance Change: XC becomes XC/2.
4. Final Result: Doubling frequency halves capacitive reactance.

30. Why does a lamp glow with AC but not DC in series with a capacitor?

The lamp glows with AC because the capacitor charges and discharges repeatedly. With DC, current stops after charging.

1. DC Case: Capacitor charges once.
2. After Charging: Current becomes zero.
3. AC Case: Charge reverses every half cycle.
4. Final Result: AC produces continuous alternating current through the circuit.

Capacitive Reactance Class 12 Questions

Capacitive reactance depends inversely on capacitance and frequency.
A smaller capacitor gives more opposition to AC in the same circuit.
These capacitive reactance Class 12 questions support direct numerical and reasoning practice.

31. How does capacitance affect capacitive reactance?

Capacitive reactance decreases when capacitance increases. The formula XC = 1/(ωC) shows inverse relation.

1. Formula: XC = 1/(ωC).
2. Larger C: Smaller XC.
3. Current Effect: I = V/XC increases.
4. Final Result: Higher capacitance allows larger AC current.

32. Why is average power zero in a pure capacitor?

Average power is zero because the capacitor returns stored energy to the source. Its current leads voltage by π/2.

1. Phase Difference: φ = 90°.
2. Power Factor: cos 90° = 0.
3. Average Power: P = VI cos φ.
4. Final Result: A pure capacitor consumes no average power.

33. Why does a capacitor block DC after charging?

A capacitor blocks DC after charging because it reaches source voltage. No further charge transfer continues.

1. Initial State: Current flows during charging.
2. Final State: Capacitor voltage equals source voltage.
3. Circuit Current: Current becomes zero.
4. Final Result: A charged capacitor blocks steady DC.

Series LCR Circuit Class 12 Questions

A series LCR circuit combines resistance, inductive reactance, and capacitive reactance.
Voltage drops across L and C can cancel because they stay opposite in phase.
These series LCR circuit Class 12 questions cover impedance, phase, and resonance preparation.

34. What is impedance in a series LCR circuit?

Impedance is the total opposition offered by a series LCR circuit to AC. Its formula is Z = √[R² + (XL − XC)²].

1. Resistance: R.
2. Inductive Reactance: XL.
3. Capacitive Reactance: XC.
4. Formula Used:
   Z = √[R² + (XL − XC)²]
5. Final Result: Impedance combines resistance and reactance.

35. What is phase angle in a series LCR circuit?

The phase angle tells whether current leads or lags voltage. Its tangent is (XC − XL)/R.

1. Formula Used:
   tan φ = (XC − XL)/R
2. Capacitive Case: XC > XL gives current lead.
3. Inductive Case: XL > XC gives current lag.
4. Final Result: Phase depends on XC − XL.

36. Find impedance if R = 3 Ω, XL = 8 Ω, and XC = 4 Ω.

The impedance is 5 Ω. Use Z = √[R² + (XL − XC)²].

1. Given Data:
   R = 3 Ω
   XL = 8 Ω
   XC = 4 Ω
2. Formula Used:
   Z = √[R² + (XL − XC)²]
3. Calculation:
   Z = √[3² + (8 − 4)²]
   Z = √(9 + 16)
   Z = 5 Ω
4. Final Result: Z = 5 Ω.

37. Why can voltages across R and C exceed source voltage in an RC circuit?

They can exceed source voltage algebraically because their phases differ. Total voltage needs phasor addition.

1. Resistor Voltage: In phase with current.
2. Capacitor Voltage: Lags current by 90°.
3. Total Voltage: V = √(VR² + VC²).
4. Final Result: AC voltages require phase-aware addition.

38. In an RC circuit, VR = 151 V and VC = 160.3 V. Find source voltage.

The source voltage is 220 V. Use phasor addition.

1. Given Data:
   VR = 151 V
   VC = 160.3 V
2. Formula Used:
   V = √(VR² + VC²)
3. Calculation:
   V = √(151² + 160.3²)
   V ≈ 220 V
4. Final Result: Source voltage = 220 V.

Resonance in LCR Circuit Class 12 Questions

Resonance occurs when inductive and capacitive reactances become equal.
The impedance becomes minimum, and current becomes maximum in a series LCR circuit.
These resonance in LCR circuit Class 12 questions cover frequency, current, and applications.

39. What is resonance in a series LCR circuit?

Resonance is the condition where XL = XC in a series LCR circuit. Current amplitude becomes maximum.

1. Condition: XL = XC.
2. Impedance: Z = R.
3. Current: im = vm/R.
4. Final Result: Series resonance gives maximum current.

40. What is the resonant angular frequency of an LCR circuit?

The resonant angular frequency is ω0 = 1/√LC. It occurs when XL equals XC.

1. Condition: XL = XC.
2. Equation: ωL = 1/ωC.
3. Result:
   ω0 = 1/√LC
4. Final Result: Resonant angular frequency = 1/√LC.

41. Why can resonance not occur in only RL or RC circuits?

Resonance cannot occur in only RL or RC circuits because both L and C must be present. Their reactances cancel at resonance.

1. RL Circuit: No capacitive reactance.
2. RC Circuit: No inductive reactance.
3. Required Cancellation: XL = XC.
4. Final Result: Resonance needs both L and C.

42. An LCR circuit has R = 20 Ω and V = 200 V at resonance. Find average power.

The average power is 2000 W. At resonance, Z = R and current equals V/R.

1. Given Data:
   R = 20 Ω
   V = 200 V
2. Current:
   I = V/R
   I = 200/20
   I = 10 A
3. Power:
   P = I²R
   P = 10² × 20
   P = 2000 W
4. Final Result: Average power = 2000 W.

43. Why does a radio use resonance for tuning?

A radio uses resonance to select one station frequency. The tuning circuit gives maximum current at the chosen frequency.

1. Antenna: Receives many frequencies.
2. Tuning Circuit: Adjusts capacitance.
3. Resonance: Current becomes maximum for one frequency.
4. Final Result: Resonance selects the required radio station.

44. How does a metal detector use resonance?

A metal detector uses resonance in an AC circuit. Metal changes the circuit impedance and changes current.

1. Detector Coil: Forms part of an AC resonant circuit.
2. Metal Object: Changes impedance.
3. Electronic Circuit: Detects current change.
4. Final Result: Metal detection uses resonance shift.

Power Factor Class 12 Physics Questions

Power factor measures how effectively an AC circuit uses supplied current.
Only the resistor dissipates average power in ideal RLC circuits.
These power factor Class 12 Physics questions cover wattless current and resonance power.

45. What is power factor in AC circuits?

Power factor is the cosine of phase difference between voltage and current. Its formula is cos φ.

1. Average Power: P = VI cos φ.
2. Power Factor: cos φ.
3. Range: 0 to 1.
4. Final Result: Power factor measures useful power fraction.

46. What is wattless current?

Wattless current is current that produces no average power. It occurs in pure inductive or capacitive circuits.

1. Phase Difference: φ = 90°.
2. Power Factor: cos φ = 0.
3. Power: P = VI cos φ = 0.
4. Final Result: Wattless current has zero average power.

47. Why is power maximum at resonance?

Power is maximum at resonance because phase angle becomes zero. The power factor becomes 1.

1. Resonance Condition: XL = XC.
2. Impedance: Z = R.
3. Phase Angle: φ = 0.
4. Final Result: At resonance, P = I²R is maximum.

48. Why does low power factor cause large transmission loss?

Low power factor requires larger current for the same power. Larger current increases I²R loss.

1. Power Formula: P = VI cos φ.
2. Small cos φ: I must increase.
3. Line Loss: Ploss = I²R.
4. Final Result: Low power factor increases transmission loss.

49. How can a capacitor improve power factor?

A capacitor improves power factor by reducing lagging wattless current. It supplies leading current.

1. Inductive Load: Current lags voltage.
2. Capacitor Current: Leads voltage.
3. Net Effect: Wattless components cancel partly.
4. Final Result: Capacitors improve lagging power factor.

Transformer Class 12 Physics Questions

A transformer changes AC voltage using mutual induction between two coils.
It works only with changing magnetic flux, so it needs AC input.
These transformer Class 12 Physics questions cover step-up, step-down, losses, and power transmission.

50. What is the principle of a transformer?

A transformer works on mutual induction. A changing current in the primary coil induces emf in the secondary coil.

1. Primary Coil: Receives AC input.
2. Core Flux: Changes with time.
3. Secondary Coil: Gets induced emf.
4. Final Result: Transformer action depends on mutual induction.

51. What is the transformer voltage ratio?

The transformer voltage ratio is Vs/Vp = Ns/Np. Voltage depends on turns ratio.

1. Primary Turns: Np.
2. Secondary Turns: Ns.
3. Formula Used:
   Vs/Vp = Ns/Np
4. Final Result: Turns ratio decides voltage ratio.

52. What is the current ratio in an ideal transformer?

The current ratio is Ip/Is = Ns/Np. Higher voltage side carries lower current.

1. Ideal Power Relation: VpIp = VsIs.
2. Voltage Ratio: Vs/Vp = Ns/Np.
3. Current Relation: Ip/Is = Ns/Np.
4. Final Result: Current changes inversely with voltage.

53. What is a step-up transformer?

A step-up transformer increases voltage. Its secondary coil has more turns than the primary coil.

1. Condition: Ns > Np.
2. Voltage Result: Vs > Vp.
3. Current Result: Is < Ip.
4. Final Result: Step-up transformer increases AC voltage.

54. What is a step-down transformer?

A step-down transformer decreases voltage. Its secondary coil has fewer turns than the primary coil.

1. Condition: Ns < Np.
2. Voltage Result: Vs < Vp.
3. Current Result: Is > Ip.
4. Final Result: Step-down transformer decreases AC voltage.

55. A transformer has Np = 100 and Ns = 200. Find output voltage for 220 V input.

The output voltage is 440 V. Use Vs/Vp = Ns/Np.

1. Given Data:
   Np = 100
   Ns = 200
   Vp = 220 V
2. Formula Used:
   Vs = Vp(Ns/Np)
3. Calculation:
   Vs = 220 × (200/100)
   Vs = 440 V
4. Final Result: Vs = 440 V.

56. What are the main energy losses in transformers?

The main losses are flux leakage, winding resistance, eddy currents, and hysteresis. Good design reduces these losses.

1. Flux Leakage: Some flux misses the secondary coil.
2. Winding Resistance: Copper wires heat up.
3. Eddy Currents: Laminated core reduces them.
4. Hysteresis: Soft magnetic core reduces it.
5. Final Result: Transformer efficiency improves by reducing core and copper losses.

NCERT Class 12 Physics Chapter 7 Questions

NCERT questions usually combine rms values, reactance, impedance, resonance, and transformers.
Students should convert mH, μF, and peak values before calculation.
These NCERT Class 12 Physics Chapter 7 questions match the 2026 exercise pattern.

57. A 100 Ω resistor connects to 220 V, 50 Hz AC. Find rms current and power.

The rms current is 2.2 A, and power is 484 W. Use I = V/R and P = VI.

1. Given Data:
   R = 100 Ω
   V = 220 V
2. Current:
   I = V/R
   I = 220/100
   I = 2.2 A
3. Power:
   P = VI
   P = 220 × 2.2
   P = 484 W
4. Final Result: I = 2.2 A and P = 484 W.

58. A 44 mH inductor connects to 220 V, 50 Hz AC. Find rms current.

The rms current is 15.9 A. Use XL = 2πfL and I = V/XL.

1. Given Data:
   L = 44 mH = 44 × 10^-3 H
   f = 50 Hz
   V = 220 V
2. Inductive Reactance:
   XL = 2 × 3.14 × 50 × 44 × 10^-3
   XL = 13.8 Ω
3. Current:
   I = 220/13.8
   I = 15.9 A
4. Final Result: I = 15.9 A.

59. A 60 μF capacitor connects to 110 V, 60 Hz AC. Find rms current.

The rms current is 2.49 A. Use XC = 1/(2πfC) and I = V/XC.

1. Given Data:
   C = 60 μF = 60 × 10^-6 F
   f = 60 Hz
   V = 110 V
2. Capacitive Reactance:
   XC = 1/(2 × 3.14 × 60 × 60 × 10^-6)
   XC = 44.2 Ω
3. Current:
   I = 110/44.2
   I = 2.49 A
4. Final Result: I = 2.49 A.

60. What is the net power absorbed by a pure inductor or capacitor over one cycle?

The net power absorbed is zero. Ideal inductors and capacitors return stored energy to the source.

1. Inductor: Stores magnetic energy.
2. Capacitor: Stores electric energy.
3. Average Power: P = VI cos 90° = 0.
4. Final Result: Average power over one cycle is zero.

61. A 30 μF capacitor connects to a 27 mH inductor. Find angular frequency of free oscillations.

The angular frequency is 1.11 × 10^3 rad/s. Use ω = 1/√LC.

1. Given Data:
   L = 27 mH = 27 × 10^-3 H
   C = 30 μF = 30 × 10^-6 F
2. Formula Used:
   ω = 1/√LC
3. Calculation:
   ω = 1/√(27 × 10^-3 × 30 × 10^-6)
   ω = 1.11 × 10^3 rad/s
4. Final Result: ω = 1.11 × 10^3 rad/s.

Class 12 Physics Chapter-Wise Important Questions