Alternating Current (प्रत्यावर्ती धारा) – AC Voltage, LCR Circuit, Power Factor, Transformers | RBSE Class 12 Physics

📅 Friday, 16 January 2026 📖 3-5 min read
Alternating Current – Class 12 Physics | Chapter 7 NCERT

Alternating Current

This chapter strictly follows NCERT Class 12 Physics textbook and examination pattern for CBSE, RBSE and other state boards

Alternating Current (AC), जिसे हिंदी में प्रत्यावर्ती धारा कहा जाता है। In contrast to direct current (DC), which flows only in one direction, AC periodically reverses its direction. AC is the form in which electric power is delivered to homes and businesses. This chapter explores the behavior of AC circuits containing resistors, inductors, and capacitors.

Notation Convention

Vector quantities are represented in bold typeface. Magnitudes are in italics. Instantaneous values use lowercase (v, i), while peak/RMS values use uppercase (V, I).

1. Introduction

Most electrical power systems in the world use alternating current. AC has several advantages over DC for power transmission:

  • Voltage can be easily stepped up or down using transformers
  • More efficient for long-distance transmission
  • Simpler generation using electromagnetic induction
  • Easier to interrupt using circuit breakers
DC vs AC Comparison

Direct Current (DC): Current flows in one direction with constant magnitude. Example: Battery-powered devices.

Alternating Current (AC): Current periodically reverses direction. Magnitude varies sinusoidally with time. Example: Household electricity supply.

1.1 AC Voltage and Current

An alternating voltage can be represented mathematically as:

\\[v(t) = V_m \\sin(\\omega t + \\phi)\\]
Where:
• v(t) = Instantaneous voltage at time t
• Vm = Peak (maximum) voltage
• ω = Angular frequency = 2πf (rad/s)
• f = Frequency (Hz)
• φ = Phase constant (initial phase)
• T = Time period = 1/f = 2π/ω

Similarly, alternating current:

\\[i(t) = I_m \\sin(\\omega t + \\phi)\\]
Where:
• i(t) = Instantaneous current
• Im = Peak (maximum) current

1.2 RMS (Root Mean Square) Values

The effective or RMS value of AC is defined as the equivalent DC value that would produce the same heating effect in a resistor.

Derivation of RMS Value

For sinusoidal AC: i(t) = Im sin(ωt)

Power in resistor R:
\\[P = i^2 R = I_m^2 \\sin^2(\\omega t) R\\]

Average power over one cycle:
\\[P_{avg} = \\frac{1}{T} \\int_0^T I_m^2 \\sin^2(\\omega t) R \\, dt\\]

Using \\(\\langle \\sin^2(\\omega t) \\rangle = \\frac{1}{2}\\):

\\[P_{avg} = \\frac{I_m^2 R}{2} = I_{rms}^2 R\\]

Important: This derivation and the resulting expression Irms = Im/√2 are valid only for sinusoidal alternating quantities. For non-sinusoidal waveforms, the RMS value must be calculated using the general definition.

\\[I_{rms} = \\frac{I_m}{\\sqrt{2}} = 0.707 I_m\\] \\[V_{rms} = \\frac{V_m}{\\sqrt{2}} = 0.707 V_m\\]
Example: Household AC in India: 220V RMS means peak voltage is 220√2 ≈ 311V

2. AC Voltage Applied to a Resistor

When an AC voltage is applied across a pure resistor, the current through it is also AC with the same frequency.

V = Vm sin ωt R i
Figure 2.1: Pure resistor in AC circuit
Analysis of Pure Resistive AC Circuit

Applied voltage: v(t) = Vm sin(ωt)

By Ohm's law:
\\[i(t) = \\frac{v(t)}{R} = \\frac{V_m \\sin(\\omega t)}{R}\\]

\\[i(t) = I_m \\sin(\\omega t)\\]
Where: Im = Vm/R

Key Point: Voltage and current are in phase (φ = 0°)
ωt v, i v i π 2π
Figure 2.2: Voltage and current waveforms in phase for pure resistor

3. Representation of AC Current and Voltage by Rotating Vectors - Phasors

A phasor is a rotating vector that represents a sinusoidally varying quantity. The projection of a rotating phasor on the chosen reference axis represents the instantaneous value.

Phasor Representation

A sinusoidal quantity v(t) = Vm sin(ωt + φ) can be represented by a phasor:

  • Length: Equal to peak value Vm
  • Direction: Makes angle (ωt + φ) with reference axis
  • Rotation: Rotates counterclockwise with angular velocity ω
Ref Vm ωt+φ v(t) ω
Figure 3.1: Phasor representation rotating with angular velocity ω

3.1 Phasor Diagram for Pure Resistor

For a pure resistor, voltage and current are in phase:

V I Phase difference φ = 0°
Figure 3.2: Phasor diagram for resistor - V and I in phase

4. AC Voltage Applied to an Inductor

An inductor opposes changes in current through it by inducing a back emf according to Faraday's law.

V = Vm sin ωt L
Figure 4.1: Pure inductor in AC circuit
Analysis of Pure Inductive AC Circuit

Applied voltage: v(t) = Vm sin(ωt)

Self-induced emf in inductor:
\\[\\varepsilon = -L \\frac{di}{dt}\\]

According to Kirchhoff's voltage law, the applied voltage balances the induced emf in the inductor. Therefore:
\\[V_m \\sin(\\omega t) = L \\frac{di}{dt}\\]

Integrating:
\\[i(t) = -\\frac{V_m}{\\omega L} \\cos(\\omega t) = \\frac{V_m}{\\omega L} \\sin(\\omega t - 90°)\\]

\\[i(t) = I_m \\sin(\\omega t - \\frac{\\pi}{2})\\]
Where: \\(I_m = \\frac{V_m}{\\omega L} = \\frac{V_m}{X_L}\\)

XL = ωL is called Inductive Reactance
Unit: Ohm (Ω)

Key Point: Current lags voltage by 90° (π/2 radians)
v i 90° lag
Figure 4.2: Current lags voltage by 90° in inductor

4.1 Phasor Diagram for Pure Inductor

V I 90° I lags V by 90°
Figure 4.3: Phasor diagram for inductor

5. AC Voltage Applied to a Capacitor

A capacitor stores electrical energy in an electric field between its plates. In an AC circuit, it continuously charges and discharges.

Analysis of Pure Capacitive AC Circuit

Applied voltage: v(t) = Vm sin(ωt)

Charge on capacitor: q = Cv = CVm sin(ωt)

Current (rate of charge flow):
\\[i(t) = \\frac{dq}{dt} = C \\frac{d}{dt}[V_m \\sin(\\omega t)]\\]

\\[i(t) = \\omega C V_m \\cos(\\omega t) = \\omega C V_m \\sin(\\omega t + 90°)\\]

\\[i(t) = I_m \\sin(\\omega t + \\frac{\\pi}{2})\\]
Where: \\(I_m = \\omega C V_m = \\frac{V_m}{X_C}\\)

XC = 1/(ωC) is called Capacitive Reactance
Unit: Ohm (Ω)

Key Point: Current leads voltage by 90° (π/2 radians)
Common Mistake

Wrong: "Inductor leads current, capacitor lags current"

Correct: In inductor, current lags voltage. In capacitor, current leads voltage.

- In Capacitor I leads V, In Inductor V leads I

6. AC Voltage Applied to a Series LCR Circuit

A series LCR circuit contains a resistor (R), inductor (L), and capacitor (C) connected in series with an AC source.

V = Vm sin ωt R VR L VL C VC i(t)
Figure 6.1: Series LCR circuit

6.1 Phasor Diagram Solution

In a series circuit, the same current flows through all components. We use current as reference phasor.

Voltage Across Each Component

Resistor: VR = IR (in phase with current)

Inductor: VL = IXL (leads current by 90°)

Capacitor: VC = IXC (lags current by 90°)

I VR VL VC VL-VC V φ
Figure 6.2: Phasor diagram for series LCR circuit (VL > VC case shown)
Impedance and Phase Angle

From phasor diagram (using Pythagoras):

\\[V^2 = V_R^2 + (V_L - V_C)^2\\]

\\[V^2 = (IR)^2 + (IX_L - IX_C)^2 = I^2[R^2 + (X_L - X_C)^2]\\]

\\[Z = \\sqrt{R^2 + (X_L - X_C)^2}\\] \\[\\tan \\phi = \\frac{X_L - X_C}{R}\\]
Z = Impedance (total opposition to AC)
φ = Phase angle between V and I

• If XL > XC: Circuit is inductive (φ positive, I lags V)
• If XC > XL: Circuit is capacitive (φ negative, I leads V)
• If XL = XC: Circuit is resonant (φ = 0, V and I in phase)

6.2 Resonance in Series LCR Circuit

Resonance occurs when inductive and capacitive reactances are equal, causing them to cancel out.

\\[X_L = X_C\\] \\[\\omega_0 L = \\frac{1}{\\omega_0 C}\\] \\[\\omega_0 = \\frac{1}{\\sqrt{LC}}\\]
Resonant frequency: \\(f_0 = \\frac{\\omega_0}{2\\pi} = \\frac{1}{2\\pi\\sqrt{LC}}\\)

At resonance:
• Impedance is minimum: Z = R
• Current is maximum: I = V/R
• Power factor is unity: cos φ = 1
• Circuit behaves purely resistive
ω I ω0 Imax Sharp resonance (high Q)
Figure 6.3: Resonance curve - Current vs frequency
Quality Factor (Q-factor)

The sharpness of resonance is measured by quality factor:

\\[Q = \\frac{\\omega_0 L}{R} = \\frac{1}{\\omega_0 CR} = \\frac{1}{R}\\sqrt{\\frac{L}{C}}\\]

Higher Q: Sharper resonance, more selective
Lower Q: Broader resonance, less selective

7. Power in AC Circuit: The Power Factor

In AC circuits, power depends not only on voltage and current magnitudes but also on the phase difference between them.

Average Power in AC Circuit

Conditions: The following derivation applies to linear AC circuits with sinusoidal sources.

Instantaneous power:
\\[p(t) = v(t) \\cdot i(t) = V_m \\sin(\\omega t) \\cdot I_m \\sin(\\omega t - \\phi)\\]

Using trigonometric identity:
\\[p(t) = \\frac{V_m I_m}{2}[\\cos \\phi - \\cos(2\\omega t - \\phi)]\\]

Average over complete cycle:
Second term averages to zero, so:

\\[P_{avg} = V_{rms} I_{rms} \\cos \\phi\\] \\[P_{avg} = I_{rms}^2 R\\]
cos φ = Power Factor

• cos φ = 1 (φ = 0°): Pure resistive, maximum power
• cos φ = 0 (φ = 90°): Pure reactive (L or C), zero average power
• 0 < cos φ < 1: Practical AC circuits

Q48. The power factor of a purely inductive circuit is:

(a) 1
(b) 0
(c) 0.5
(d) Depends on frequency

Answer: (b) 0
Explanation: In a pure inductor, the current lags the voltage by 90° (φ = 90°). The power factor is given by cos φ = cos 90° = 0. Therefore, no real power is consumed in a purely inductive circuit; all power is reactive.
Types of Power in AC

1. Active/Real Power (P): P = VI cos φ (Watts)
Power actually consumed/dissipated

2. Reactive Power (Q): Q = VI sin φ (VAR - Volt-Ampere Reactive)
Power oscillating between source and reactive elements

3. Apparent Power (S): S = VI (VA - Volt-Ampere)
Product of RMS voltage and current

Relationship: S² = P² + Q²

8. Applications of Alternating Current

Practical Applications

Alternating current finds widespread application in various fields due to its unique properties and ease of voltage transformation.

1. Power Transmission and Distribution: AC is preferred for long-distance power transmission because voltage can be stepped up to reduce transmission losses (I²R losses), then stepped down for safe domestic use. High-voltage AC transmission is more economical than DC.

2. Transformers: These devices work exclusively on AC principle using electromagnetic induction. They enable efficient voltage conversion for different applications - from power stations (hundreds of kV) to household appliances (220V) to electronic devices (5V-12V).

3. Induction Motors: Most industrial and household motors operate on AC. The rotating magnetic field produced by AC drives the rotor without requiring physical contact, making these motors reliable and low-maintenance.

4. Choke Coils: Inductors in AC circuits control current without dissipating power as heat. Used in fluorescent lamps, they limit current while maintaining high power factor, unlike resistors which waste energy.

5. Radio and Communication: Series LCR circuits at resonance form the basis of radio tuning. By varying capacitance or inductance, the resonant frequency matches the desired station frequency, selecting one signal from many.

6. Welding and Heating: AC arc welding provides stable arc with less equipment complexity. Induction heating uses eddy currents induced by AC magnetic fields for metal treatment and cooking (induction cooktops).

7. X-ray Machines and Medical Equipment: High-voltage AC, stepped up through transformers, powers X-ray tubes and various diagnostic equipment in hospitals.

9. Important Constants and Units

Quantity Symbol SI Unit Formula
Inductive Reactance XL Ohm (Ω) ωL = 2πfL
Capacitive Reactance XC Ohm (Ω) 1/(ωC) = 1/(2πfC)
Impedance Z Ohm (Ω) √[R² + (XL - XC)²]
Power Factor cos φ Dimensionless R/Z or cos(angle between V and I)
Quality Factor Q Dimensionless ω₀L/R or (1/R)√(L/C)
Resonant Frequency f₀ Hertz (Hz) 1/(2π√(LC))

Key Formulas Summary:

  • RMS values: Vrms = Vm/√2, Irms = Im/√2
  • Pure R: V and I in phase
  • Pure L: I lags V by 90°
  • Pure C: I leads V by 90°
  • Series LCR: Z = √[R² + (XL - XC)²]
  • Resonance: ω₀ = 1/√(LC), Zmin = R
  • Power: P = VI cos φ = I²R
Chapter Connections

Previous Chapter: Electromagnetic Induction (Faraday's Law, Lenz's Law, Inductance)

Current Chapter: Alternating Current (AC Circuits, Resonance, Power Factor)

Next Chapter: Electromagnetic Waves (Wave propagation, Electromagnetic spectrum)

Understanding the principles of electromagnetic induction from the previous chapter provides the foundation for AC generation and transformer operation covered here.

Summary

Summary:
  • Phasor diagrams for R, L, C, and LCR circuits
  • Impedance derivation for series LCR circuit
  • Resonance condition and Q-factor
  • Power factor and average power derivation
  • Numerical problems on reactance, impedance, resonance
  • Phase relationships - for conceptual questions
  • RMS vs peak values - in MCQs

From the analysis of phase relationships, it is observed that in a capacitive circuit the current leads the voltage by π/2 radians, whereas in an inductive circuit the current lags the voltage by π/2 radians. In a purely resistive circuit, the current and voltage remain in phase.

SVG Diagram Gallery (Reference Only)

Figure Index

Diagrams in this chapter:

  • Figure 2.1: Pure resistor in AC circuit
  • Figure 2.2: Voltage and current waveforms for resistor
  • Figure 3.1: Phasor representation
  • Figure 3.2: Phasor diagram for resistor
  • Figure 4.1: Pure inductor in AC circuit
  • Figure 4.2: Current lagging voltage in inductor
  • Figure 4.3: Phasor diagram for inductor
  • Figure 6.1: Series LCR circuit
  • Figure 6.2: Phasor diagram for LCR circuit
  • Figure 6.3: Resonance curve

Practice Questions

Exercise Structure

This chapter includes practice exercises divided into four sections:

  • Section A: Multiple Choice Questions (49 questions)
  • Section B: Very Short Answer Questions (48 questions)
  • Section C: Short Answer Questions (48 questions)
  • Section D: Long Answer Questions (5 questions)

Answer Key and Solutions: Detailed answers to all questions are provided in the "Answers and Solutions" section following the questions.

Section A: Multiple Choice Questions (49 questions)

Instructions: Choose the correct option.

Note: Answers and detailed solutions to all questions are provided at the end of this section.

Q1. The frequency of AC in India is:

(a) 50 Hz
(b) 60 Hz
(c) 100 Hz
(d) 220 Hz

Answer: (a) 50 Hz
Standard AC frequency in India is 50 Hz (50 cycles per second).

Q2. RMS value of AC current with peak value Im is:

(a) Im/2
(b) Im/√2
(c) Im/√3
(d) Im

Answer: (b) Im/√2
Irms = Im/√2 = 0.707 Im

Q3. In pure resistive AC circuit, current and voltage are:

(a) In phase
(b) 90° out of phase
(c) 180° out of phase
(d) 45° out of phase

Answer: (a) In phase
In resistor, V and I have zero phase difference.

Q4. Inductive reactance is:

(a) ωL
(b) 1/(ωL)
(c) ωC
(d) 1/(ωC)

Answer: (a) ωL
XL = ωL = 2πfL ohms

Q5. In pure inductor, current:

(a) Leads voltage by 90°
(b) Lags voltage by 90°
(c) Is in phase with voltage
(d) Leads voltage by 45°

Answer: (b) Lags voltage by 90°
In inductor, current lags voltage by π/2 radians.

Q6. Capacitive reactance is:

(a) ωC
(b) 1/(ωC)
(c) ωL
(d) 1/(ωL)

Answer: (b) 1/(ωC)
XC = 1/(ωC) = 1/(2πfC) ohms

Q7. In pure capacitor, current:

(a) Leads voltage by 90°
(b) Lags voltage by 90°
(c) Is in phase with voltage
(d) Lags voltage by 45°

Answer: (a) Leads voltage by 90°
In capacitor, current leads voltage by π/2 radians.

Q8. Impedance of series LCR circuit is:

(a) R + XL + XC
(b) √[R² + (XL + XC)²]
(c) √[R² + (XL - XC)²]
(d) R + (XL - XC)

Answer: (c) √[R² + (XL - XC)²]
Impedance from phasor addition: Z = √[R² + (XL - XC)²]

Q9. At resonance in series LCR circuit:

(a) XL = XC
(b) XL = 0
(c) XC = 0
(d) R = 0

Answer: (a) XL = XC
Resonance condition: Inductive reactance equals capacitive reactance.

Q10. Resonant frequency of LCR circuit is:

(a) 1/(2πLC)
(b) 1/(2π√(LC))
(c) 2π√(LC)
(d) √(LC)/2π

Answer: (b) 1/(2π√(LC))
f₀ = 1/(2π√(LC)) Hz

Q11. Peak voltage 311V. RMS voltage is:

(a) 220V
(b) 311V
(c) 440V
(d) 156V

Answer: (a) 220V
V_rms = V_m/√2 = 311/1.414 ≈ 220V

Q12. AC frequency 50Hz. Time period is:

(a) 0.01s
(b) 0.02s
(c) 0.05s
(d) 0.1s

Answer: (b) 0.02s
T = 1/f = 1/50 = 0.02 seconds

Q13. Household AC 220V means:

(a) Peak value
(b) Average value
(c) RMS value
(d) Instantaneous value

Answer: (c) RMS value
Standard notation is RMS value

Q14. Inductive reactance at f=0 (DC) is:

(a) Zero
(b) Infinite
(c) ωL
(d) Maximum

Answer: (a) Zero
X_L = ωL = 2πfL = 0 when f=0

Q15. Capacitive reactance at f=0 (DC) is:

(a) Zero
(b) ωC
(c) 1/(ωC)
(d) Infinite

Answer: (d) Infinite
X_C = 1/(ωC) = ∞ when ω=0 (open circuit)

Q16. Frequency doubles, X_L becomes:

(a) Half
(b) Double
(c) Four times
(d) Same

Answer: (b) Double
X_L = ωL = 2πfL, so X_L ∝ f

Q17. Frequency doubles, X_C becomes:

(a) Half
(b) Double
(c) Four times
(d) Same

Answer: (a) Half
X_C = 1/(ωC), so X_C ∝ 1/f

Q18. In pure L, average power is:

(a) Maximum
(b) VI
(c) Zero
(d) V²/L

Answer: (c) Zero
Wattless current, cos φ = 0

Q19. In pure C, average power is:

(a) Maximum
(b) VI
(c) V²C
(d) Zero

Answer: (d) Zero
Wattless current, cos φ = 0

Q20. Only component consuming power:

(a) Resistor
(b) Inductor
(c) Capacitor
(d) All

Answer: (a) Resistor
Only R dissipates power as heat

Q21. If X_L = X_C in LCR, circuit is:

(a) Resistive
(b) Inductive
(c) Capacitive
(d) At resonance

Answer: (d) At resonance
Resonance condition: X_L = X_C

Q22. At resonance, impedance equals:

(a) Zero
(b) R
(c) X_L
(d) X_C

Answer: (b) R
Z_min = R at resonance

Q23. At resonance, current is:

(a) Zero
(b) Minimum
(c) Maximum
(d) V/L

Answer: (c) Maximum
I_max = V/R at resonance

Q24. Phase angle at resonance:

(a) 90°
(b) 45°
(c) 0°
(d) 180°

Answer: (c) 0°
φ = 0 at resonance, purely resistive

Q25. If X_L > X_C, circuit is:

(a) Inductive
(b) Capacitive
(c) Resistive
(d) Resonant

Answer: (a) Inductive
Inductive nature dominates

Q26. If X_C > X_L, current:

(a) Leads V
(b) Lags V
(c) In phase
(d) Zero

Answer: (a) Leads V
Capacitive circuit, I leads V

Q27. Impedance unit is:

(a) Henry
(b) Farad
(c) Ohm
(d) Watt

Answer: (c) Ohm
Same as resistance, reactance

Q28. tan φ in LCR circuit:

(a) R/Z
(b) (X_L-X_C)/R
(c) Z/R
(d) R/(X_L-X_C)

Answer: (b) (X_L-X_C)/R
tan φ = (X_L - X_C)/R

Q29. Power factor range is:

(a) 0 to 1
(b) -1 to +1
(c) 0 to ∞
(d) Any value

Answer: (a) 0 to 1
cos φ ranges from 0 to 1

Q30. Unity power factor means:

(a) φ = 90°
(b) φ = 0°
(c) φ = 45°
(d) φ = 180°

Answer: (b) φ = 0°
cos φ = 1 when φ = 0

Q31. Resonant frequency formula:

(a) 2π√(LC)
(b) 1/(2π√(LC))
(c) √(LC)/2π
(d) 2π/√(LC)

Answer: (b) 1/(2π√(LC))
f_0 = 1/(2π√(LC))

Q32. Q-factor measures:

(a) Power
(b) Energy
(c) Sharpness
(d) Frequency

Answer: (c) Sharpness
Quality factor = sharpness of resonance

Q33. High Q means:

(a) Broad resonance
(b) Sharp resonance
(c) Low selectivity
(d) High resistance

Answer: (b) Sharp resonance
High Q = sharp, selective

Q34. Q = ω_0L/R also equals:

(a) ω_0CR
(b) 1/(ω_0CR)
(c) ω_0C/R
(d) R/(ω_0L)

Answer: (b) 1/(ω_0CR)
Q = ω_0L/R = 1/(ω_0CR)

Q35. Bandwidth of resonance:

(a) f_0Q
(b) f_0/Q
(c) Qf_0
(d) Q/f_0

Answer: (b) f_0/Q
BW = f_0/Q

Half-Power Points: These are the frequencies at which the power dissipated in the circuit is half the maximum power. At these points, the current amplitude is 1/√2 times the maximum current.

Q36. Wattless current exists in:

(a) Pure R
(b) Pure L or C
(c) LCR at resonance
(d) Transformer

Answer: (b) Pure L or C
φ = 90° in pure L or C

Q37. Apparent power unit:

(a) Watt
(b) VAR
(c) VA
(d) Joule

Answer: (c) VA
Volt-Ampere (VA)

Q38. Reactive power unit:

(a) Watt
(b) VAR
(c) VA
(d) Ohm

Answer: (b) VAR
VAR (Volt-Ampere Reactive)

Q39. Real power formula:

(a) VI
(b) VI sin φ
(c) VI cos φ
(d) I²X_L

Answer: (c) VI cos φ
P = VI cos φ watts

Q40. cos φ = 0 means:

(a) Maximum power
(b) Zero power
(c) Resistive
(d) Resonance

Answer: (b) Zero power
Zero average power, purely reactive

Q41. Choke coil is used to:

(a) Increase current
(b) Decrease current
(c) Store energy
(d) Dissipate power

Answer: (b) Decrease current
Reduces current without power loss

Q42. Choke coil should have:

(a) High R, low L
(b) Low R, high L
(c) High R, high L
(d) Low R, low L

Answer: (b) Low R, high L
Minimize copper loss, maximize reactance

Q43. AC ammeter measures:

(a) Peak value
(b) Average value
(c) RMS value
(d) Instantaneous

Answer: (c) RMS value
Calibrated to show RMS

Q44. Hot wire instrument works on:

(a) Magnetic effect
(b) Heating effect
(c) Chemical effect
(d) EMI

Answer: (b) Heating effect
Uses I²R heating

Q45. L=0.1H, C=100μF, resonant frequency:

(a) 50Hz
(b) 159Hz
(c) 318Hz
(d) 100Hz

Answer: (b) 159Hz
f_0 = 1/(2π√(0.1×100×10⁻⁶)) ≈ 159Hz

Q46. R=3Ω, X_L=4Ω, X_C=0, impedance:

(a) 3Ω
(b) 4Ω
(c) 5Ω
(d) 7Ω

Answer: (c) 5Ω
Z = √(9+16) = 5Ω

Q47. V=200V, I=2A, φ=60°, power:

(a) 400W
(b) 200W
(c) 100W
(d) 346W

Answer: (b) 200W
P = 200×2×cos(60°) = 200W

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