Class 12 Physics Notes – Chapter 7: Alternating Current

1. Introduction to Alternating Current

An alternating voltage (AC voltage) is a voltage that varies sinusoidally with time. When such a voltage is applied to a circuit, the current driven by it is called an alternating current (AC).

Voltage Equation: v = v_m sin(ωt)

Current Equation: i = i_m sin(ωt)

Where v_m and i_m are the amplitude (peak values) of the oscillating potential difference and current, and ω is the angular frequency.

2. AC Voltage Applied to a Resistor

When an AC voltage is applied to a pure resistor R, the current and voltage are in the same phase (there is no phase difference).

Current amplitude: i_m = v_m / R

RMS Values (Root Mean Square)

To express AC power in the same form as DC power, we use effective values or RMS values.

• RMS Current: I = i_m / √2 = 0.707 i_m
• RMS Voltage: V = v_m / √2 = 0.707 v_m

Average power dissipated in a resistor: P = I² R = V I

3. AC Voltage Applied to an Inductor

When an AC voltage is applied to a pure inductor (L), the current lags the voltage by a phase angle of π/2 (or 90°).

Equation: i = i_m sin(ωt – π/2)

Inductive Reactance (X_L)

The opposition offered by the inductor to the flow of AC is called inductive reactance. Its SI unit is Ohm (Ω).

Formula: X_L = ω L

Amplitude of current: i_m = v_m / X_L

The average power supplied to a pure inductor over one complete cycle is zero.

4. AC Voltage Applied to a Capacitor

When an AC voltage is applied to a pure capacitor (C), the current leads the voltage by a phase angle of π/2 (or 90°).

Equation: i = i_m sin(ωt + π/2)

Capacitive Reactance (X_C)

The opposition offered by the capacitor to the flow of AC is called capacitive reactance. Its SI unit is also Ohm (Ω).

Formula: X_C = 1 / (ω C)

Amplitude of current: i_m = v_m / X_C

The average power supplied to a pure capacitor over one complete cycle is also zero.

5. AC Voltage Applied to a Series LCR Circuit

When a resistor (R), inductor (L), and capacitor (C) are connected in series to an AC source, the total opposition to the current is called Impedance (Z).

Impedance Formula: Z = √[R² + (X_C – X_L)²]

Current amplitude: i_m = v_m / Z

Phase angle (φ): tan φ = (X_C – X_L) / R

6. Resonance in LCR Circuit

Resonance occurs in a series LCR circuit when the capacitive reactance becomes equal to the inductive reactance (X_C = X_L). At this point, the impedance is minimum (Z = R) and the current amplitude is maximum.

Resonant Frequency (ω₀): ω₀ = 1 / √(LC)

7. Power in AC Circuit: The Power Factor

The average power dissipated in an AC circuit depends on the phase angle φ between voltage and current.

Average Power (P) = V I cos φ

• The term cos φ is called the Power Factor (cos φ = R / Z).
• For a pure resistor, φ = 0, so Power Factor = 1 (Maximum power).
• For a pure inductor or capacitor, φ = 90°, so Power Factor = 0 (Wattless current).

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Multiple Choice Questions (MCQs)

1. The relation between peak voltage (v_m) and RMS voltage (V) is:
a) V = v_m × 2
b) V = v_m / √2
c) V = v_m × √2
d) V = v_m / 2
Answer: (b) V = v_m / √2

2. In an AC circuit containing only a pure inductor, the current:
a) Leads the voltage by 90°
b) Lags the voltage by 90°
c) Is in phase with the voltage
d) Leads the voltage by 180°
Answer: (b) Lags the voltage by 90°

3. The SI unit of Inductive Reactance (X_L) and Capacitive Reactance (X_C) is:
a) Henry
b) Farad
c) Ohm
d) Hertz
Answer: (c) Ohm

4. At resonance in an LCR series circuit, which of the following conditions is true?
a) X_L > X_C
b) X_L < X_C
c) X_L = X_C
d) R = 0
Answer: (c) X_L = X_C

5. The power factor of a purely resistive circuit is:
a) 0
b) 0.5
c) 1
d) Infinity
Answer: (c) 1

6. The formula for the impedance (Z) of a series LCR circuit is:
a) Z = R + X_L + X_C
b) Z = √[R² + (X_C – X_L)²]
c) Z = √(R² + X_L²)
d) Z = R² + (X_L – X_C)²
Answer: (b) Z = √[R² + (X_C – X_L)²]

7. The average power dissipated in a pure capacitor over one complete cycle is:
a) Zero
b) V I
c) I²R
d) V² / C
Answer: (a) Zero

8. What is the value of power factor for a purely inductive circuit?
a) 1
b) -1
c) 0
d) 0.5
Answer: (c) 0

9. The resonant frequency (ω₀) of a series LCR circuit is given by:
a) 1 / (LC)
b) √(LC)
c) 1 / √(LC)
d) L / C
Answer: (c) 1 / √(LC)

10. The current which does not contribute to any power dissipation in the circuit is called:
a) Direct current
b) Alternating current
c) Induced current
d) Wattless current
Answer: (d) Wattless current

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Textbook Exercise Solutions

Q 7.1: A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Solution:
Given: R = 100 Ω, V_rms = 220 V, ν = 50 Hz.
(a) RMS Current I_rms = V_rms / R = 220 / 100 = 2.2 A.
(b) Net power consumed P = V_rms × I_rms = 220 × 2.2 = 484 W.

Q 7.2: (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Solution:
(a) Given Peak voltage v_m = 300 V.
RMS Voltage V_rms = v_m / √2 = 300 / 1.414 = 212.1 V.
(b) Given RMS current I_rms = 10 A.
Peak current i_m = I_rms × √2 = 10 × 1.414 = 14.14 A.

Q 7.3: A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Solution:
Given: L = 44 mH = 44 × 10^-3 H, V_rms = 220 V, ν = 50 Hz.
Inductive reactance X_L = 2πνL = 2 × 3.14 × 50 × 44 × 10^-3 = 13.82 Ω.
RMS current I_rms = V_rms / X_L = 220 / 13.82 = 15.9 A.

Q 7.4: A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Solution:
Given: C = 60 μF = 60 × 10^-6 F, V_rms = 110 V, ν = 60 Hz.
Capacitive reactance X_C = 1 / (2πνC) = 1 / (2 × 3.14 × 60 × 60 × 10^-6)
X_C = 1 / (0.0226) = 44.2 Ω.
RMS current I_rms = V_rms / X_C = 110 / 44.2 = 2.49 A.

Q 7.5: In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Solution:
The net power absorbed by both circuits is Zero.
Explanation: In a purely inductive circuit (Q 7.3) and a purely capacitive circuit (Q 7.4), the phase difference between current and voltage is 90° (π/2).
Since Power P = V I cos(90°) and cos(90°) = 0, the average power consumed over a complete cycle is zero.

Q 7.6: A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Solution:
Given: C = 30 μF = 30 × 10^-6 F, L = 27 mH = 27 × 10^-3 H.
Angular frequency of free oscillations ω_r = 1 / √(LC)
ω_r = 1 / √(27 × 10^-3 × 30 × 10^-6)
ω_r = 1 / √(810 × 10^-9) = 1 / (9 × 10^-4)
ω_r = 10000 / 9 = 1.11 × 10³ rad s^-1.

Q 7.7: A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Solution:
Given: R = 20 Ω, L = 1.5 H, C = 35 μF, V_rms = 200 V.
When the frequency equals the natural frequency (at resonance), X_L = X_C. Thus, the impedance Z becomes equal to the resistance R.
So, Z = R = 20 Ω.
Current I_rms = V_rms / Z = 200 / 20 = 10 A.
Average power transferred P = I_rms² × R = (10)² × 20 = 100 × 20 = 2000 W.

Q 7.8: Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 Ω.
Solution:
(Note: Based on the given data, the most common calculations are finding resonant frequency and impedance at resonance.)
Resonant angular frequency ω_r = 1 / √(LC) = 1 / √(5.0 × 80 × 10^-6)
ω_r = 1 / √(400 × 10^-6) = 1 / 0.02 = 50 rad s^-1.
At resonance, Impedance Z = R = 40 Ω.