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NCERT Class 12 Physics Chapter 7 Alternating Current
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Alternating Current
Chapter: 7
| Part – I |
EXERCISE
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?
Ans: Given:
R = 100 Ω
Vrms = 220 V
Ⅰrms = Vrms/ R
= 220 V / 100 Ω
= 2.2 A.
(b) What is the net power consumed over a full cycle?
Ans: Given:
R = 100 Ω
Vrms = 220 V
P = Ⅰrms2 × R
= (2.2)2 × 100
= 484 W.
2. (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
Ans: V = Vpeak / √2
Vpeak = 300 V
Vrms = 300 V / √2
= 300V / 1.414
= 212 V.
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Ans: Given:
Ⅰrms = 10 A
Ⅰpeak = Ⅰrms × √2
= 10 A × √2
= 10 A × 1.414
= 14.14 A.
3. A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Ans: Ⅰrms = Vrms / Z
Z = ωL
Given:
ω = 2𝜋𝑓
L = 44 mH = 44 × 10-3 H
Vrms = 220 V
𝑓 = 50 Hz
ω = 2𝜋 × 50 Hz = 100𝜋 rad/s
Z = ωL = 100 𝜋 × 44𝜋 × 10-3 H.
= 4.4𝜋 (4.4 × 22 / 7)
= 13.82.
Ⅰrms = 220V / 13.82
= 15.92 A.
4. A 60 mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Ans: Given:
Ⅰrms = Vrms / Z
Z = Xc = 1/ωC
C = 60 mF = 6 × 10-3 F.
Vrms = 110 V
𝑓 = 60 Hz
ω = 2𝜋 × 60 Hz = 120𝜋 rad/s
Xc = 1/ωC
= 1/ (120𝜋 × 60 × 10-3 F)
= 1 / 22.62
Ⅰrms = 110 V / 0.0442
= 2489 A.
5. In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Ans: The average power in an AC circuit is given by:
P = Vrms × Ⅰrms × cos Ø
In the case of an inductor, the phase difference between the current and the voltage is 90° degrees. Since cos 90°, the average power becomes:
P = Vrms × Ⅰrms × cos 90° = 0
This means that no net power is consumed by the inductor over time.
Similarly, for a capacitor, the phase difference between the current and the voltage is also 90° degrees.
Therefore, the average power is:
P = Vrms × Ⅰrms × cos 90° = 0
This indicates that, like the inductor, the capacitor also does not consume any net power over time.
6. A charged 30 mF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Ans: Given:
C = 30 × 10-6 F.
L = 27 × 10-3 H.
= 1/9 × 104 rad s-1
= 1.1 × 103 rad s-1.
7. A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 mF 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?
Ans: Given:
L = 1.5 H
C = 35 mF = 35 × 10-3 F
𝑓0 = 1 / 2𝜋√LC
Z = R = 20 Ω
P = V2rms / R
Vrms = 200 V
R = 20 Ω
P = (200)2 / 20
= 40000/20
= 2000 W.
8. Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μ F, R = 40 Ω.
(a) Determine the source frequency which drives the circuit in resonance.
Ans: L = 5.0 H
C = 80 μF = 80 × 10-6 F
R = 40 Ω
Vrms = 230 V
𝑓0 = 1/ (2𝜋√(LC))
= 1 / (2𝜋√(5.0 × 80 × 10-6))
= 7.95 Hz.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
Ans: Z = R = 40Ω
Ⅰ0 = Vrms / Z
= 230 V / 40Ω
= 5.75 A.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Ans: Resistor:
Vrms R = Ⅰ0 × R
= 5.75 A × 40Ω
= 230 V.
Inductor:
Vrms R = Ⅰ0 × XL
At resonance, XL = XC, so VrmsL = 0V
Capacitor:
Vrms R = Ⅰ0 × XC
XL = XC VrmsC = 0V.
