NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 7: Alternating Current
NCERT Solutions for Class 12th Physics, Chapter 7: Alternating Current available here. You can also download these solutions with the help of download link given at the end of this article. Some important topics of this chapter are AC voltage applied to a resistor, representation of AC current and voltage by rotating vectors (Phasor), AC voltage applied to an inductor, AC voltage applied to a capacitor, AC voltage applied to a series LCR circuit, Resonance, Sharpness of resonance, LC oscillations, Transformers. Most of the questions given in this chapter of NCERT textbook are based on these topics.
NCERT Solutions for Class 12 Physics, Chapter 7: Alternating Current are given below:
Question7.1: 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?
R = 100 ohms
V = 220 V
f = 50 Hz
(a) We know
Irms = Vrms / R
Substituting the values
Irms = 220 / 100 = 2.2 A
(b) Power = V.I
Or Power = 220 x 2.2
Or Power = 484 W
(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?
(a) We know
Vrms = Vpeak / 1.414
Vrms = 300 / 1.414
Or Vrms = 212.13 V
(b) Using above identity for current
Ipeak = 1.414 x Irms
Or Ipeak = 1.414 x 10 = 14.14 A.
Question7.3. A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
L = 44 mH
V = 220 V
f = 50 Hz
Irms is given by = V / XL
Determining inductive reactance
XL = 2 × 3.14 × 50 × 44 × 10‒3
XL = 13.82 ohms
Irms = 220 / 13.82
Or Irms = 15.92 A.
Question7.4: A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
C = 60 microfarads
V = 110 volts
f = 60 hertzs
Irms = V / Xc
Xc = 1 / (2 x 3.14 x 60 x 60 x 10-6)
Xc = 44.248 ohms
Irms = 110 / 44.248 = 2.488 A.
Question7.5: In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Zero. Power is absorbed only by resistance in the circuit.
Question7.6: Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?
L = 2 H
C = 32 microF
R = 10 ohms
Resonant frequency, ωr = 1/(LC)1/2
ωr = 125 s‒1
Now, Q-value = (ωrL)/R
Putting the desired values gives us
Q-value = 25.
Question7.7: A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
C = 30 microF
L = 27 mH
Angular frequency of free oscillations = 1/(LC)1/2
Angular frequency = 1111.11 s‒1.
Question7.8: Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Initial energy, Ui = (qm)2/2C
Ui = 0.6 J
The energy will remain constant at all times.
Question7.9: 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?
R = 20 ohms
L = 1.5 henries
C = 35 micro farads
V = 200 volts
Natural frequency = 1/(LC)1/2 = 138 s‒1.
At natural frequency,
Z = R
So, I = V / R = 200 / 20 = 10 A
P = I2R
Or P = 10 × 10 × 20 = 2000 W.
Question7.10: A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor? [Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.]
By using the relation, 1/(LC)1/2 = 2πf
Solving for C, C = 1/(4π2 f2 L)
For f = 800 kHz, C’ = 197.8 pF
For f = 1200 kHz, C’’= 87.9 pF
Range: 88 pF to 198 pF.
Question7.11: Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 Ω.
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(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.
V = 230 V
L = 5 H
C = 80 μF
R = 40 ohms
(a) Source frequency at resonance = 1/(LC)1/2
Putting respective values, the frequency is 50 rad/s.
(b) At resonance,
Impedance (Z) = Resistance (R)
⇒ Z = R = 40 ohms
Now rms value of current, I = V / R
Or I = 230 / 40
Hence, I = 5.75 A
Amplitude of this value of current = 1.414 × I = 1.414 × 5.75 = 8.13 A.
(c) Now taking into consideration the rms potential drops
Across Resistance, VR = IR = 5.75 × 40 = 230 V.
Across Capacitance, VC = IXC = 1437.5 V.
Across Inductance, VL = IXL = 5.75 × 50 × 5 = 1437.5 V.
Across LC combination, VLC = I (XL – XC) = 0 (at resonating frequency)