NCERT Solutions Class 12 Physics Alternating Current
NCERT Solutions Class 12 Physics Alternating Current Chapter 7 is a part of NCERT Solutions Class 12 Physics. Here we have given NCERT Solutions Class 12 Physics Alternating Current Chapter 7 which will help you to understand and for preparation of CBSE examination and other competitive examinations. Prepare for exams with NCERT Solutions Class 12 Physics Alternating Current Chapter 7 and refer to your friends also.
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NCERT Solutions Class 12 Physics Electromagnetic Induction
Question : 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?
Show Answer :
Answer :
Given:
Resistance of the resistor, R = 100 Ω
Supply voltage, V = 220 V
Frequency, ν = 50 Hz
(a) The rms value of current in the circuit is given as
I= (V/R)
= (220)/ (100)
= 2.20 A
(b) The net power consumed over a full cycle is given as:
P = VI
= 220 × 2.2 = 484 W
Question : 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?
Show Answer :
Answer :
- Peak voltage of the ac supply, Vo = 300 V
Rms voltage is given as:
V = (Vo/√2)
= (300 /√2)
= 212.1 V
- The rms value of current is given as:
I = 10 A
Now, peak current is given as:
Io = √2I
= (√2 × 10) = 14.1 A
Question : 3
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Show Answer :
Answer :
Given:
Inductance of inductor, L = 44 mH = 44 × 10−3 H
Supply voltage, V = 220 V
Frequency, ν = 50 Hz
Angular frequency, ω = 2πν
Inductive reactance, XL = ω L
= (2πν) L
= (2π × 50 × 44 × 10−3) Ω
RMS value of current is given as:
I = (V/XL)
= (220)/ (2π × 50 × 44 × 10−3) = 15.92 A
Hence, the rms value of current in the circuit is 15.92 A.
Question : 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.
Show Answer :
Answer :
Given:
Capacitance of capacitor, C = 60 μF = 60 × 10−6 F
Supply voltage, V = 110 V
Frequency, ν = 60 Hz
Angular frequency, ω = 2πν
Capacitive reactance, XC = (1/ωC)
= (1)/ (2πνC)
= (1)/ (2π × 60 × 60 × 10−6) Ω
RMS value of current is given as:
I = (V)/ (XC)
= (220) / (2π × 60 × 60 × 10−6) = 2.49 A
Hence, the rms value of current is 2.49 A.
Mddle block 1
NCERT Solutions Class 12 Physics Electromagnetic Induction
Question : 5
In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Show Answer :
Answer :
In the inductive circuit,
Given:
Rms value of current, I = 15.92 A
Rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos Φ
Where,
Φ = Phase difference between V and I.
For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit,
Given:
RMS value of current, I = 2.49 A
RMS value of voltage, V = 110 V
Hence, the net power absorbed can be obtained as:
P = VI Cos Φ
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ = 90°.
Hence, P = 0 i.e., the net power is zero.
Question : 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?
Show Answer :
Answer :
Given:
Inductance, L = 2.0 H
Capacitance, C = 32 μF = 32 × 10−6 F
Resistance, R = 10 Ω
Resonant frequency is given by the relation,
ωr=(1√LC)
= (1)/ (√2 × 32 × 10−6)
= (1) / (8 × 10−3) = 125 rad/s.
Now, Q-value of the circuit is given as:
Q = (1/R) (√ (L/C))
= (1/10) (√ (2)/ (32 × 10−6))
= (1) / (10 × 4 × 10−3) = 25
Hence, the Q-Value of this circuit is 25.
Question : 7
A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Show Answer :
Answer :
Given:
Capacitance, C = 30μF
= (30×10−6) F
Inductance, L = 27 mH = 27 × 10−3 H
Angular frequency is given as:
ωr = (1)/(√LC)
= (1/)/ (√27 × 10−3 × 30 × 10−6)
= (1)/ (9 × 10−4)
= 1.11 × 103 rad/s
Hence, the angular frequency of free oscillations of the circuit is:
(1.11 × 103) rad/s.
Question : 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?
Show Answer :
Answer :
Given:
Capacitance of the capacitor, C = 30 μF = (30×10−6) F
Inductance of the inductor, L = 27 mH = (27 × 10−3) H
Charge on the capacitor, Q = 6 mC = (6 × 10−3) C
Total energy stored in the capacitor can be calculated as:
E = (1/2) Q2C
= (1/2) (6 × 10−3)2 / (30 × 10−6)
= (6/10)
E = 0.6 J
Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.
NCERT Solutions Class 12 Physics Electromagnetic Induction
Question : 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?
Show Answer :
Answer :
At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.
Given:
Resistance, R = 20 Ω
Inductance, L = 1.5 H
Capacitance, C = 35 μF = (30 × 10−6) F
AC supply voltage to the LCR circuit, V = 200 V
Impedance of the circuit is given by the relation,
Z = (√R2 + (XL − XC) 2)
At resonance, XL = XC
∴ Z = R = 20 Ω
Current in the circuit can be calculated as:
I = (V/Z) = (200/20) = 10 A
Hence, the average power transferred to the circuit in one complete cycle:
VI = (200 × 10) = 2000 W.
Question : 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.]
Show Answer :
Answer :
The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.
Given:
Lower tuning frequency, ν1 = 800 kHz = 800 × 103 Hz
Upper tuning frequency, ν2 = 1200 kHz = 1200 × 103 Hz
Effective inductance of circuit L = 200 μH = 200 × 10−6 H
Capacitance of variable capacitor for ν1 is given as:
C1 = (1)/ (ω12L)
Where,
ω1 = Angular frequency for capacitor C1
= 2πν1
= (2π × 800 × 103) rad/s
∴ C1 = (1)/ (2π× 800 × 103)2× 200 × 10−6)
= (1.9809 × 10−10) F = 198 pF
Capacitance of variable capacitor for ν2 is given as:
C2 = (1)/ (ω22L)
Where,
ω2 = Angular frequency for capacitor C2
= 2πν2
= (2π × 1200 × 103) rad/s
∴ C2 = (1)/ (2π× 1200 × 103)2× (200 × 10−6)
= 90.8804 × 10−10) F = 88 pF
Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.
Question : 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.
Show Answer :
Answer :
Given:
Inductance of the inductor, L = 5.0 H
Capacitance of the capacitor, C = 80 μH = 80 × 10−6 F
Resistance of the resistor, R = 40 Ω
Potential of the variable voltage source, V = 230 V
(a) Resonance angular frequency is given as:
ωr = (1)/(√LC)
= (1)/√ (5 × 80 × 10−6)
= (103/20)
ωr = 50 rad/s
Hence, the circuit will come in resonance for a source frequency of 50 rad/s.
(b) Impedance of the circuit is given by the relation:
Z = (√R2 + (XL − XC) 2)
At resonance, XL = XC⇒ Z = R = 40 Ω
Amplitude of the current at the resonating frequency is given as:
Io = (Vo/Z)
Where,
Vo = Peak voltage = √2 V
∴ Io = (√2 V)/ (Z)
= (√2 × 230)/40 = 8.13 A
Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.
(c) RMS potential drop across the inductor,
(VL)rms = I × ωrL
Where,
Irms = (Io /√2)
= (√2 V)/ (√2 Z)
= (230/40) = (23/4) A
∴ (VL) RMS = (23/4) x (50×5) = 1437.5 V
Potential drop across the capacitor:
∴ (VC) RMS = (I × (1/ωrC))
= (23/4) × ((1/50) × 80 × 10−6) = 1437.5 V
Potential drop across the resistor:
(VR)RMS = IR = ((23/4) × 40) = 230 V
Potential drop across the LC combination:
VLC = (I (XL − XC))
At resonance, XL = XC⇒ VLC = 0
Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.
NCERT Solutions Class 12 Physics Electromagnetic Induction
NCERT Solutions Class 12 Physics Electromagnetic Induction
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Multiple Choice Questions Class 12 Physics
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