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Alternating Current
Physics XII - Chapter 07: Complete NCERT Exercise Solutions
Master Alternating Current with NCERT solutions for AC circuits, impedance, resonance, power factor, and transformers.
Question 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?
Answer & Explanation:
(a) RMS Current: \( I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{220}{100} = 2.2 \, \text{A} \)
(b) Net Power: \( P = I_{\text{rms}}^2 R = (2.2)^2 \times 100 = 484 \, \text{W} \)
Answer: (a) 2.2 A, (b) 484 W
(b) Net Power: \( P = I_{\text{rms}}^2 R = (2.2)^2 \times 100 = 484 \, \text{W} \)
Answer: (a) 2.2 A, (b) 484 W
Question 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?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Answer & Explanation:
(a) RMS Voltage: \( V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{300}{\sqrt{2}} \approx 212.1 \, \text{V} \)
(b) Peak Current: \( I_0 = I_{\text{rms}} \times \sqrt{2} = 10 \times \sqrt{2} \approx 14.14 \, \text{A} \)
Answer: (a) ~212.1 V, (b) ~14.14 A
(b) Peak Current: \( I_0 = I_{\text{rms}} \times \sqrt{2} = 10 \times \sqrt{2} \approx 14.14 \, \text{A} \)
Answer: (a) ~212.1 V, (b) ~14.14 A
Question 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.
Answer & Explanation:
Inductive reactance: \( X_L = 2 \pi f L = 2 \pi \times 50 \times 44 \times 10^{-3} \approx 13.82 \, \Omega \)
RMS Current: \( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L} = \frac{220}{13.82} \approx 15.92 \, \text{A} \)
Answer: ~15.92 A
RMS Current: \( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L} = \frac{220}{13.82} \approx 15.92 \, \text{A} \)
Answer: ~15.92 A
Question 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.
Answer & Explanation:
Capacitive reactance: \( X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 60 \times 60 \times 10^{-6}} \approx 44.21 \, \Omega \)
RMS Current: \( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = \frac{110}{44.21} \approx 2.49 \, \text{A} \)
Answer: ~2.49 A
RMS Current: \( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = \frac{110}{44.21} \approx 2.49 \, \text{A} \)
Answer: ~2.49 A
Question 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.
Answer & Explanation:
For pure inductor and pure capacitor:
Power absorbed over a full cycle = 0.
Reason: Phase difference between voltage and current is \( \pi/2 \), so average power \( P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi = 0 \).
Answer: Zero in both cases.
Power absorbed over a full cycle = 0.
Reason: Phase difference between voltage and current is \( \pi/2 \), so average power \( P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi = 0 \).
Answer: Zero in both cases.
Question 7.6
Obtain the resonant frequency \(\omega_r\) of a series LCR circuit with \(L = 2.0 \text{H}\), \(C = 32 \, \mu\text{F}\) and \(R = 10 \, \Omega\). What is the Q‑value of this circuit?
Answer & Explanation:
Resonant angular frequency: \( \omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2.0 \times 32 \times 10^{-6}}} \approx 125 \, \text{rad/s} \)
Q‑factor: \( Q = \frac{\omega_r L}{R} = \frac{125 \times 2.0}{10} = 25 \)
Answer: \( \omega_r \approx 125 \, \text{rad/s} \), Q = 25
Q‑factor: \( Q = \frac{\omega_r L}{R} = \frac{125 \times 2.0}{10} = 25 \)
Answer: \( \omega_r \approx 125 \, \text{rad/s} \), Q = 25
Question 7.7
A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer & Explanation:
Angular frequency: \( \omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}} \approx 1.11 \times 10^3 \, \text{rad/s} \)
Answer: \( \approx 1.11 \times 10^3 \, \text{rad/s} \)
Answer: \( \approx 1.11 \times 10^3 \, \text{rad/s} \)
Question 7.8
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?
Answer & Explanation:
Initial energy: \( U = \frac{Q^2}{2C} = \frac{(6 \times 10^{-3})^2}{2 \times 30 \times 10^{-6}} = 0.6 \, \text{J} \)
Later time: In ideal LC circuit, total energy remains constant (0.6 J) as it oscillates between C and L.
Answer: Initially 0.6 J, constant thereafter.
Later time: In ideal LC circuit, total energy remains constant (0.6 J) as it oscillates between C and L.
Answer: Initially 0.6 J, constant thereafter.
Question 7.9
A series LCR circuit with \(R = 20 \, \Omega\), \(L = 1.5 \, \text{H}\) and \(C = 35 \, \mu\text{F}\) 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?
Answer & Explanation:
At resonance, \( Z = R \).
\( I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{200}{20} = 10 \, \text{A} \)
Average power: \( P_{\text{avg}} = I_{\text{rms}}^2 R = 10^2 \times 20 = 2000 \, \text{W} \)
Answer: 2000 W
\( I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{200}{20} = 10 \, \text{A} \)
Average power: \( P_{\text{avg}} = I_{\text{rms}}^2 R = 10^2 \times 20 = 2000 \, \text{W} \)
Answer: 2000 W
Question 7.10
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 radiowave.]
[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 radiowave.]
Answer & Explanation:
Formula: \( f = \frac{1}{2\pi\sqrt{LC}} \) → \( C = \frac{1}{(2\pi f)^2 L} \)
For \( f_1 = 800 \, \text{kHz} \): \( C_1 \approx 198.9 \, \text{pF} \)
For \( f_2 = 1200 \, \text{kHz} \): \( C_2 \approx 88.4 \, \text{pF} \)
Answer: ~88.4 pF to ~198.9 pF
For \( f_1 = 800 \, \text{kHz} \): \( C_1 \approx 198.9 \, \text{pF} \)
For \( f_2 = 1200 \, \text{kHz} \): \( C_2 \approx 88.4 \, \text{pF} \)
Answer: ~88.4 pF to ~198.9 pF
Question 7.11
Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. \(L = 5.0 \, \text{H}\), \(C = 80 \, \mu\text{F}\), \(R = 40 \, \Omega\).
(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.
(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.
Answer & Explanation:
(a) \( f_r = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{5 \times 80 \times 10^{-6}}} \approx 7.96 \, \text{Hz} \)
(b) At resonance, \( Z = R = 40 \, \Omega \)
Current amplitude \( I_0 = \frac{V_0}{R} = \frac{230\sqrt{2}}{40} \approx 8.13 \, \text{A} \)
(c) \( V_R = I_{\text{rms}} R = \frac{230}{40} \times 40 = 230 \, \text{V} \)
\( V_L = I_{\text{rms}} X_L \), \( V_C = I_{\text{rms}} X_C \) → At resonance \( X_L = X_C \), so \( V_L = V_C \) and they cancel.
Answer: (a) ~7.96 Hz, (b) Z=40 Ω, \( I_0 \approx 8.13 \, \text{A} \), (c) \( V_R=230 \, \text{V} \), \( V_L=V_C \) (equal and opposite).
(b) At resonance, \( Z = R = 40 \, \Omega \)
Current amplitude \( I_0 = \frac{V_0}{R} = \frac{230\sqrt{2}}{40} \approx 8.13 \, \text{A} \)
(c) \( V_R = I_{\text{rms}} R = \frac{230}{40} \times 40 = 230 \, \text{V} \)
\( V_L = I_{\text{rms}} X_L \), \( V_C = I_{\text{rms}} X_C \) → At resonance \( X_L = X_C \), so \( V_L = V_C \) and they cancel.
Answer: (a) ~7.96 Hz, (b) Z=40 Ω, \( I_0 \approx 8.13 \, \text{A} \), (c) \( V_R=230 \, \text{V} \), \( V_L=V_C \) (equal and opposite).
Question 7.12
An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be \( t = 0 \).
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored
(i) completely electrical (i.e., stored in the capacitor)? (ii) completely magnetic (i.e., stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and the capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored
(i) completely electrical (i.e., stored in the capacitor)? (ii) completely magnetic (i.e., stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and the capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Answer & Explanation:
(a) \( U = \frac{Q^2}{2C} = \frac{(0.01)^2}{2 \times 50 \times 10^{-6}} = 1 \, \text{J} \); Yes, conserved in ideal LC.
(b) \( f = \frac{1}{2\pi\sqrt{LC}} \approx 159.2 \, \text{Hz} \)
(c) (i) \( t = 0, T/2, T, \dots \) (ii) \( t = T/4, 3T/4, \dots \)
(d) When \( U_C = U_L = U/2 \), occurs at \( t = T/8, 3T/8, 5T/8, \dots \)
(e) Entire initial energy (1 J) is eventually dissipated as heat.
Answer: As above.
(b) \( f = \frac{1}{2\pi\sqrt{LC}} \approx 159.2 \, \text{Hz} \)
(c) (i) \( t = 0, T/2, T, \dots \) (ii) \( t = T/4, 3T/4, \dots \)
(d) When \( U_C = U_L = U/2 \), occurs at \( t = T/8, 3T/8, 5T/8, \dots \)
(e) Entire initial energy (1 J) is eventually dissipated as heat.
Answer: As above.
Question 7.13
A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
Answer & Explanation:
(a) Impedance \( Z = \sqrt{R^2 + (2\pi f L)^2} \approx \sqrt{100^2 + (157)^2} \approx 186.2 \, \Omega \)
\( I_0 = \frac{V_0}{Z} = \frac{240\sqrt{2}}{186.2} \approx 1.82 \, \text{A} \)
(b) Phase lag \( \phi = \tan^{-1}\left(\frac{X_L}{R}\right) \approx \tan^{-1}(1.57) \approx 57.5^\circ \)
Time lag \( \Delta t = \frac{\phi}{\omega} = \frac{57.5 \times \pi/180}{2\pi \times 50} \approx 3.2 \, \text{ms} \)
Answer: (a) ~1.82 A, (b) ~3.2 ms
\( I_0 = \frac{V_0}{Z} = \frac{240\sqrt{2}}{186.2} \approx 1.82 \, \text{A} \)
(b) Phase lag \( \phi = \tan^{-1}\left(\frac{X_L}{R}\right) \approx \tan^{-1}(1.57) \approx 57.5^\circ \)
Time lag \( \Delta t = \frac{\phi}{\omega} = \frac{57.5 \times \pi/180}{2\pi \times 50} \approx 3.2 \, \text{ms} \)
Answer: (a) ~1.82 A, (b) ~3.2 ms
Question 7.14
Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Answer & Explanation:
At 10 kHz:
\( X_L = 2\pi f L = 2\pi \times 10^4 \times 0.5 \approx 31416 \, \Omega \)
\( Z \approx X_L \) (since \( X_L \gg R \))
\( I_0 \approx \frac{240\sqrt{2}}{31416} \approx 0.0108 \, \text{A} \) → very small.
Interpretation: High \( X_L \) blocks AC → open circuit. In DC steady state, inductor acts as short circuit (zero resistance).
Answer: \( I_0 \approx 0.0108 \, \text{A} \), time lag ~ small.
\( X_L = 2\pi f L = 2\pi \times 10^4 \times 0.5 \approx 31416 \, \Omega \)
\( Z \approx X_L \) (since \( X_L \gg R \))
\( I_0 \approx \frac{240\sqrt{2}}{31416} \approx 0.0108 \, \text{A} \) → very small.
Interpretation: High \( X_L \) blocks AC → open circuit. In DC steady state, inductor acts as short circuit (zero resistance).
Answer: \( I_0 \approx 0.0108 \, \text{A} \), time lag ~ small.
Question 7.15
A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Answer & Explanation:
(a) \( X_C = \frac{1}{2\pi f C} \approx 26.53 \, \Omega \)
\( Z = \sqrt{R^2 + X_C^2} \approx 47.7 \, \Omega \)
\( I_0 = \frac{V_0}{Z} = \frac{110\sqrt{2}}{47.7} \approx 3.26 \, \text{A} \)
(b) Phase lead \( \phi = \tan^{-1}\left(\frac{X_C}{R}\right) \approx 33.7^\circ \)
Time lead \( \Delta t = \frac{\phi}{\omega} \approx 1.56 \, \text{ms} \)
Answer: (a) ~3.26 A, (b) ~1.56 ms
\( Z = \sqrt{R^2 + X_C^2} \approx 47.7 \, \Omega \)
\( I_0 = \frac{V_0}{Z} = \frac{110\sqrt{2}}{47.7} \approx 3.26 \, \text{A} \)
(b) Phase lead \( \phi = \tan^{-1}\left(\frac{X_C}{R}\right) \approx 33.7^\circ \)
Time lead \( \Delta t = \frac{\phi}{\omega} \approx 1.56 \, \text{ms} \)
Answer: (a) ~3.26 A, (b) ~1.56 ms
Question 7.16
Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
Answer & Explanation:
At 12 kHz:
\( X_C = \frac{1}{2\pi \times 12000 \times 100 \times 10^{-6}} \approx 0.133 \, \Omega \)
\( Z \approx R = 40 \, \Omega \)
\( I_0 \approx \frac{110\sqrt{2}}{40} \approx 3.89 \, \text{A} \)
Interpretation: Very low \( X_C \) at high frequency → capacitor acts like short circuit (conductor). In DC steady state, capacitor acts as open circuit.
Answer: \( I_0 \approx 3.89 \, \text{A} \), phase lag negligible.
\( X_C = \frac{1}{2\pi \times 12000 \times 100 \times 10^{-6}} \approx 0.133 \, \Omega \)
\( Z \approx R = 40 \, \Omega \)
\( I_0 \approx \frac{110\sqrt{2}}{40} \approx 3.89 \, \text{A} \)
Interpretation: Very low \( X_C \) at high frequency → capacitor acts like short circuit (conductor). In DC steady state, capacitor acts as open circuit.
Answer: \( I_0 \approx 3.89 \, \text{A} \), phase lag negligible.
Question 7.17
Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.
Answer & Explanation:
At resonance \( X_L = X_C \), so currents in L and C branches are equal and opposite → cancel each other in total current.
Net current = current through R only → minimum.
For Ex 7.11: \( I_R = \frac{230}{40} = 5.75 \, \text{A} \), \( I_L = I_C = \frac{230}{X_L} \) with \( X_L = 2\pi f_r L \).
Answer: Minimum total current; \( I_R=5.75 \, \text{A} \), \( I_L=I_C \) (calculated).
Net current = current through R only → minimum.
For Ex 7.11: \( I_R = \frac{230}{40} = 5.75 \, \text{A} \), \( I_L = I_C = \frac{230}{X_L} \) with \( X_L = 2\pi f_r L \).
Answer: Minimum total current; \( I_R=5.75 \, \text{A} \), \( I_L=I_C \) (calculated).
Question 7.18
A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit? [Average implies 'averaged over one cycle'.]
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit? [Average implies 'averaged over one cycle'.]
Answer & Explanation:
(a) \( X_L = 25.13 \, \Omega \), \( X_C = 53.05 \, \Omega \), net \( X = |X_L - X_C| = 27.92 \, \Omega \)
\( I_0 = \frac{V_0}{X} = \frac{230\sqrt{2}}{27.92} \approx 11.65 \, \text{A} \), \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \approx 8.24 \, \text{A} \)
(b) \( V_L = I_{\text{rms}} X_L \approx 207 \, \text{V} \), \( V_C = I_{\text{rms}} X_C \approx 437 \, \text{V} \)
(c) & (d) Zero (pure L or C)
(e) Zero
Answer: (a) \( I_0 \approx 11.65 \, \text{A} \), (b) \( V_L \approx 207 \, \text{V} \), \( V_C \approx 437 \, \text{V} \), (c)–(e) 0.
\( I_0 = \frac{V_0}{X} = \frac{230\sqrt{2}}{27.92} \approx 11.65 \, \text{A} \), \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \approx 8.24 \, \text{A} \)
(b) \( V_L = I_{\text{rms}} X_L \approx 207 \, \text{V} \), \( V_C = I_{\text{rms}} X_C \approx 437 \, \text{V} \)
(c) & (d) Zero (pure L or C)
(e) Zero
Answer: (a) \( I_0 \approx 11.65 \, \text{A} \), (b) \( V_L \approx 207 \, \text{V} \), \( V_C \approx 437 \, \text{V} \), (c)–(e) 0.
Question 7.19
Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Answer & Explanation:
Now \( Z = \sqrt{R^2 + (X_L - X_C)^2} \approx \sqrt{15^2 + 27.92^2} \approx 31.6 \, \Omega \)
\( I_{\text{rms}} = \frac{230}{31.6} \approx 7.28 \, \text{A} \)
Power in R: \( P_R = I_{\text{rms}}^2 R \approx 795 \, \text{W} \)
Power in L and C: 0
Total power = 795 W
Answer: \( P_R \approx 795 \, \text{W} \), others 0, total 795 W.
\( I_{\text{rms}} = \frac{230}{31.6} \approx 7.28 \, \text{A} \)
Power in R: \( P_R = I_{\text{rms}}^2 R \approx 795 \, \text{W} \)
Power in L and C: 0
Total power = 795 W
Answer: \( P_R \approx 795 \, \text{W} \), others 0, total 795 W.
Question 7.20
A series LCR circuit with \(L = 0.12 \, \text{H}\), \(C = 480 \, \text{nF}\), \(R = 23 \, \Omega\) is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the \(Q\)-factor of the given circuit?
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the \(Q\)-factor of the given circuit?
Answer & Explanation:
(a) \( f_r = \frac{1}{2\pi\sqrt{LC}} \approx 663 \, \text{Hz} \), \( I_{\text{max}} = \frac{V_{\text{rms}}\sqrt{2}}{R} \approx 14.14 \, \text{A} \)
(b) Power max at resonance, \( P_{\text{max}} = \frac{V_{\text{rms}}^2}{R} \approx 2300 \, \text{W} \)
(c) Half‑power frequencies: \( f = f_r \pm \frac{R}{4\pi L} \) → calculate.
Current amplitude = \( \frac{I_{\text{max}}}{\sqrt{2}} \approx 10 \, \text{A} \)
(d) \( Q = \frac{\omega_r L}{R} \approx 21.7 \)
Answer: As above.
(b) Power max at resonance, \( P_{\text{max}} = \frac{V_{\text{rms}}^2}{R} \approx 2300 \, \text{W} \)
(c) Half‑power frequencies: \( f = f_r \pm \frac{R}{4\pi L} \) → calculate.
Current amplitude = \( \frac{I_{\text{max}}}{\sqrt{2}} \approx 10 \, \text{A} \)
(d) \( Q = \frac{\omega_r L}{R} \approx 21.7 \)
Answer: As above.
Question 7.21
Obtain the resonant frequency and \(Q\)-factor of a series LCR circuit with \(L = 3.0 \, \text{H}\), \(C = 27 \, \mu\text{F}\), and \(R = 7.4 \, \Omega\). It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
Answer & Explanation:
\( f_r = \frac{1}{2\pi\sqrt{LC}} \approx 17.7 \, \text{Hz} \)
\( Q = \frac{\omega_r L}{R} \approx 45.1 \)
To reduce FWHM by half → double Q → reduce R by half (if possible) or increase L/decrease C.
Answer: \( f_r \approx 17.7 \, \text{Hz} \), \( Q \approx 45.1 \); reduce R to ~3.7 Ω.
\( Q = \frac{\omega_r L}{R} \approx 45.1 \)
To reduce FWHM by half → double Q → reduce R by half (if possible) or increase L/decrease C.
Answer: \( f_r \approx 17.7 \, \text{Hz} \), \( Q \approx 45.1 \); reduce R to ~3.7 Ω.
Question 7.22
Answer the following questions:
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across \(C\) and the ac signal across \(L\).
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across \(C\) and the ac signal across \(L\).
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
Answer & Explanation:
(a) Instantaneous: Yes (Kirchhoff’s loop rule). RMS: No (due to phase differences).
(b) Capacitor interrupts DC quickly → high induced emf in secondary.
(c) DC: L acts as short, C blocks → voltage across C. AC: C short at high f, L blocks → voltage across L.
(d) DC: iron core increases L but DC resistance unchanged → brightness same. AC: increased L reduces current → lamp dims.
(e) Choke limits current without power loss (low resistance). Resistor would waste power as heat.
Answer: Conceptual.
(b) Capacitor interrupts DC quickly → high induced emf in secondary.
(c) DC: L acts as short, C blocks → voltage across C. AC: C short at high f, L blocks → voltage across L.
(d) DC: iron core increases L but DC resistance unchanged → brightness same. AC: increased L reduces current → lamp dims.
(e) Choke limits current without power loss (low resistance). Resistor would waste power as heat.
Answer: Conceptual.
Question 7.23
A power transmission line feeds input power at 2300 V to a step‑down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
Answer & Explanation:
Transformer relation: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
\( N_s = N_p \times \frac{V_s}{V_p} = 4000 \times \frac{230}{2300} = 400 \)
Answer: 400 turns
\( N_s = N_p \times \frac{V_s}{V_p} = 4000 \times \frac{230}{2300} = 400 \)
Answer: 400 turns
Question 7.24
At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 \(\text{m}^3\text{s}^{-1}\). If the turbine generator efficiency is 60%, estimate the electric power available from the plant (\(g = 9.8 \, \text{ms}^{-2}\)).
Answer & Explanation:
Hydraulic power: \( P_{\text{hyd}} = \rho g h Q \)
\( = 1000 \times 9.8 \times 300 \times 100 = 2.94 \times 10^8 \, \text{W} \)
Electric power: \( P_{\text{elec}} = \eta \times P_{\text{hyd}} = 0.6 \times 2.94 \times 10^8 = 1.764 \times 10^8 \, \text{W} \)
Answer: ~176.4 MW
\( = 1000 \times 9.8 \times 300 \times 100 = 2.94 \times 10^8 \, \text{W} \)
Electric power: \( P_{\text{elec}} = \eta \times P_{\text{hyd}} = 0.6 \times 2.94 \times 10^8 = 1.764 \times 10^8 \, \text{W} \)
Answer: ~176.4 MW
Question 7.25
A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 \(\Omega\) per km. The town gets power from the line through a 4000‑220 V step‑down transformer at a sub‑station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step up transformer at the plant.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step up transformer at the plant.
Answer & Explanation:
(a) Total line resistance \( R = 0.5 \times 30 = 15 \, \Omega \) (go and return).
Current in line: \( I = \frac{P}{V} = \frac{800 \times 10^3}{4000} = 200 \, \text{A} \) (at 4000 V after step‑up).
Line loss \( = I^2 R = 200^2 \times 15 = 600,000 \, \text{W} = 600 \, \text{kW} \)
(b) Plant supply = Town demand + Loss = 800 + 600 = 1400 kW.
(c) Step‑up transformer at plant: 440 V to 4000 V.
Answer: (a) 600 kW, (b) 1400 kW, (c) 440 V → 4000 V.
Current in line: \( I = \frac{P}{V} = \frac{800 \times 10^3}{4000} = 200 \, \text{A} \) (at 4000 V after step‑up).
Line loss \( = I^2 R = 200^2 \times 15 = 600,000 \, \text{W} = 600 \, \text{kW} \)
(b) Plant supply = Town demand + Loss = 800 + 600 = 1400 kW.
(c) Step‑up transformer at plant: 440 V to 4000 V.
Answer: (a) 600 kW, (b) 1400 kW, (c) 440 V → 4000 V.
Question 7.26
Do the same exercise as above with the replacement of the earlier transformer by a 40,000‑220 V step‑down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?
Answer & Explanation:
Now transmission voltage = 40,000 V.
Current in line \( I = \frac{800 \times 10^3}{40,000} = 20 \, \text{A} \)
Line loss \( = 20^2 \times 15 = 6,000 \, \text{W} = 6 \, \text{kW} \)
Plant supply = 800 + 6 = 806 kW.
Explanation: Higher voltage drastically reduces current, hence \( I^2R \) losses become very small → efficient transmission.
Answer: Loss = 6 kW, Supply = 806 kW.
Current in line \( I = \frac{800 \times 10^3}{40,000} = 20 \, \text{A} \)
Line loss \( = 20^2 \times 15 = 6,000 \, \text{W} = 6 \, \text{kW} \)
Plant supply = 800 + 6 = 806 kW.
Explanation: Higher voltage drastically reduces current, hence \( I^2R \) losses become very small → efficient transmission.
Answer: Loss = 6 kW, Supply = 806 kW.
📘 Exam Preparation Tip:
These exercise questions will help you understand AC circuit analysis and power transmission. You'll learn to analyze RLC circuits with AC sources, calculate impedance and phase relationships, determine resonant frequencies, understand power factor concepts, and solve problems involving transformers (step-up/step-down). Essential for understanding electrical power systems and household electricity