Source: https://www.toppersbulletin.com/ncert-solutions-for-class-12-chapter-7-alternating-current/
Timestamp: 2019-04-25 06:37:30+00:00

Document:
Q1 :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?
Q2 :(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?
Q3 :A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Hence, the rms value of current in the circuit is 15.92 A.
Q4 :A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Hence, the rms value of current is 2.49 A.
Q5 :In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
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.
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.
Q6 :Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?
Hence, the Q-Value of this circuit is 25.
Q7 :A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Hence, the angular frequency of free oscillations of the circuit is 1.11 × 103 rad/s.
Q8 :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?
Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.
Q9 :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?
At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.
= 200 × 10 = 2000 W.
Q10 :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?
The range of frequency (v) of a radio is 800 kHz to 1200 kHz.
Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.
(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.
Hence, the circuit will come in resonance for a source frequency of 50 rad/s.
Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.
Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.
Q12 :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?
(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?
Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.
Hence, the natural frequency of the circuit is 103 rad/s.
For energy stored is electrical, we can write Q’ = Q.
(ii) Magnetic energy is the maximum when electrical energy, Q ‘ is equal to 0.
(d) Q1 = Charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time t.
When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor =1/2 (maximum energy).
(e) If a resistor is inserted in the circuit, then total initial energy is dissipated as heat energy in the circuit. The resistance damps out the LC oscillation.
Q13 :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?
At time, t = 0.
Hence, the time lag between maximum voltage and maximum current is Φ/ω.
Hence, the time lag between maximum voltage and maximum current is 3.2 ms.
Q14 :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?
It can be observed that I0 is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.
In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.
Q15 :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?
Hence, the time lag between maximum current and maximum voltage is 1.55 ms.
Q16 :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.
Hence, Φ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor.
In a dc circuit, after the steady state is achieved, ω = 0. Hence, capacitor C amounts to an open circuit.
Q17 :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.
Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum.
Q18 :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?
(c) Average power consumed by the inductor is zero as actual voltage leads the current by π/2.
(d) Average power consumed by the capacitor is zero as voltage lags current by π/2.
(e) The total power absorbed (averaged over one cycle) is zero.
Q19 :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.
Hence, the total power absorbed by the circuit is 788.44 W.
Q20 :A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω 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?
(c) The power transferred to the circuit is half the power at resonant frequency.
Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half.
Hence, the Q-factor of the given circuit is 21.74.
Q21 :Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4Ω. 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.
(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?
It is true that in any ac circuit, the applied voltage is equal to the average sum of the instantaneous voltages across the series elements of the circuit. However, this is not true for rms voltage because voltages across different elements may not be in phase.
(b) High induced voltage is used to charge the capacitor.
A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.
(c) The dc signal will appear across capacitor C because for dc signals, the impedance of an inductor (L) is negligible while the impedance of a capacitor (C) is very high (almost infinite). Hence, a dc signal appears across C. For an ac signal of high frequency, the impedance of L is high and that of C is very low. Hence, an ac signal of high frequency appears across L.
(d) If an iron core is inserted in the choke coil (which is in series with a lamp connected to the ac line), then the lamp will glow dimly. This is because the choke coil and the iron core increase the impedance of the circuit.
(e) A choke coil is needed in the use of fluorescent tubes with ac mains because it reduces the voltage across the tube without wasting much power. An ordinary resistor cannot be used instead of a choke coil for this purpose because it wastes power in the form of heat.
Q23 :A power transmission line feeds input power at 2300 V to a stepdown 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?
Hence, there are 400 turns in the second winding.
Q24 :At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3 s-1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g= 9.8 m s-2).
Q25 :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 Ω 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 step-down transformer of rating 4000 – 220 V is used in the sub-station.
Also, the power generated is 440 V.
Hence, the rating of the step-up transformer situated at the power plant is 440 V – 7000 V.
Q26 :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?
The rating of a step-down transformer is 40000 V – 220 V.
(b) Assuming that the power loss is negligible due to the leakage of current.
The power is being generated in the plant at 440 V.
440 V – 40300 V.
In the previous exercise, the power loss due to the same reason is . Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.

References: V. 
 V. 
 V.

 V.

 V.

 V.

 V.