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Free NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current solved by Expert Teachers as per NCERT (CBSE) Book guidelines and brought to you by Toppers Bulletin. These Alternating Current Exercise Questions with Solutions for Class 12 Physics covers all questions of Chapter Alternating Current 12 and help you to revise complete Syllabus and Score More marks as per CBSE Board guidelines from the latest NCERT book for class 12 Physics. You can read and download NCERT Book Solution to get a better understanding of all topics and concepts 7.1 – INTRODUCTION 7.2 – AC VOLTAGE APPLIED TO A RESISTOR 7.3 – REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS 7.4 – AC VOLTAGE APPLIED TO AN INDUCTOR 7.5 – AC VOLTAGE APPLIED TO A CAPACITOR 7.6 – AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT 7.6.1 – Phasor-diagram solution 7.6.2 – Analytical solution 7.6.3 – Resonance 7.8 – LC OSCILLATIONS 7.9 – TRANSFORMERS. Alternating Current NCERT Solutions – Class 12 PhysicsExercise : Solutions of Questions on Page Number : 266 (a) The rms value of current in the circuit is given as:  (b) The net power consumed over a full cycle is given as: P = VI = 220 × 2.2 = 484 W Q2 :(a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) Therms value of current is given as: I = 10 A Now, peak current is given as:
Q3 :A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. Inductive reactance, XL = ωL
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. Capacitive reactance
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. Hence, P = 0 i.e., the net power is zero. 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? Capacitance, C = 32 μF = 32 × 10-6 F Resistance, R = 10 Ω Resonant frequency is given by the relation,
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? 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,
At resonance,
= 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? Lower tuning frequency, v1 = 800 kHz = 800 × 103 Hz Upper tuning frequency, v2 = 1200 kHz = 1200× 103 Hz Effective inductance of circuit L = 200 μH = 200 × 10-6 H Capacitance of variable capacitor for v1 is given as: C1
ω1 = Angular frequency for capacitor C1 Capacitance of variable capacitor for v2, C2
ω2 = Angular frequency for capacitor C2 Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF. Q11 :Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40Ω 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: (b) Impedance of the circuit is given by the relation,
Amplitude of the current at the resonating frequency is given as: V0 = Peak voltage 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, I = rms current
At resonance, ∴VLC= 0 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. (b)Natural frequency of the circuit is given by the relation,
Hence, the natural frequency of the circuit is 103 rad/s. (c) (i) For time period (T Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, t = Hence, it can be inferred that the energy stored in the capacitor is completely magnetic at time, (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).
Hence, total energy is equally shared between the inductor and the capacity at time, (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) Peak voltage is given as:
(b) Equation for voltage is given as: V = V0 cos ωt Equation for current is given as: I = I0 cos (ωt – Φ) Where, Φ = Phase difference between voltage and current At time, t = 0.V = V0(voltage is maximum) For ωt – Φ = 0 i.e., at time, t = Φ/ωI = I0 (current is maximum) Hence, the time lag between maximum voltage and maximum current is Φ/ω. Now, phase angle Φ is given by the relation,
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? Frequency of the supply,v= 10 kHz = 104 Hz Angular frequency, ω = 2πv= 2π × 104 rad/s (a) Peak voltage, Maximum current, (b) For phase differenceΦ, we have the relation: 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) Frequency of oscillations, v= 60 Hz Angular frequency,ω = 2πv = 2π x 60 rad/s For a RC circuit, we have the relation for impedance as:
Peak voltage, V0 =
(b) In a capacitor circuit, the voltage lags behind the current by a phase angle of Φ. This angle is given by the relation: 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. Frequency of the supply, v = 12 kHz = 12 × 103 Hz Angular Frequency,ω = 2 πv= 2 × π × 12 × 10303 = 24π × 103 rad/s Peak voltage, Maximum current,
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. C = 80 μF = 80 × 10-6 F R = 40 Ω Potential of the voltage source, V = 230 V Impedance (Z) of the given parallel LCR circuit is given as:
At resonance,
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. Capacitance, C = 60 μF = 60 × 10-6 F Supply voltage, V = 230 V Frequency, v= 50 Hz Angular frequency,ω = 2πv= 100 π rad/sPeak voltage, V0= (a) Maximum current is given as: Amplitude of maximum current, Hence, rms value of current, (b) Potential difference across the inductor, VL=I ×ωL = 8.22 × 100 π × 80 × 10-3 = 206.61 V Potential difference across 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. Inductance of inductor, L = 80 mH = 80 × 10-3 H Capacitance of capacitor, C = 60 μF = 60 × 10-6 F Resistance of resistor, R = 15 Ω Potential of voltage supply, V = 230 V Frequency of signal, v = 50 Hz Angular frequency of signal, ω = 2πv= 2π × (50) = 100π rad/s The elements are connected in series to each other. Hence, impedance of the circuit is given as:
Current flowing in the circuit, PR= I2R = (7.25)2 × 15 = 788.44 W Average power transferred to capacitor, PC = Average power transferred to inductor, PL = 0 Total power absorbed by the circuit: = PR + PC + PL = 788.44 + 0 + 0 = 788.44 W 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. Capacitance, C = 480 nF = 480 × 10-9 F Resistance, R = 23 Ω Supply voltage, V = 230 V Peak voltage is given as:V0 = (a) Current flowing in the circuit is given by the relation, I0 = maximum at resonance At resonance, we have
ωR = Resonance angular frequency ∴Resonant frequency, And, maximum current (b) Maximum average power absorbed by the circuit is given as: Hence, resonant frequency (VR) is 663.48 Hz (c) The power transferred to the circuit is half the power at resonant frequency. Frequencies at which power transferred is half, =
Hence, change in frequency, ∴ And,
(d) Q-factor of the given circuit can be obtained using the relation, 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. Capacitance, C = 27 μF = 27 × 10-6 F Resistance, R = 7.4 Ω At resonance, angular frequency of the source for the given LCR series circuit is given as:
Resistance = Q22 :Answer the following questions: (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). Volume of water flow per second, V = 100 m3/s Efficiency of turbine generator, n = 60% = 0.6Acceleration due to gravity, g = 9.8 m/s2 Density of water,ρ= 103 kg/m3 Electric power available from the plant = η x hρgV = 0.6 x 300 x 103 x 9.8 x 100 = 176.4 x 106 W = 176.4 MW 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. Input voltage, V1 = 4000 V Output voltage, V2 = 220 V Rms current in the wire lines is given as:
(a) Line power loss = I2R = (200)2 × 15 = 600 × 103 W = 600 kW (b) Assuming that the power loss is negligible due to the leakage of the current: Total power supplied by the plant = 800 kW + 600 kW = 1400 kW(c) Voltage drop in the power line = IR = 200 × 15 = 3000 V Hence, total voltage transmitted from the plant = 3000 + 4000 = 7000 V 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? Input voltage, V1 = 40000 V Output voltage, V2 = 220 V Total electric power required, P = 800 kW = 800 × 103 W Source potential, V = 220 V Voltage at which the electric plant generates power, V’ = 440 V Distance between the town and power generating station, d = 15 km Resistance of the two wire lines carrying power = 0.5 Ω/km Total resistance of the wire lines, R = (15 + 15)0.5 = 15 Ω P = V1I Rms current in the wire line is given as:
I=P/V1 (a) Line power loss = I2R = (20)2 × 15 = 6 kW (b) Assuming that the power loss is negligible due to the leakage of current. Hence, power supplied by the plant = 800 kW + 6kW = 806 kW(c) Voltage drop in the power line = IR = 20 × 15 = 300 V Hence, voltage that is transmitted by the power plant = 300 + 40000 = 40300 V The power is being generated in the plant at 440 V. Hence, the rating of the step-up transformer needed at the plant is 440 V – 40300 V.Hence, power loss during transmission = In the previous exercise, the power loss due to the same reason is |
In a circuit carrying an ideal coil with negligible resistance the power dissipated is
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