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Additional Exercises
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? Answer 7.12: Inductance of the inductor, L = 20 mH = 20 × 10−^3 H Capacitance of the capacitor, C = 50 μF = 50 × 10−^6 F Initial charge on the capacitor, Q = 10 mC = 10 × 10−^3 C (a) Total energy stored initially in the circuit is given as:
Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.
(b) Natural frequency of the circuit is given by the relation,
Natural angular frequency,
Hence, the natural frequency of the circuit is 10 3 rad/s.
(c) (i) For time period (T ), total charge on the capacitor at time t,
For energy stored is electrical, we can write Q’ = Q. Hence, it can be inferred that the energy stored in the capacitor is completely
electrical at time, t =
(ii) Magnetic energy is the maximum when electrical energy, Q′ is equal to
Hence, it can be inferred that the energy stored in the capacitor is completely
magnetic at time, (d) Q^1 = Charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time t.
Answer 7.13: Inductance of the inductor, L = 0.50 H Resistance of the resistor, R = 100 Ω Potential of the supply voltage, V = 240 V Frequency of the supply, ν = 50 Hz (a) Peak voltage is given as:
Angular frequency of the supply, ω = 2 πν = 2π × 50 = 100 π rad/s Maximum current in the circuit is given as:
(b) Equation for voltage is given as: V = V 0 cos ωt Equation for current is given as: I = I 0 cos (ωt − Φ) Where, Φ = Phase difference between voltage and current At time, t = 0. V = V 0 (voltage is maximum)
For ωt − Φ = 0 i.e., at time , I = I 0 (current is maximum)
Hence, the time lag between maximum voltage and maximum current is.
Now, phase angle Φ is given by the relation,
Hence, the time lag between maximum voltage and maximum current is 3. 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 7.14: Inductance of the inductor, L = 0.5 Hz Resistance of the resistor, R = 100 Ω Potential of the supply voltages, V = 240 V Frequency of the supply, ν = 10 kHz = 10^4 Hz
Angular frequency, ω = 2πν= 2π × 10^4 rad/s
Answer 7.15: Capacitance of the capacitor, C = 100 μF = 100 × 10−^6 F Resistance of the resistor, R = 40 Ω Supply voltage, V = 110 V (a) Frequency of oscillations, ν= 60 Hz Angular frequency, For a RC circuit, we have the relation for impedance as:
𝑍 = √𝑅 2 + (^) 𝜔 21 𝐶 2
Peak voltage, 𝑉 0 = 𝑉√2 = 110√
Maximum current is given as:
(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. 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 7.16: Capacitance of the capacitor, C = 100 μF = 100 × 10−^6 F Resistance of the resistor, R = 40 Ω Supply voltage, V = 110 V Frequency of the supply, ν = 12 kHz = 12 × 10^3 Hz Angular Frequency, ω = 2 πν= 2 × π × 12 × 10303 = 24π × 10^3 rad/s
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 7.17: An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where, L = 5.0 H 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:
Where, ω = Angular frequency
Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum. rms current flowing through inductor L is given as:
At resonance,
rms current flowing through capacitor C is given as:
rms current flowing through resistor R is given as:
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’.] Answer 7.18: Inductance, L = 80 mH = 80 × 10−^3 H Capacitance, C = 60 μF = 60 × 10−^6 F
(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.
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 7.19: Average power transferred to the resistor = 788.44 W Average power transferred to the capacitor = 0 W Total power absorbed by the circuit = 788.44 W 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, ν = 50 Hz Angular frequency of signal, ω = 2πν= 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,
Average power transferred to resistance is given as: PR= I^2 R = (7.25)^2 × 15 = 788.44 W Average power transferred to capacitor, P (^) C = Average power transferred to inductor, P (^) L = 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.
Question 7.20: 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.
And, maximum current (b) Maximum average power absorbed by the circuit is given as:
(c) The power transferred to the circuit is half the power at resonant frequency. Frequencies at which power transferred is half, =
Where,
∴Resonant frequency,
Hence, resonant frequency ( ) is
Hence, change in frequency,
∴
And, Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half. At these frequencies, current amplitude can be given as:
(d) Q-factor of the given circuit can be obtained using the relation,
Hence, the Q-factor of the given circuit is 21.74.
Question 7.21: Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ω. 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 7.21: Inductance, L = 3.0 H 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:
Answer 7.22: (a) Yes; the statement is not true for rms voltage 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.
Question 7.23: 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? Answer 7.23: Input voltage, V 1 = 2300 Number of turns in primary coil, n 1 = 4000 Output voltage, V 2 = 230 V Number of turns in secondary coil = n (^2) Voltage is related to the number of turns as:
Hence, there are 400 turns in the second winding.
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 m 3 s−^1. If the turbine generator
efficiency is 60%, estimate the electric power available from the plant (g =
9.8 m s−^2 ).
Answer 7.24: Height of water pressure head, h = 300 m Volume of water flow per second, V = 100 m^3 /s
