173 + Interview Questions in Resonance & Magnetically Coupled Circuit in Network Theory Page 3 InterviewSolution

101.	For the parallel resonant circuit given below, the value of the equivalent impedance of the circuit is ________(a) 56.48 kΩ(b) 78.58 kΩ(c) 89.12 kΩ(d) 26.35 kΩThis question was posed to me in an online quiz.Question is taken from Problems of Parallel Resonance Involving Quality Factor in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right CHOICE is (b) 78.58 kΩ Easiest EXPLANATION: f = \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}\) = \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\) = \(\frac{1}{799×10^{-9}}\) = 1.25 MHz. Inductive Reactance, XL = 2πfL = (6.28) (1.25×10^6)(100×10^-6) = 785.394 Ω Q = \(\frac{X_L}{R} = \frac{785.394}{7.85}\) = 100.05 ∴ ZEQ = QXL = (100.05)(785.394) = 78.58 kΩ.

Discussion

102.	For the parallel resonant circuit given below, the value of the inductive and capacitive reactance is _________(a) XL = 654.289 Ω; XC = 458.216 Ω(b) XL = 985.457 Ω; XC = 875.245 Ω(c) XL = 785.394 Ω; XC = 785.417 Ω(d) XL = 125.354 Ω; XC = 657.215 ΩThe question was posed to me in examination.The above asked question is from Problems of Parallel Resonance Involving Quality Factor topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct choice is (c) XL = 785.394 Ω; XC = 785.417 Ω To EXPLAIN: Resonant Frequency, F = \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}\) = \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\) = \(\frac{1}{799×10^{-9}}\) = 1.25 MHz. Inductive REACTANCE, XL = 2πfL = (6.28) (1.25×10^6)(100×10^-6) = 785.394 Ω Capacitive Reactance, XC = \(\frac{1}{2πfC} = \frac{1}{(6.28)(1.25×10^6)(162.11×10^{-12})}\) = \(\frac{1}{1.273×10^{-3}}\) = 785.417 Ω.

Discussion

103.	For the parallel resonant circuit shown below, the value of the resonant frequency is _________(a) 1.25 MHz(b) 2.5 MHz(c) 5 MHz(d) 1.5 MHzThis question was addressed to me by my school principal while I was bunking the class.This interesting question is from Problems of Parallel Resonance Involving Quality Factor in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT choice is (a) 1.25 MHZ The best explanation: Resonant Frequency, FR = \(\FRAC{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(100×10^{-6})(162.11×10^{-12})}}\) = \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\) = \(\frac{1}{799×10^{-9}}\) = 1.25 MHz.

Discussion

104.	A circuit with a resistor, inductor and capacitor in series is resonant with frequency f Hz. If all the component values are now doubled, then the new resonant frequency will be?(a) 2 f(b) f(c) \(\frac{f}{4}\)(d) \(\frac{f}{2}\)I got this question during a job interview.My query is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct answer is (d) \(\frac{f}{2}\) The BEST I can explain: Resonance FREQUENCY = \(\frac{1}{2π\sqrt{LC}} = \frac{1}{2π2\sqrt{LC}}\) Hence, \(\frac{f}{f_o}= \frac{\frac{1}{2π\sqrt{LC}}}{\frac{1}{2π2\sqrt{LC}}}\) = 2 ∴ f = 2fo.

Discussion

105.	In a series RLC circuit for lower frequency and for higher frequency, power factors are respectively ______________(a) Leading, Lagging(b) Lagging, Leading(c) Independent of Frequency(d) Same in both casesThe question was posed to me in class test.My enquiry is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct CHOICE is (a) Leading, Lagging The best explanation: A Leading power FACTOR MEANS that the current in the circuit leads the APPLIED voltage. This condition occurs in capacitive circuits. On the other hand, a lagging power factor indicates that the current lags the voltage and this condition happens in an INDUCTIVE circuit.

Discussion

106.	In a series RLC circuit R = 10 kΩ, L = 0.5 H and C = \(\frac{1}{250}\) μF. Calculate the frequency when the circuit is in a state of resonance.(a) 4.089 × 10^4 Hz(b) \(\frac{11111.1}{π}\)Hz(c) 4.089π × 10^4 Hz(d) \(\frac{1}{π}\) × 10^4 HzI got this question in a job interview.This interesting question is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT choice is (b) \(\FRAC{11111.1}{π}\)Hz Easiest explanation: Resonance frequency = \(\frac{1}{2π\sqrt{LC}}\) = \(\frac{1}{2π\sqrt{\frac{1}{250} × 10^{-6} × 0.5}} = \frac{11111.1}{π}\)Hz.

Discussion

107.	A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 50. If R and L are doubled and C is kept same, the new Q of the circuit is?(a) 25.52(b) 35.35(c) 45.45(d) 20.02The question was posed to me in a job interview.My doubt stems from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT answer is (b) 35.35 Easy explanation: Quality FACTOR Q of the series RLC circuit is given by, Q = \(\frac{1}{R} \sqrt{\frac{L}{C}}\) Qnew = \(\frac{1}{2R} \sqrt{\frac{2L}{C}} = \frac{1}{2} ×\frac{1}{R} \sqrt{\frac{2L}{C}} = \frac{1}{2} × \sqrt{2} × Q\) = 35.35.

Discussion

108.	For the circuit given below, if the frequency of the source is 50 Hz, then a value of to which results in a transient free response is?(a) 0(b) 1.78 ms(c) 7.23 ms(d) 9.21 msThis question was posed to me in a job interview.My question comes from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct choice is (B) 1.78 ms To explain I would SAY: For the ideal case, transient response will DIE out with time constant. T = \(\frac{L}{R} = \frac{0.01}{5}\) = 0.002 s = 2 ms Practically, T will be less than 2 ms.

Discussion

109.	In the circuit given below, the magnitudes of VL and VC are twice that of VK. Calculate the inductance of the coil, given that f = 50.50 Hz.(a) 6.41 mH(b) 5.30 mH(c) 3.18 mH(d) 2.31 mHI have been asked this question during a job interview.The origin of the question is Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» CORRECT option is (c) 3.18 mH To explain I would say: VL = VC = 2 VR ∴ Q = \(\frac{V_L}{V_R}\)= 2 But we know, Q = \(\frac{ωL}{R} = \frac{1}{ωCR}\) ∴ 2 = \(\frac{2πf × L}{5}\) Or, L = 3.18 mH.

Discussion

110.	A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, the Q factor of the circuit is?(a) 82.63(b) 47.115(c) 27.38(d) 92.38This question was addressed to me by my college professor while I was bunking the class.Question is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct ANSWER is (b) 47.115 The best explanation: We KNOW that, Q = \(\frac{ωL}{R}\) Or, Q = \(\frac{2π×50×15×10^{-3}}{100×10^{-3}}\) = 47.115.

Discussion

111.	A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, what is the value of the capacitor?(a) 676 μF(b) 575 μF(c) 591 μF(d) 610 μFI got this question during an interview for a job.The above asked question is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct answer is (a) 676 μF Explanation: We KNOW that fo = \(\frac{1}{2π\sqrt{LC}}\) Or, 50 = \(\frac{1}{2π\sqrt{5×10^{-3}×C}}\) Or, C = \(\frac{1}{4π^2×50^2×15×10^{-3}}\) ≈ 676 μF.

Discussion

112.	At what frequency will the current lead the voltage by 30° in a series circuit with R = 10Ω and C = 25 μF?(a) 11.4 kHz(b) 4.5 kHz(c) 1.1 kHz(d) 24.74 kHzI had been asked this question during an internship interview.My question is based upon Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right choice is (a) 11.4 kHz The explanation: From the IMPEDANCE diagram, 10 – jXC = Z∠-30° ∴ – XC = 10 TAN (-30°) = -5.77 Ω ∴ XC = 5.77 Ω Then, XC = \(\frac{1}{2πfC}\) or, f = \(\frac{1}{2πX_C C}\) = 1103 HZ = 1.1 kHz.

Discussion

113.	A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, the half power frequencies of the circuit are?(a) 50.53 Hz, 49.57 Hz(b) 52.12 HZ, 49.8 Hz(c) 55.02 Hz, 48.95 Hz(d) 50 HZ, 49 HzThis question was addressed to me in an interview.My enquiry is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct option is (a) 50.53 HZ, 49.57 Hz Explanation: Bandwidth, BW = \(\FRAC{f_o}{Q} = \frac{50}{47.115}\) = 1.061 Hz f2, higher half power frequency = f0 + \(\frac{BW}{2}\) ∴ f2 = 50 + \(\frac{1.061}{2}\) =50.53 Hz F1, lower half power frequency = f0 – \(\frac{BW}{2}\) ∴ f1 = 100 – \(\frac{1.59}{2}\) = 49.47 Hz.

Discussion

114.	Two series resonant filters are shown below. Let the 3 dB bandwidth of filter 1 be B1 and that of filter 2 be B2. The value of \(\frac{B_1}{B_2}\)is ____________(a) 0.25(b) 1(c) 0.5(d) 0.75I got this question by my school principal while I was bunking the class.This interesting question is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct OPTION is (a) 0.25 To elaborate: For series resonant circuit, 3dB BANDWIDTH is \(\frac{R}{L}\) B1 = \(\frac{R}{L_1}\) B2 = \(\frac{R}{L_2}= \frac{4R}{L_1}\) HENCE, \(\frac{B_1}{B_2}\) = 0.25.

Discussion

115.	Given, R = 10 Ω, L = 100 mH and C = 10 μF. Selectivity is?(a) 10(b) 1.2(c) 0.15(d) 0.1This question was posed to me in an internship interview.Query is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» CORRECT answer is (d) 0.1 To explain: SELECTIVITY = \(\FRAC{1}{Q}\) Q = \(\frac{ωL}{R} = \frac{1000×100×10^{-3}}{10}\) ∴ Q = 10 So, Selectivity = 0.1.

Discussion

116.	Given, R = 10 Ω, L = 100 mH and C = 10 μF. The values of ω0, ω1 and ω2 respectively are?(a) ω0 = 1000 rad/s, ω1 = 951.25 rad/s, ω2 = 1075.54 rad/s(b) ω0 = 1000 rad/s, ω1 = 975.25 rad/s, ω2 = 1051.25 rad/s(c) ω0 = 1000 rad/s, ω1 = 951.25 rad/s, ω2 = 1051.25 rad/s(d) ω0 = 1000 rad/s, ω1 = 825.30 rad/s, ω2 = 1075.54 rad/sI have been asked this question in an interview for internship.Question is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct OPTION is (c) ω0 = 1000 rad/s, ω1 = 951.25 rad/s, ω2 = 1051.25 rad/s To explain I would SAY: ω0 = \(\frac{1}{\sqrt{100×10^{-3}×10×10^{-6}}}\) = 1000 rad/s Now, ω1 = \(\frac{-R + \sqrt{R^2 + \frac{4L}{C}}}{2L} = \frac{-10 + \frac{\sqrt{100 + 4 × 100 × 10^{-3}}}{10×10^{-6}}}{2×100×10^{-3}}\) = 951.25 rad/s And, ω2 = \(\frac{+R+ \sqrt{R^2+ 4L/C}}{2L}\) = 1051.25 rad/s.

Discussion

117.	A series RLC circuit has R = 50 Ω, L = 0.01 H and C = 0.04 × 10^-6 F. System voltage is 100 V. The frequency, at which the maximum voltage appears across the capacitor, is?(a) 5937.81 Hz(b) 7370 Hz(c) 7937.81 Hz(d) 981 HzThis question was addressed to me in an online quiz.The origin of the question is Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT option is (C) 7937.81 Hz Explanation: Frequency FC at which VC is maximum fc = \(\frac{1}{2π}\Big[\frac{1}{0.01×0.01×10^{-6}} –\frac{50×50}{2×0.01×0.01}\Big]^{0.5}\) So, fc = 7937.81 Hz.

Discussion

118.	A series RLC circuit has R = 50 Ω, L = 0.01 H and C = 0.04 × 10^-6 F. System voltage is 100 V. At this frequency, the circuit impedance is?(a) 50 – j2.5 Ω(b) 50 – j2.8 Ω(c) 50 + j2.5 Ω(d) 50 + j2.8 ΩThe question was posed to me during an interview.My query is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct option is (a) 50 – j2.5 Ω Easy EXPLANATION: Z = R + J (ωL + \(\frac{1}{ωC}\)) = 50 + j \(\left(2π × 7937.81 × 0.01 – \frac{1}{2π×7937.81×0.04×10^{-6}}\right)\) = 50 – j2.5 Ω.

Discussion

119.	A coil (which can be modelled as a series RL circuit) has been designed for a high Quality Factor (Q) performance. The voltage is rated at and a specified frequency. If the frequency of operation is increased 10 times and the coil is operated at the same rated voltage. The new value of Q factor and the active power P will be?(a) P is doubled and Q is halved(b) P is halved and Q is doubled(c) P remains constant and Q is doubled(d) P decreases 100 times and Q is increased 10 timesThe question was posed to me during an internship interview.This question is from Problems of Series Resonance Involving Quality Factor topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» CORRECT option is (d) P DECREASES 100 times and Q is increased 10 times The best explanation: ω2 L = 10 ω1 LR will remain constant ∴ Q2 = \(\frac{10 ω_1L}{R}\) = 10 Q1 That is Q is increased 10 times. Now, I1 = \(\frac{V}{ω_1L} \) For a high Q coil, ωL >> R, I2 = \(\frac{V}{10 ω_1L} = \frac{I_1}{10}\) ∴ P2 = R \((\frac{I_1}{10})^2 = \frac{P_1}{100}\) Thus, P decreases 100 times and Q is increased 10 times.

Discussion

120.	A 50 Hz, bar primary CT has a secondary with 800 turns. The secondary supplies 7 A current into a purely resistive burden of 2 Ω. The magnetizing ampere-turns are 300. The phase angle is?(a) 3.1°(b) 85.4°(c) 94.6°(d) 175.4°This question was addressed to me in an online quiz.I'd like to ask this question from Problems of Series Resonance Involving Quality Factor in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right CHOICE is (a) 3.1° Easy explanation: Secondary burden is purely resistive and the RESISTANCE of burden is EQUAL to the resistance of the secondary winding; the resistance of secondary winding = 1Ω. The voltage induced in secondary × resistance of secondary winding = 7 × 2 = 14V. Secondary power factor is unity as the load is purely resistive. The loss component of no-load current is to be neglected i.e. Ie = 0. IM = 300 A. Secondary winding current IS = 7 A Reflected secondary winding current = n IS = 5600 A ∴ tan θ = \(\frac{I_M}{nI_S} \). So, θ = 3.1°.

Discussion

121.	A 50 μF capacitor, when connected in series with a coil having a resistance of 40Ω, resonates at 1000 Hz. The inductance of the coil for the resonant circuit is ____________(a) 2.5 mH(b) 1.2 mH(c) 0.5 mH(d) 0.1 mHThe question was asked in an international level competition.This key question is from Problems of Series Resonance Involving Quality Factor in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct choice is (c) 0.5 mH To explain: At RESONANCE, XL = XC Or, 2πfL = \(\FRAC{1}{2πfC} \) ∴ L = \(\frac{1}{4π^2 f^2 C} \) = \(\frac{1}{4π^2 × (1000)^2 × 50 × 10^{-6}} \) = \(\frac{1}{39.46×50} \) = 0.0005 So, L = 0.5 mH.

Discussion

122.	In a series resonance type BPF, C = 1.8 pf, L = 25 mH, RF = 52 Ω and RL = 9 kΩ. The Bandwidth is ___________(a) 331 kHz(b) 575 kHz(c) 331 Hz(d) 575 HzThe question was posed to me during an interview.The origin of the question is Problems of Series Resonance Involving Quality Factor in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct option is (c) 331 Hz To elaborate: R = \(\frac{R_f×R_L}{R_f+R_L}\) ≈ 52 Ω ∴ Qfactor = \(\frac{ω_r L}{R} = \frac{2π×751×25×10^{-3}}{52} \) = \(\frac{π×37580}{52}\) = 2270 Now, BW = \(\frac{f_r}{Q} = \frac{751 × 10^3}{2270}\) ∴ BW = 331 Hz.

Discussion

123.	In a series resonance type BPF, C = 1.8 pf, L = 25 mH, RF = 52 Ω and RL = 9 kΩ. The Resonant frequency f is __________(a) 75.1 kHz(b) 751 kHz(c) 575 kHz(d) 57.5 kHzThe question was posed to me in a national level competition.I want to ask this question from Problems of Series Resonance Involving Quality Factor topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT ANSWER is (b) 751 kHz For explanation: F = \(\frac{1}{2π\SQRT{LC}} = \frac{1}{2π\sqrt{0.025×1.8×10^{-12}}}\) = \(\frac{1}{2π} × \frac{10^6}{\sqrt{0.45}}\) = \(\frac{1}{2π} × \frac{10^6}{0.67}\) = 751000 ∴ f = 751 kHz.

Discussion

124.	For the series circuit given below, the value of the cut-off frequencies are ____________(a) 78.235 kHz; 16.215 kHz(b) 13.135 kHz; 81.531 kHz(c) 16.716 kHz; 15.124 kHz(d) 50.561 kHz; 18.686 kHzThe question was asked by my college director while I was bunking the class.This is a very interesting question from Problems of Series Resonance Involving Quality Factor in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right option is (c) 16.716 kHz; 15.124 kHz Best explanation: Resonant Frequency, FR = \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}\) = \(\frac{1}{6.28\sqrt{10^{-10}}}\) = \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz. Inductive REACTANCE, XL = 2πfL = (6.28) (15.92 × 10^3)(5 × 10^-6) = 0.5 kΩ ∴ QUALITY factor Q = \(\frac{X_L}{R} = \frac{0.5 kΩ}{50}\) = 10 ∴ ∆F = \(\frac{f}{Q} = \frac{15.9 kHz}{10}\) = 1.592 kHz ∴ BANDWIDTH = \(\frac{∆F}{2}\) = ±796 Hz. Therefore, f2 = f + \(\frac{∆F}{2}\) = 16.716 kHz and f1 = f – \(\frac{∆F}{2}\) = 15.124 kHz.

Discussion

125.	For the series resonant circuit given below, the bandwidth is ____________(a) ±351 Hz(b) ±796 Hz(c) ±531 Hz(d) ±225 HzThe question was asked in an online quiz.This is a very interesting question from Problems of Series Resonance Involving Quality Factor topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct option is (b) ±796 Hz Easiest explanation: Resonant Frequency, FR = \(\FRAC{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}\) = \(\frac{1}{6.28\sqrt{10^{-10}}}\) = \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz. Inductive Reactance, XL = 2πfL = (6.28) (15.92 × 10^3)(5 × 10^-6) = 0.5 kΩ ∴ Quality factor Q = \(\frac{X_L}{R} = \frac{0.5 kΩ}{50}\) = 10 ∴ ∆F = \(\frac{f}{Q} = \frac{15.9 kHz}{10}\) = 1.592 kHz ∴ Bandwidth = \(\frac{∆F}{2}\) = ±796 Hz.

Discussion

126.	For the series resonant circuit given below, the value of the quality factor is ___________(a) 15(b) 36(c) 25(d) 10This question was addressed to me by my school principal while I was bunking the class.Query is from Problems of Series Resonance Involving Quality Factor topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct choice is (d) 10 The explanation: Resonant FREQUENCY, FR = \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}\) = \(\frac{1}{6.28\sqrt{10^{-10}}}\) = \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz. Inductive Reactance, XL = 2πfL = (6.28) (15.92 × 10^3)(5 × 10^-6) = 0.5 kΩ ∴ Quality factor Q = \(\frac{X_L}{R} = \frac{0.5 kΩ}{50}\) = 10.

Discussion

127.	A 50 Hz voltage is measured with a moving iron voltmeter and a rectifier type AC voltmeter connected in parallel. If the meter readings are Va and Vb respectively. Then the form factor may be estimated as?(a) \(\frac{V_a}{V_b} \)(b) \(\frac{1.11V_a}{V_b}\)(c) \(\frac{\sqrt{2} V_a}{V_b}\)(d) \(\frac{πV_a}{V_b}\)I have been asked this question in an online interview.My question comes from Problems of Series Resonance Involving Quality Factor in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT choice is (b) \(\frac{1.11V_a}{V_b}\) Explanation: Form FACTOR of the wave = \(\frac{RMS \,value}{Mean \,value}\) MOVING iron instrument will show rms value. Rectifier voltmeter is calibrated to READ rms value of sinusoidal voltage that is, with form factor of 1.11. ∴ Mean value of the applied voltage = \(\frac{V_b}{1.11}\) ∴ Form factor = \(\frac{V_a}{V_b/1.11} = \frac{1.11V_a}{V_b}\).

Discussion

128.	For the series resonant circuit shown below, the value of the resonant frequency is _________(a) 36.84 kHz(b) 40.19 kHz(c) 25.28 kHz(d) 15.9 kHzThe question was posed to me in an international level competition.This intriguing question comes from Problems of Series Resonance Involving Quality Factor in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct choice is (d) 15.9 KHZ Best explanation: RESONANT Frequency, FR = \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}\) = \(\frac{1}{6.28\sqrt{10^{-10}}}\) = \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz.

Discussion

129.	For the resonant circuit given below, the value of the quality factor of the circuit is __________(a) 5.6(b) 7.1(c) 8.912(d) 12.6This question was addressed to me in an interview.This interesting question is from Problems of Series Resonance Involving Quality Factor in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right CHOICE is (B) 7.1 To EXPLAIN I would say: f = \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(10 × 10^{-6})(200 × 10^{-12})}}\) = \(\frac{1}{6.28\sqrt{2 × 10^{-15}}}\) = \(\frac{1}{888 × 10^{-9}}\) = 1.13 MHz Inductive Reactance, XL = 2πfL = (6.28) (1.13 × 10^6)(10 × 10^-6) = 70.96 Ω ∴ Q = \(\frac{X_L}{R} = \frac{70.96}{10}\) = 7.1.

Discussion

130.	Consider a circuit consisting of two capacitors C1 and C2. Let R be the resistance and L be the inductance which are connected in series. Let Q1 and Q2 be the quality factor for the two capacitors. While measuring the Q value by the Series Connection method, the value of the Q factor is?(a) Q = \(\frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\)(b) Q = \(\frac{(C_2 – C_1) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\)(c) Q = \(\frac{(C_1 – C_2) Q_1 Q_2}{Q_2 C_2-Q_1 C_1}\)(d) Q = \(\frac{(C_2 – C_1) C_1 C_2}{Q_1 C_1-Q_2 C_2}\)I got this question in an online quiz.The question is from Problems of Series Resonance Involving Quality Factor topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT answer is (a) Q = \(\FRAC{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\) Best EXPLANATION: ωL = \(\frac{1}{ωC}\) and Q1 = \(\frac{ωL}{R} = \frac{1}{ωC_1 R}\) XS = \(\frac{C_1 – C_2}{ωC_1 C_2}\), RS = \(\frac{Q_1 C_1 – Q_2 C_2}{ωC_1 C_2 Q_1 Q_2}\) QX = \(\frac{X_S}{R_S}= \frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\).

Discussion

131.	Consider a circuit consisting of two capacitors C1 and C2. Let R be the resistance and L be the inductance which are connected in series. Let Q1 and Q2 be the quality factor for the two capacitors. While measuring the Q value by the Parallel Connection method, the value of the Q factor is?(a) Q = \(\frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\)(b) Q = \(\frac{(C_2 – C_1) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\)(c) Q = \(\frac{(C_1 – C_2) Q_1 Q_2}{Q_2 C_2-Q_1 C_1}\)(d) Q = \(\frac{(C_2 – C_1) C_1 C_2}{Q_1 C_1-Q_2 C_2}\)This question was posed to me during an interview.This is a very interesting question from Problems of Series Resonance Involving Quality Factor topic in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT choice is (b) Q = \(\frac{(C_2 – C_1) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\) BEST explanation: \(\frac{1}{R_P}= \frac{ωC_1}{Q_2}– \frac{1}{RQ_1^2}\),XP = \(\frac{1}{ω(C_2-C_1)}\) Q = \(\frac{(C_2 – C_1) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\).

Discussion

132.	A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Q factor is?(a) 50(b) 100(c) 150(d) 200I had been asked this question in an online interview.This interesting question is from Problems of Series Resonance Involving Quality Factor in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT option is (c) 150 Easiest EXPLANATION: Q = \(\FRAC{ω}{ω1 – ω2} = \frac{f}{F2-f1}\) Here, f = 1.5 × 10^6 Hz f1 = (1.5 × 10^6 – 5 × 10^3) f2 = (1.5 × 10^6 + 5 × 10^3) So, f2 –f1 = 10 × 10^3 Hz ∴ Q = \(\frac{1.5 × 10^6}{10 × 10^3}\) = 150.

Discussion

133.	A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Effective resistance of the circuit is?(a) 2 Ω(b) 3 Ω(c) 5.5 Ω(d) 4.7 ΩThe question was asked in an online interview.The question is from Problems of Series Resonance Involving Quality Factor in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT OPTION is (d) 4.7 Ω Easy explanation: R =\(\FRAC{F2-f1}{2πf^2 L}\) Here, f = 1.5 × 10^6 Hz f1 = (1.5 × 10^6 – 5 × 10^3) f2 = (1.5 × 10^6 + 5 × 10^3) So, f2 – f1 = 10 × 10^3 Hz ∴ R = 4.7 Ω.

Discussion

134.	Find the value of capacitance (µF) in the circuit shown below.(a) 10(b) 20(c) 30(d) 40I have been asked this question in exam.My doubt is from Resonant Frequency for a Tank Circuit in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct ANSWER is (d) 40 The explanation is: ω^2r = 1/LC. L = 0.1H, ωr = 500 rad/sec. On SOLVING, C = 40 µF. In order to TUNE a PARALLEL circuit to a lower frequency the capacitance must be increased.

Discussion

135.	In the circuit shown in the figure, an inductance of 0.1H having a Q of 5 is in parallel with a capacitor. Determine the value coil resistance (Ω) of at a resonant frequency of 500 rad/sec.(a) 10(b) 20(c) 30(d) 40I have been asked this question in class test.This interesting question is from Resonant Frequency for a Tank Circuit topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right choice is (a) 10 The explanation is: Quality FACTOR Q = ωrL/R. L = 0.1H, Q = 5, ωr = 500 rad/sec. On solving, R = 10Ω. While plotting the voltage and current variation with frequency, at resonant frequency, the current is maximum.

Discussion

136.	The quality factor is defined as?(a) I(b) IC(c) I/IC(d) IC/II have been asked this question in an interview for job.My question is taken from Resonant Frequency for a Tank Circuit topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right choice is (d) IC/I The EXPLANATION: The quality factor is DEFINED as IC/I. Quality factor = IC/I.As the FREQUENCY goes above resonance CAPACITIVE REACTANCE dominates and impedance decreases.

Discussion

137.	The expression of quality factor is?(a) IL/I(b) I/IL(c) IL(d) II had been asked this question in homework.This intriguing question comes from Resonant Frequency for a Tank Circuit topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT ANSWER is (a) IL/I The explanation is: The expression of quality factor is IL/I. Quality factor = IL/I. At the the point XL = XC, the impedance is at its MAXIMUM.

Discussion

138.	The maximum energy stored in a capacitor is?(a) CV^2(b) CV^2/2(c) CV^2/4(d) CV^2/8This question was addressed to me in my homework.I'm obligated to ask this question of Resonant Frequency for a Tank Circuit topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right option is (b) CV^2/2 Best explanation: The maximum energy STORED in a CAPACITOR is CV^2/2. Maximum energy = CV^2/2. As frequency increases the impedance also increases and the inductive reactance DOMINATES until the resonant frequency is reached.

Discussion

139.	The quality factor in case of parallel resonant circuit is?(a) C(b) ωrRC(c) ωrC(d) 1/ωrRCThis question was addressed to me in examination.The doubt is from Resonant Frequency for a Tank Circuit in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right choice is (d) 1/ωrRC The best I can explain: The quality factor in CASE of parallel resonant circuit is Q = 1/ωrRC. The impedance of parallel resonant circuit is MAXIMUM at the resonant frequency and decreases at lower and higher FREQUENCIES.

Discussion

140.	The quality factor is the product of 2π and the ratio of ______ to _________(a) maximum energy stored, energy dissipated per cycle(b) energy dissipated per cycle, maximum energy stored(c) maximum energy stored per cycle, energy dissipated(d) energy dissipated, maximum energy stored per cycleI had been asked this question during an online exam.I'd like to ask this question from Resonant Frequency for a Tank Circuit topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct option is (a) maximum energy stored, energy dissipated per cycle The EXPLANATION is: At low frequencies XL is very small and XC is very LARGE so the total impedance is essentially inductive. The quality factor is the PRODUCT of 2π and the ratio of maximum energy stored to energy dissipated per cycle.

Discussion

141.	The expression of bandwidth for the parallel resonant circuit is?(a) 1/RC(b) RC(c) 1/R(d) 1/CI have been asked this question during an interview for a job.My question is from Resonant Frequency for a Tank Circuit topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct ANSWER is (a) 1/RC To explain: The expression of BANDWIDTH for parallel resonant circuit is BW = 1/RC. the circuit is SAID to be in a resonant condition when the SUSCEPTANCE PART of admittance is zero.

Discussion

142.	The expression of ωr in a parallel resonant circuit is?(a) 1/(2√LC)(b) 1/√LC(c) 1/(π√LC)(d) 1/(2π√LC)The question was posed to me in an interview for job.I need to ask this question from Resonant Frequency for a Tank Circuit in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct answer is (B) 1/√LC Easy explanation: The stored energy is transferred back and forth between the capacitor and coil and vice-versa. The expression of ωr in PARALLEL RESONANT circuit is ωr = 1/√LC.

Discussion

143.	For the tank circuit shown below, find the resonant frequency.(a) 157.35(b) 158.35(c) 159.35(d) 160.35The question was posed to me during a job interview.My question is from Resonant Frequency for a Tank Circuit in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right answer is (b) 158.35 To explain I WOULD say: The parallel resonant circuit is generally called a tank circuit because of the FACT that the circuit STORES energy in the magnetic FIELD of the coil and in the electric field of the capacitor. The resonant frequency FR = (1/2π) √((1/LC)-(R^2/L^2)).

Discussion

144.	If in a circuit, if Q value is decreased then it will cause?(a) small bandwidth(b) no effect on bandwidth(c) first increases and then decreases(d) large bandwidthThis question was addressed to me in an international level competition.I'd like to ask this question from Parallel Resonance in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct OPTION is (d) large bandwidth The explanation is: If in a CIRCUIT, if the Q value is decreased then bandwidth INCREASES and the bandwidth do not decrease.

Discussion

145.	If the value of Q of the circuit is high, then its effect on bandwidth is?(a) large bandwidth(b) small bandwidth(c) no effect on bandwidth(d) first increases and then decreasesThis question was posed to me by my college director while I was bunking the class.This interesting question is from Parallel Resonance topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct ANSWER is (b) small BANDWIDTH For explanation I would say: If the value of Q of the CIRCUIT is HIGH, then small bandwidth because bandwidth is inversely proportional to the quality factor.

Discussion

146.	The value of ωr in parallel resonant circuit is?(a) 1/(2√LC)(b) 1/√LC(c) 1/(π√LC)(d) 1/(2π√LC)This question was addressed to me during an interview for a job.My question is from Parallel Resonance topic in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right option is (B) 1/√LC To explain: Basically parallel resonance occurs when XL = XL. The frequency at which the resonance occurs is called the RESONANT frequency. The VALUE of ωr in parallel resonant circuit is ωr = 1/√LC.

Discussion

147.	Find the resonant frequency in the ideal parallel LC circuit shown in the figure.(a) 7.118(b) 71.18(c) 711.8(d) 7118The question was posed to me by my school principal while I was bunking the class.This interesting question is from Parallel Resonance topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» CORRECT choice is (d) 7118 For explanation I WOULD say: The expression for resonant frequency is fr = 1/(2π√LC). GIVEN L = 50mH and C = 0.01uF. On substituting the given values in the equation we GET the resonant frequency = 1/(2π√(50×10^-3)×0.01×10^-6))=7117.6 Hz.

Discussion

148.	The expression of resonant frequency for parallel resonant circuit is?(a) 1/(2π√LC)(b) 1/(π√LC)(c) 1/(2√LC)(d) 1/√LCThis question was posed to me during a job interview.My question is taken from Parallel Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right choice is (a) 1/(2π√LC) To explain: The CONDITION for resonance occurs when XL = XL. The EXPRESSION of resonant FREQUENCY for PARALLEL resonant circuit is fr = 1/(2π√LC).

Discussion

149.	Considering the voltage across the capacitor, the magnification in resonance is?(a) VC(b) V x VC(c) VC/V(d) V/VCI have been asked this question in final exam.My question is taken from Parallel Resonance topic in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct answer is (c) VC/V For explanation: The ratio of VOLTAGE across capacitor to the voltage applied at RESONANCE can be DEFINED as magnification. Considering the voltage across the capacitor, the magnification in resonance is Q = VC/V.

Discussion

150.	The magnification in resonance considering the voltage across inductor is?(a) V/VL(b) VL/V(c) V x VL(d) VLThis question was addressed to me in an interview for internship.My question comes from Parallel Resonance topic in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT answer is (b) VL/V Easiest EXPLANATION: The ratio of voltage ACROSS inductor to the voltage applied at RESONANCE can be defined as magnification. The magnification in resonance considering the voltage across inductor is Q = VL/V.

Discussion

Explore topic-wise InterviewSolutions in .

Find the value of capacitance (µF) in the circuit shown below.(a) 10(b) 20(c) 30(d) 40I have been asked this question in exam.My doubt is from Resonant Frequency for a Tank Circuit in chapter Resonance & Magnetically Coupled Circuit of Network Theory

The quality factor is defined as?(a) I(b) IC(c) I/IC(d) IC/II have been asked this question in an interview for job.My question is taken from Resonant Frequency for a Tank Circuit topic in section Resonance & Magnetically Coupled Circuit of Network Theory

The expression of quality factor is?(a) IL/I(b) I/IL(c) IL(d) II had been asked this question in homework.This intriguing question comes from Resonant Frequency for a Tank Circuit topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory

The maximum energy stored in a capacitor is?(a) CV^2(b) CV^2/2(c) CV^2/4(d) CV^2/8This question was addressed to me in my homework.I'm obligated to ask this question of Resonant Frequency for a Tank Circuit topic in division Resonance & Magnetically Coupled Circuit of Network Theory

The quality factor in case of parallel resonant circuit is?(a) C(b) ωrRC(c) ωrC(d) 1/ωrRCThis question was addressed to me in examination.The doubt is from Resonant Frequency for a Tank Circuit in division Resonance & Magnetically Coupled Circuit of Network Theory

The expression of bandwidth for the parallel resonant circuit is?(a) 1/RC(b) RC(c) 1/R(d) 1/CI have been asked this question during an interview for a job.My question is from Resonant Frequency for a Tank Circuit topic in section Resonance & Magnetically Coupled Circuit of Network Theory

For the tank circuit shown below, find the resonant frequency.(a) 157.35(b) 158.35(c) 159.35(d) 160.35The question was posed to me during a job interview.My question is from Resonant Frequency for a Tank Circuit in section Resonance & Magnetically Coupled Circuit of Network Theory

The value of ωr in parallel resonant circuit is?(a) 1/(2√LC)(b) 1/√LC(c) 1/(π√LC)(d) 1/(2π√LC)This question was addressed to me during an interview for a job.My question is from Parallel Resonance topic in portion Resonance & Magnetically Coupled Circuit of Network Theory

The expression of resonant frequency for parallel resonant circuit is?(a) 1/(2π√LC)(b) 1/(π√LC)(c) 1/(2√LC)(d) 1/√LCThis question was posed to me during a job interview.My question is taken from Parallel Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory

Considering the voltage across the capacitor, the magnification in resonance is?(a) VC(b) V x VC(c) VC/V(d) V/VCI have been asked this question in final exam.My question is taken from Parallel Resonance topic in portion Resonance & Magnetically Coupled Circuit of Network Theory

The magnification in resonance considering the voltage across inductor is?(a) V/VL(b) VL/V(c) V x VL(d) VLThis question was addressed to me in an interview for internship.My question comes from Parallel Resonance topic in portion Resonance & Magnetically Coupled Circuit of Network Theory