173 + Interview Questions in Resonance & Magnetically Coupled Circuit in Network Theory Page 2 InterviewSolution

51.	An air capacitor of capacitance 0.005 μF is connected to a direct voltage of 500 V. It is disconnected and then immersed in oil with a relative permittivity of 2.5. The energy after immersion is?(a) 275 μJ(b) 250 μJ(c) 225 μJ(d) 625 μJThis question was posed to me during an internship interview.Asked question is from Problems Involving Dot Conventions topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right OPTION is (b) 250 μJ The explanation: E = \(\frac{1}{2}\) CV^2 Or, C = \(\frac{ε_0 ε_r A}{d} = ϵ_r (\frac{ε_0 A}{d})\) = 2.5 × 0.005 × 10^-6 ∴ CNEW = 12.5 × 10^-9 F Now, q = CV = 0.005 × 10^-6 × 500 = 2.5 × 10^-6 VNEW = \(\frac{q} {C_{NEW}}\) = \(\frac{2.5 × 10^{-6}}{12.5 × 10^{-9}}\) VNEW = 200 E = \(\frac{1}{2}\) CV^2 = \(\frac{1}{2}\) × 12.5 × 10^-9 × (200)^2 = 250 μJ.

Discussion

52.	The resistance of copper motor winding at t=20°C is 3.42 Ω. After extended operation at full load, the motor windings measures 4.22 Ω. If the temperature coefficient is 0.0426, what is the rise in temperature?(a) 60°C(b) 45.2°C(c) 72.9°C(d) 10.16°CThe question was posed to me by my school principal while I was bunking the class.I want to ask this question from Problems Involving Dot Conventions in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right option is (d) 10.16°C Explanation: GIVEN that, R1 = 3.42 Ω T1 = 20° and α = 0.0426 R2 = 4.22 Ω Now, \(\frac{R_1}{1 + αT_1} = \frac{R_2}{1 + αT_2}\) Or, \(\frac{3.42}{1 + 0.0426 × 20} = \frac{4.22}{1 + 0.0426 T_2}\) ∴ RISE in TEMPERATURE = T2 – T1 = 30.16 – 20 = 10.16°C.

Discussion

53.	When a DC voltage is applied to an inductor, the current through it is found to build up in accordance with I = 20(1-e^-50t). After the lapse of 0.02 s, the voltage is equal to 2 V. What is the value of inductance?(a) 2 mH(b) 5.43 mH(c) 1.54 mH(d) 0.74 mHI had been asked this question by my school principal while I was bunking the class.Asked question is from Problems Involving Dot Conventions in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right CHOICE is (b) 5.43 mH The best EXPLANATION: VL = L\(\frac{dI}{dt}\) Where, I = 20(1-e^-50t) Therefore, VL = L\(\frac{d 20(1-e^{-50t})}{dt}\) = L × 20 × 50e^-50t At t = 0.02 s, VL = 2 V ∴ L = \(\frac{2}{20 × 50 × e^{-50×0.02}}\) = 5.43 μH.

Discussion

54.	In the figure given below, the voltage source provides the circuit with a voltage V. The number of non-planar graph of independent loop equations is______________(a) 8(b) 12(c) 7(d) 5The question was asked in examination.This intriguing question originated from Problems Involving Dot Conventions topic in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct choice is (d) 5 Explanation: The total number of independent loop equations are GIVEN by L = B – N + 1 where, L = number of loop equations B = number of branches = 12 N = number of nodes = 8 ∴ L = 12 – 8 + 1 = 5.

Discussion

55.	In the circuit given below, the equivalent capacitance is ______________(a) 3.5 μF(b) 1.2 μF(c) 2.4 μF(d) 4.05 μFI got this question by my college director while I was bunking the class.This interesting question is from Problems Involving Dot Conventions in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» CORRECT choice is (d) 4.05 μF To elaborate: The 2.5 μF capacitor is in PARALLEL with 1 μF capacitor and this COMBINATION is in series with 1.5 μF. Hence, C1 = \(\frac{1.5(2.5+1)}{1.5+2.5+1}\) = \(\frac{5.25}{5}\) = 1.05 Now, C1 is in parallel with the 3 μF capacitor. ∴ CEQ = 1.05 + 3 = 4.05 μF.

Discussion

56.	In the circuit given below, the voltage across A and B is?(a) 13.04 V(b) 17.84 V(c) 12 V(d) 10.96 VThe question was posed to me during a job interview.My query is from Problems Involving Dot Conventions topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct OPTION is (b) 17.84 V Best EXPLANATION: LOOP current I1 = \(\frac{6}{10}\) = 0.6 A I2 = \(\frac{12}{14}\) = 0.86 A VAB = (0.6) (4) + 12 + (0.86) (4) = 2.4 + 12 + 3.44 = 17.84 V.

Discussion

57.	In the circuit given below, the equivalent capacitance is ____________(a) 1.625 F(b) 1.583 F(c) 0.583 F(d) 0.615 FThe question was asked by my college professor while I was bunking the class.The above asked question is from Problems Involving Dot Conventions in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT CHOICE is (d) 0.615 F Best EXPLANATION: CCB = \(\left(\FRAC{C_2 C_3}{C_2+ C_3}\RIGHT)\) + C5 = 1.5 F Now, CAB =\(\left(\frac{C_1 C_{CB}}{C_1+ C_{CB}}\right)\) + C6 = 1.6 F CXY = \(\frac{C_{AB}× C_4}{C_{AB} + C_4}\) = 0.615 F.

Discussion

58.	The current through an electrical conductor is 1A when the temperature of the conductor 0°C and 0.7 A when the temperature is 100°C. The current when the temperature of the conductor is 1200°C is ___________(a) 0.08 A(b) 0.16 A(c) 0.32 A(d) 0.64 AThis question was addressed to me during an interview for a job.This question is from Problems Involving Dot Conventions in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right CHOICE is (b) 0.16 A To elaborate: \(\frac{1}{0.7} = \frac{R_O (1+αt)}{R_O}\) = 1 + α100 ∴α = 0.0043 PER °C Current at 1200 °C is GIVEN by, \(\frac{1}{I} = \frac{R_O (1+α1200)}{R_O}\) = 1 + α1200 = 1 + 0.0043 × 1200 = 6.16 ∴ I = \(\frac{1}{6.16}\) = 0.16 A.

Discussion

59.	In the circuit given below, the capacitor initially has a charge of 10 C. The current in the circuit at t=1 sec after the switch S is closed will be ___________(a) 14.7 A(b) 18.5 A(c) 40 A(d) 50 AI have been asked this question during an interview for a job.The doubt is from Dot Convention in Magnetically Coupled Circuits in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct answer is (a) 14.7 A Explanation: Using KVL, 100 = R\(\frac{dq}{dt} + \frac{q}{C} \) Or, 100 C = RC\(\frac{dq}{dt}\) + q Now, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} ∫_0^t dt\) Or, 100C – q = (100C – qo) e^-t/RC I = \(\frac{dq}{dt} = \frac{(100C – q_0)}{RC} e^{-1/1}\) = 40e^-1 = 14.7 A.

Discussion

60.	In the circuit given below, the switch is closed at time t=0. The voltage across the inductance just at t=0^+ is ____________(a) 2 V(b) 4 V(c) -6 V(d) 8 VThis question was addressed to me during an interview.The above asked question is from Dot Convention in Magnetically Coupled Circuits topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct ANSWER is (B) 4 V To elaborate: A t=0^+, I (0^+) = \(\frac{10}{4\|\|4+3} = \frac{10}{5}\) = 2A ∴ I2(0^+) = \(\frac{2}{2}\) = 1A VL(0^+) = 1 × 4 = 4 V.

Discussion

61.	A rectangular voltage wave of magnitude A and duration B is applied to a series combination of resistance R and capacitance C. The voltage developed across the capacitor is ____________(a) A[1 – exp(-\(\frac{B}{RC}\))](b) \(\frac{AB}{RC}\)(c) A(d) A exp(-\(\frac{B}{RC}\))I had been asked this question by my school principal while I was bunking the class.This interesting question is from Dot Convention in Magnetically Coupled Circuits in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right option is (a) A[1 – exp(-\(\FRAC{B}{RC}\))] For explanation I WOULD SAY: VC = \(\frac{1}{C} ∫Idt\) = \(\frac{1}{C} ∫_0^B \frac{A}{R} e^{-\frac{t}{RC}}\) dt VC = A[1 – exp(-\(\frac{B}{RC}\))] Hence, maximum VOLTAGE = V [1 – exp (-\(\frac{B}{RC}\))].

Discussion

62.	In the figure given below, what is the RMS value of the periodic waveform?(a) 2\(\sqrt{6}\) A(b) 6\(\sqrt{2}\) A(c) \(\sqrt{\frac{4}{3}}\) A(d) 1.5 AThis question was addressed to me in a job interview.My query is from Dot Convention in Magnetically Coupled Circuits topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct option is (a) 2\(\sqrt{6}\) A The explanation is: The rms VALUE for any waveform is = \(\sqrt{\frac{1}{T} \int_0^T f^2 (t)dt)}\) = \([\frac{1}{T}(\int_0^{\frac{T}{2}}(mt)^2 dt + \int_{\frac{T}{2}}^T 6^2 dt]^{1/2}\) = \([\frac{1}{T}(\frac{144}{T^2}× \frac{T^3}{3×8} + 36 × \frac{T}{2})]^{\frac{1}{2}}\) = \([6+18]^{\frac{1}{2}} = \sqrt{24} = 2\sqrt{6}\) A.

Discussion

63.	An ideal capacitor is charged to a voltage VO and connected at t=0 across an ideal inductor L. Ifω = \(\frac{1}{\sqrt{LC}}\), the voltage across the capacitor at time t>0 is ____________(a) VO(b) VO cos(ωt)(c) VO sin(ωt)(d) VO e^-ωtcos(ωt)I got this question in an international level competition.This is a very interesting question from Dot Convention in Magnetically Coupled Circuits topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right answer is (B) VO COS(ωt) For explanation: Voltage across capacitor will discharge through inductor up to voltage across the capacitor becomes zero. Now, inductor will start charging capacitor. Voltage across capacitor will be DECREASING from VO and periodic and is not decaying SINCE both L and C is ideal. ∴ Voltage across the capacitor at time t>0 is VO cos(ωt).

Discussion

64.	In the circuit given below, the switch S1 is initially closed and S2 is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through C during t=0^+ is _____________(a) 55 A(b) 5.5 A(c) 45 A(d) 4.5 AI had been asked this question at a job interview.This is a very interesting question from Dot Convention in Magnetically Coupled Circuits in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» CORRECT option is (d) 4.5 A The explanation is: By KCL, we get, \(\frac{V_L}{10} – 10 + \frac{V_L-10}{10}\) = 0 Hence, 2 VL = 110 ∴ VL = 55 V Or, IC = \(\frac{55-10}{10}\) = 4.5 A.

Discussion

65.	In the circuit given below, the switch S1 is initially closed and S2 is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through L during t=0^+ is _____________(a) 55 A(b) 5.5 A(c) 45 A(d) 4.5 AI had been asked this question during an online exam.I need to ask this question from Dot Convention in Magnetically Coupled Circuits in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT option is (a) 55 A Easy EXPLANATION: By KCL, we get, \(\FRAC{V_L}{10} – 10 + \frac{V_L-10}{10}\) = 0 Hence, 2 VL = 110 ∴ VL = 55 V.

Discussion

66.	In the circuit given below, for time t
Answer» Correct answer is (d) 3 V Explanation: When S1 is CLOSED and S2 is open, VC1 (0^–) = VC1 (0^+) = 3V When S1 is OPENED and S2 is closed, VC2 (0^+) = VC2 (0^+) = 3V.

Discussion

67.	For a unit step signal u (t), the response is V1 (t) = (1-e^-3t) for t>0. If a signal 3u (t) + δ(t) is applied, the response will be (considering zero initial conditions)?(a) (3-6e^-3t)u(t)(b) (3-3e^-3t)u(t)(c) 3u(t)(d) (3+3e^-3t)u(t)The question was asked in a job interview.My query is from Dot Convention in Magnetically Coupled Circuits in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct option is (C) 3u(t) Explanation: For u (t) = 1, t>0 V1 (t) = (1-e^-3t) Or, V1 (s) = \(\left(\frac{1}{s} + \frac{1}{s+3}\right) = \frac{3}{s(s+3)}\) And T(s) = \(\frac{V_1 (S)}{u(S)} = \frac{3}{s+3}\) Now, for R(s) = (\(\frac{3}{s}\) + 1) RESPONSE, H(s) = R(s) T(s) = \((\frac{3+s}{s}) (\frac{3}{s+3}) = \frac{3}{s}\) Or, h (t) = 3 u (t).

Discussion

68.	Initial voltage on capacitor VO as marked \|VO\| = 5 V, VS = 8 u (t), where u (t) is the unit step. The voltage marked V at t=0^+ is _____________(a) 1 V(b) -1 V(c) \(\frac{13}{3}\) V(d) –\(\frac{13}{3}\) VThis question was posed to me during an online exam.My question comes from Dot Convention in Magnetically Coupled Circuits in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct option is (C) \(\frac{13}{3}\) V Easiest explanation: Applying voltage divider method, we get, I = \(\frac{V}{R_{EQ}} \) = \(\frac{8}{1+1\|\|1} \) = \(\frac{8}{1+\frac{1}{2}} = \frac{16}{3}\) A I1 = \(\frac{16}{3} × \frac{1}{2} = \frac{8}{3}\) A And \(I’_2 = \frac{V}{R_{EQ}}= \frac{5}{1+1\|\|1}\) = \(\frac{5}{1+\frac{1}{2}} = \frac{10}{3}\) A Now, \(I’_1 = I’_2 × \frac{1}{1+1} \) = \(\frac{10}{3} × \frac{1}{2} = \frac{5}{3}\) A Hence, the net current in 1Ω RESISTANCE = I1 + \(I’_1\) = \(\frac{8}{3} + \frac{5}{3} = \frac{13}{3}\) A ∴ Voltage DROP across 1Ω = \(\frac{13}{3} × 1 = \frac{13}{3}\) V.

Discussion

69.	In the circuit shown below the current I(t) for t≥0^+ (assuming zero initial conditions) is ___________(a) 0.5-0.125e^-1000t A(b) 1.5-0.125e^-1000t A(c) 0.5-0.5e^-1000t A(d) 0.375e^-1000t AI got this question in unit test.This interesting question is from Dot Convention in Magnetically Coupled Circuits in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct ANSWER is (a) 0.5-0.125e^-1000t A Easy EXPLANATION: I (t) = \(\frac{1.5}{3}\) = 0.5 LEQ = 15 mH REQ = 5+10 = 15Ω L = \(\frac{L_{EQ}}{R_{EQ}}\) = \(\frac{15 × 10^3}{15} = \frac{1}{1000}\) I (t) A – (A – B) e^-t = 0.5 – (0.5-B) e^-1000t = 0.5(0.5 – 0.375) e^-1000t = 0.5 – 0.125 e^-1000t I (t) = 0.5-0.125e^-1000t.

Discussion

70.	In the circuit shown, the voltage source supplies power which is _____________(a) Zero(b) 5 W(c) 10 W(d) 100 WI had been asked this question in homework.Origin of the question is Dot Convention in Magnetically Coupled Circuits in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct choice is (a) Zero For EXPLANATION: Let the current supplied by a VOLTAGE source. Applying KVL in outer loop, 10 – (I+3) × (I+1) – (I+2) × 2 = 0 10 – 2(I+3) – 2(I-2) = 0 Or, I = 0 ∴ POWER VI = 0.

Discussion

71.	The switch in the circuit shown was on position X for a long time. The switch is then moved to position Y at time t=0. The current I(t) for t>0 is ________(a) 0.2e^-125tu(t) mA(b) 20e^-1250tu(t) mA(c) 0.2e^-1250tu(t) mA(d) 20e^-1000tu(t) mAThis question was posed to me in a job interview.This interesting question is from Dot Convention in Magnetically Coupled Circuits in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct choice is (C) 0.2e^-1250tu(t) mA The BEST explanation: CEQ = \(\frac{0.8 × 0.2}{0.8+0.2}\) = 0.16 VC (t=0^–) = 100 V At t≥0, The discharging current I (t) = \(\frac{V_O}{R} E^{-\frac{t}{RC}}\) = \(\frac{100}{5000} e^{- \frac{t}{5×10^3×0.16×10^{-6}}}\) = 0.2e^-1250tu(t) mA.

Discussion

72.	A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is _____________(a) 25(b) 50(c) 100(d) 200I got this question by my college director while I was bunking the class.Origin of the question is Dot Convention in Magnetically Coupled Circuits topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» CORRECT choice is (b) 50 Easiest explanation: Q = \(\frac{f_0}{BW}\) And f0 = 1/2π (LC)^0.5 BW = R/L Or, Q = \(\frac{1}{R}(\frac{L}{C})^{0.5}\) When R, L and C are DOUBLED, Q’ = 50.

Discussion

73.	In the circuit given below, the input impedance ZIN of the circuit is _________(a) 0.52 – j4.30 Ω(b) 0.52 + j15.70 Ω(c) 64.73 + j17.77 Ω(d) 0.3 – j33.66 ΩThe question was posed to me in quiz.My doubt stems from Dot Convention in Magnetically Coupled Circuits topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» CORRECT choice is (c) 64.73 + j17.77 Ω The BEST I can explain: ZIN = (-6J) \|\| (ZA) ZA = j10 + \(\FRAC{12^2}{(j30+j6-j2+4)}\) = 0.49 + j5.82 ZIN = \(\frac{(-j6)(0.49+j5.82)}{(-j6+0.49+j5.82)}\) = 64.73 + j17.77 Ω.

Discussion

74.	A 50 μF capacitor, when connected in series with a coil having resistance of 40Ω, resonates at 1000 Hz. The circuit is in resonating condition. The voltage across the coil is __________(a) 100.31 V(b) 200.31 V(c) 300.31 V(d) 400.31 VThe question was asked in examination.This interesting question is from Advanced Problems on Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct choice is (a) 100.31 V To explain: At resonance, \| I \| = \(\frac{V}{Z} = \frac{V}{R} \) Given that, R = 40Ω and V = 100 V ∴ \| I \| = \(\frac{100}{40}\) = 2.5 A XL = ωL = 2π × 1000 × 0.5 × 10^-3 = 2π × 0.5 = 3.14 Ω ∴ VCOIL = I Z (At resonance) = 2.5 \(\SQRT{40^2 + 3.14^2}\) = 2.5 × 40.122 = 100.31 V.

Discussion

75.	A 50 μF capacitor, when connected in series with a coil having resistance of 40Ω, resonates at 1000 Hz. The circuit current if the applied voltage is 100 V for the resonant circuit is ___________(a) 2.5 A(b) 3.5 A(c) 4.5 A(d) 0.5 AThe question was posed to me during an interview.This key question is from Advanced Problems on Resonance topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right answer is (a) 2.5 A Easiest EXPLANATION: We KNOW that the magnitude of current \|I\| for a resonating circuit is GIVEN by, \| I \| = \(\frac{V}{Z} = \frac{V}{R} \) Given that, R = 40Ω and V = 100 V ∴ \| I \| = \(\frac{100}{40}\) = 2.5 A.

Discussion

76.	An ideal transformer is rated as 2400/120 V, 9.6 kVA and has 50 turns on the secondary side. The number of turns on the primary side is ____________(a) 1000(b) 2000(c) 3000(d) 4000I had been asked this question in exam.Question is taken from Advanced Problems on Resonance in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct option is (a) 1000 Explanation: Turns ratio, N = \(\FRAC{N_2}{N_1}\) Now, the turns ratio, n = \(\frac{V_2}{V_1}= \frac{120}{2400}\) = 0.05 Also, n = \(\frac{50}{N_1}\) So, N1 = \(\frac{50}{0.05}\) = 1000 turns.

Discussion

77.	A 50 μF capacitor, when connected in series with a coil having a resistance of 40Ω, resonates at 1000 Hz. The circuit is in resonating condition. The voltage across the capacitor is __________(a) 8.04 V(b) 7.96 V(c) 5.68 V(d) 1.25 VI had been asked this question during an internship interview.My question is based upon Advanced Problems on Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct choice is (c) 5.68 V Easy explanation: At RESONANCE, \| I \| = \(\frac{V}{Z} = \frac{V}{R} \) GIVEN that, R = 40Ω and V = 100 V ∴ \| I \| = \(\frac{100}{40}\) = 2.5 A Power loss of the COIL = I^2 R = 2.5^2 × 40 = 250 W Also, VC = IXC = 2.5 × \(\frac{1}{2πfC}\) and XC = \(\frac{1}{2πfC}\) = 2.5 × \(\frac{1}{2π×1000×50×10^{-6}}\) = 2.5 × \(\frac{100}{31.41}\) = 5.68 V.

Discussion

78.	The time constant for the given circuit under resonance condition will be?(a) \(\frac{1}{9}\) s(b) \(\frac{1}{4}\) s(c) 4 s(d) 9 sThis question was posed to me in an interview.Enquiry is from Advanced Problems on Resonance topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct choice is (c) 4 s The best EXPLANATION: For finding the time constant, we neglect CURRENT source as an open circuit. Hence, the circuit becomes: So, from the above circuit, the Time constant = Req Ceq = 6 × \(\FRAC{2}{3}\) = 4 s.

Discussion

79.	The frequency for the given circuit when the circuit is in a state of resonance will be?(a) 1 rad/s(b) 2 rad/s(c) 3 rad/s(d) 4 rad/sThis question was addressed to me by my school teacher while I was bunking the class.My query is from Advanced Problems on Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT OPTION is (C) 3 rad/s Easy EXPLANATION: Z = jωL + \(\frac{R}{1+jωRC} \) ∴ Z = j0.1ω + \(\frac{1}{1+jω} × \frac{1-jω}{1-jω} \) = j0.1ω + \(\frac{1-jω}{1+ω^2} = \frac{1}{1+ω^2} + j (0.1ω – \frac{ω}{1 + ω^2}) \) At resonance, imaginary part must be zero. 0.1 ω – \(\frac{ω}{ 1+ ω^2}\) = 0 Or, 0.1 = \(\frac{1}{1+ω^2}\) Or, ω^2 + 1 = 10 ∴ ω^2 = 9 Or, ω = 3 rad/s.

Discussion

80.	The RL circuit in the below figure is fed from a constant magnitude, variable frequency sinusoidal voltage source VIN. At 50 Hz, the R and L elements each have a voltage drop Vrms. If the frequency of the source is changed to 25 Hz, the new voltage drop acrossR is ___________(a) \(\sqrt{\frac{5}{8}}\) Vrms(b) \(\sqrt{\frac{2}{3}}\) Vrms(c) \(\sqrt{\frac{8}{5}}\) Vrms(d) \(\sqrt{2}\) VrmsThis question was addressed to me in examination.Question is taken from Advanced Problems on Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT option is (c) \(\sqrt{\frac{8}{5}}\) Vrms Easiest EXPLANATION: At 50 HZ, VR = VL and R = ωL Also Vrms = \(\frac{V_{in}}{\sqrt{2}} \) At 25 Hz, Iin= \(\frac{V_{in}}{\sqrt{R^2+\frac{ω^2 L^2}{4}}}\) = \(\frac{V_{in}}{\sqrt{ω^2 L^2 + \frac{ω^2 L^2}{4}}} = \frac{2V_{in}}{\sqrt{5} ωL}\) So, Vnew = \(\frac{2V_{in}}{\sqrt{5} ωL} × R = \frac{2V_{in}}{\sqrt{5}}\) =\(\frac{2\sqrt{2} V_{rms}}{\sqrt{5}} = \sqrt{\frac{8}{5}}\) Vrms.

Discussion

81.	In a series circuit with R = 10Ω and C = 25 μF, at what frequency will the current lead the voltage by 30°?(a) 11.4 kHz(b) 4.5 kHz(c) 1.1 kHz(d) 24.74 kHzThe question was posed to me in examination.Query is from Advanced Problems on Resonance topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right option is (a) 11.4 kHz To ELABORATE: 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 = \(\frac{1}{2πX_C C} \) = \(\frac{1}{2π×5.77×25×10^{-6}}\) = \(\frac{10^6}{906.18}\) = 1103 HZ = 1.1 kHz.

Discussion

82.	An ideal capacitor is charged to a voltage V and connected at t=0 across an ideal inductor L. If we take ω = \(\frac{1}{\sqrt{LC}}\), the voltage across the capacitor at time t>0 will be?(a) V(b) V cos(ωt)(c) V sin(ωt)(d) Ve^-ωt cos(ωt)The question was asked in class test.My enquiry is from Advanced Problems on Resonance topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct answer is (b) V cos(ωt) The best EXPLANATION: Voltage across the capacitor will discharge through INDUCTOR unless the voltage across the capacitor BECOMES zero. Now, the inductor will start charging the capacitor. Voltage across the capacitor will be decreasing from V and will be periodic. But it will not be decaying with time, since both L and C are ideal. ∴ Voltage across the capacitor at time t>0 is GIVEN by V cos (ωt).

Discussion

83.	The circuit shown in the figure below will act as an ideal source with respect to terminal A and B when the frequency is ___________(a) Zero(b) 1 rad/s(c) 4 rad/s(d) 16 rad/sThe question was posed to me by my college professor while I was bunking the class.My question comes from Advanced Problems on Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT answer is (c) 4 rad/s To explain: Z(s) t = \(\frac{(Ls)\frac{1}{Cs}}{Ls + \frac{1}{Cs}} = \frac{Ls}{LCs^2+1} \) PUTTING s = jω, we GET, Z (jω) = \(\frac{jωL}{1 -LCω^2} \) For ideal current SOURCE, Z (jω) = ∞ Or, 1 – LCω^2 = 0 Or, ω = \(\frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{\frac{1}{16}×1}}\) = 4 rad/s.

Discussion

84.	For a series RLC circuit, excited by voltage V = 20 sinωt. The values of R = 2 Ω, L = 1mH and C = 0.4 µF. The quality factor and bandwidth of the circuit is ______________(a) 25, 2(b) 30, 5(c) 40, 10(d) 50, 2I have been asked this question in homework.The above asked question is from Advanced Problems on Resonance in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct ANSWER is (a) 25, 2 The explanation is: We know that, the bandwidth is, B = w2 – w1 = 2 krad/s Also, B = \(\frac{R}{L} = \frac{2}{10^{-3}}\) = 2 krad/s Also, the quality factor is GIVEN by, Q = \(\frac{w_0}{B}\) Or, Q = \(\frac{w_0}{B} = \frac{50}{2}\) = 25.

Discussion

85.	For the circuit shown in the figure below, the value of R for critical damping will be?(a) 10.5 Ω(b) 8.57 Ω(c) 3.5 Ω(d) 3 ΩI got this question by my college professor while I was bunking the class.My question is based upon Advanced Problems on Resonance in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct answer is (b) 8.57 Ω Easiest explanation: For a parallel RLC circuit, \(\frac{d^2 V}{dt^2} + \frac{1}{RC} \frac{dV}{dt} + \frac{V}{LC}\) = 0 Solving the 2nd order DIFFERENTIAL equation, we GET, 2rωn = \(\frac{1}{RC}\) and ωn = \(\frac{1}{\sqrt{LC}}\) As, R=1, ∴ R = \(\frac{1}{2} \sqrt{\frac{L}{C}} \) = \(\frac{1}{2} \sqrt{\frac{7}{\frac{1}{42}}} = \frac{1}{2}\) × 17.15 = 8.57 Ω.

Discussion

86.	An ideal transformer is rated as 2400/120 V, 9.6 kVA and has 50 turns on the secondary side. The turns ratio is ____________(a) 0.04(b) 0.05(c) 0.06(d) 0.07The question was posed to me during an interview.The query is from Advanced Problems on Resonance in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct answer is (b) 0.05 Best EXPLANATION: The given transformer is a STEP down transformer. So, V1 = 2400 V But it is greater than V2 which is equal to 120V So, the turns ratio, n = \(\frac{V_2}{V_1}= \frac{120}{2400}\) = 0.05.

Discussion

87.	In the circuit shown in the figure, if ω = 40 rad/s, then Zin is _____________(a) 4.77 – j1.15 Ω(b) 2.96 – j0.807 Ω(c) 2.26 – j3.48 Ω(d) 2.26 – j3.48 ΩI got this question during a job interview.The above asked question is from Advanced Problems on Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» CORRECT choice is (b) 2.96 – j0.807 Ω Easy explanation: Z = Z11 + \(\FRAC{ω^2 M^2}{Z_{22}} \) Or, Z = 2 + j (40) \((\frac{1}{10}) + \frac{40^2 (\frac{1}{4})^2}{4+j(40)\frac{1}{2}}\) = 2 + j (4) + \(\frac{100}{4+j (20)} \) = 2 + j (4) + \(\frac{100}{4+j (20)} × \frac{4-j (20)}{4-j (20)}\) = 2 + j (4) + \(\frac{400-j (2000)}{416}\) = 2.96 – j0.807 Ω.

Discussion

88.	For a series RLC circuit, excited by voltage V = 20 sinωt. The values of R = 2 Ω, L = 1mH and C = 0.4 µF. The resonant frequency of the circuit is ______________(a) 20 krad/s(b) 30 krad/s(c) 40 krad/s(d) 50 krad/sThis question was posed to me in an international level competition.This intriguing question comes from Advanced Problems on Resonance in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT answer is (d) 50 krad/s To explain: We know that the resonant frequency w0 = 1/(LC)^0.5 Given R = 2 Ω, L = 1mH and C = 0.4 µF. So, w0 = \(\frac{1}{\sqrt{10^{-3}×0.4×10^{-6}}}\) Or, w0 = 50 krad/s.

Discussion

89.	A circuit excited by voltage V has a resistance R which is in series with an inductor and capacitor, which are connected in parallel. The voltage across the resistor at the resonant frequency is ___________(a) 0(b) \(\frac{V}{2}\)(c) \(\frac{V}{3}\)(d) VThis question was posed to me in unit test.I need to ask this question from Problems of Parallel Resonance Involving Quality Factor in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct answer is (a) 0 The EXPLANATION: Dynamic resistance of the tank circuit, ZDY = L/(RLC) But GIVEN that RL = 0 So, ZDY = L/(0XC) = ∞ Therefore current through circuit, I = \(\frac{V}{∞}\) = 0 ∴ VD = 0.

Discussion

90.	For the series resonant circuit given below, the value of the quality factor is ___________(a) 35.156(b) 56.118(c) 50.294(d) 46.128This question was addressed to me in final exam.The above asked question is from Problems of Parallel Resonance Involving Quality Factor in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right choice is (d) 46.128 For explanation: Resonant Frequency, \(\FRAC{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}\) = \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\) = \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz Inductive Reactance, XL = 2πfL = (6.28) (73.142 × 10^3)(4.7 × 10^-6) = 2.168 kΩ Quality factor Q = \(\frac{X_L}{R} = \frac{2.168 kΩ}{47}\) = 46.128.

Discussion

91.	For the circuit given below, the nature of the circuit is ____________(a) Inductive(b) Capacitive(c) Resistive(d) Both inductive as well as capacitiveThe question was asked by my school teacher while I was bunking the class.This interesting question is from Problems of Parallel Resonance Involving Quality Factor in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Correct option is (c) Resistive The BEST explanation: θ = 0° SINCE XL and XC are cancelling, which means at resonance the CIRCUIT is purely resistive.

Discussion

92.	For a parallel RLC circuit, the incorrect statement among the following is _____________(a) The bandwidth of the circuit decreases if R is increased(b) The bandwidth of the circuit remains same if L is increased(c) At resonance, input impedance is a real quantity(d) At resonance, the magnitude of input impedance attains its minimum valueI got this question in a national level competition.Asked question is from Problems of Parallel Resonance Involving Quality Factor topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct choice is (d) At resonance, the magnitude of input impedance attains its minimum VALUE The explanation is: BW = 1/RC It is clear that the BANDWIDTH of a parallel RLC circuit is independent of L and decreases if R is increased. At resonance, imaginary part of input impedance is zero. Hence, at resonance input impedance is a REAL quantity. In parallel RLC circuit, the admittance is minimum, at resonance. Hence the magnitude of input impedance attains its maximum value at resonance.

Discussion

93.	For the series circuit given below, the value of the voltage across the capacitor and inductor are _____________(a) VC = 16.306 V; VL = 16.306 V(b) VC = 11.268 V; VL = 11.268 V(c) VC = 16.306 V; VL = 16.306 V(d) VC = 14.441 V; VL = 14.441 VI have been asked this question at a job interview.This intriguing question comes from Problems of Parallel Resonance Involving Quality Factor in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right option is (c) VC = 16.306 V; VL = 16.306 V Best explanation: Resonant Frequency, \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}\) = \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\) = \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz Inductive REACTANCE, XL = 2πfL = (6.28) (73.142 × 10^3)(4.7 × 10^-6) = 2.168 kΩ Capacitive Reactance, XC = \(\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}\) = \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ ZEQ = R = 47 Ω IT = \(\frac{V_{in}}{Z_{EQ}}= \frac{V_{in}}{R} = \frac{0.3535}{47}\) = 7.521 mA ∴ Voltage ACROSS the capacitor, VC = XCIT = (2.168 kΩ)(7.521 mA) = 16.306 V ∴ Voltage across the INDUCTOR, VL = XLIT = (2.168 kΩ)(7.521 mA) = 16.306 V.

Discussion

94.	For the series resonant circuit given below, the value of the total current flowing through the circuit is ____________(a) 7.521 mA(b) 6.327 mA(c) 2.168 mA(d) 9.136 mAThis question was addressed to me during an interview for a job.This question is from Problems of Parallel Resonance Involving Quality Factor in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right OPTION is (a) 7.521 mA Easy explanation: Resonant Frequency, \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}\) = \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\) = \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz Inductive Reactance, XL = 2πfL = (6.28) (73.142 × 10^3)(4.7 × 10^-6) = 2.168 kΩ Capacitive Reactance, XC = \(\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}\) = \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ ZEQ = R = 47 Ω IT = \(\frac{V_{in}}{Z_{EQ}}= \frac{V_{in}}{R} = \frac{0.3535}{47}\) = 7.521 mA.

Discussion

95.	For the series resonant circuit given below, the value of the inductive and capacitive reactance is _________(a) XL = 5.826 kΩ; XC = 5.826 kΩ(b) XL = 2.168 kΩ; XC = 2.168 kΩ(c) XL = 6.282 kΩ; XC = 6.282 kΩ(d) XL = 10.682 kΩ; XC = 10.682 kΩI have been asked this question in my homework.I'm obligated to ask this question of Problems of Parallel Resonance Involving Quality Factor topic in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT OPTION is (c) XL = 6.282 kΩ; XC = 6.282 kΩ The BEST I can explain: Resonant Frequency, \(\FRAC{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}\) = \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\) = \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz Inductive Reactance, XL = 2πfL = (6.28) (73.142 × 10^3)(4.7 × 10^-6) = 2.168 kΩ Capacitive Reactance, XC = \(\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}\) = \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ.

Discussion

96.	For the series resonant circuit given below, the value of the equivalent impedance of the circuit is ________(a) 55 Ω(b) 47 Ω(c) 64 Ω(d) 10 ΩI got this question in a national level competition.I want to ask this question from Problems of Parallel Resonance Involving Quality Factor topic in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right answer is (b) 47 Ω Easy explanation: Resonant Frequency, \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}\) = \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\) = \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz Inductive Reactance, XL = 2πfL = (6.28) (73.142 × 10^3)(4.7 × 10^-6) = 2.168 kΩ Capacitive Reactance, XC = \(\frac{1}{2πfC} = \frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})}\) = \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ We see that, XC = XL are equal, along with being 180° out of phase. Hence the net reactance is ZERO and the total impedance equal to the resistor. ∴ ZEQ = R = 47 Ω.

Discussion

97.	For the series resonant circuit shown below, the value of the resonant frequency is _________(a) 10.262 kHz(b) 44.631 kHz(c) 50.288 kHz(d) 73.412 kHzI had been asked this question in my homework.My doubt stems from Problems of Parallel Resonance Involving Quality Factor in portion Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The correct option is (d) 73.412 kHz Easiest explanation: Resonant FREQUENCY, FR = \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(4.7×10^{-3})(0.001×10^{-6})}}\) = \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\) = \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz.

Discussion

98.	For the parallel resonant circuit given below, the cut-off frequencies are ____________(a) ∆f1 = 2.389 MHz; ∆f2 = 441.124 MHz(b) ∆f1 = 1.256 MHz; ∆f2 = 1.244 MHz(c) ∆f1 = 5.658 MHz; ∆f2 = 6.282 MHz(d) ∆f1 = 3.656 MHz; ∆f2 = 8.596 MHzThis question was posed to me by my school teacher while I was bunking the class.The query is from Problems of Parallel Resonance Involving Quality Factor topic in division Resonance & Magnetically Coupled Circuit of Network Theory
Answer» The CORRECT option is (b) ∆f1 = 1.256 MHz; ∆f2 = 1.244 MHz Easiest explanation: 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 Q = \(\frac{X_L}{R} = \frac{785.394}{7.85}\) = 100.05 ∴ ∆F = \(\frac{f}{Q} = \frac{1.25 × 10^6}{100.05}\) = 12.5 kHz Hence, BANDWIDTH = \(\frac{∆F}{2}\) = 6.25 kHz ∴ ∆f1 = f + \(\frac{∆F}{2}\) = 1.25 MHz + 6.25 kHz = 1.256 MHz ∴ ∆f1 = f – \(\frac{∆F}{2}\) = 1.25 MHz – 6.25 kHz = 1.244 MHz.

Discussion

99.	For the parallel resonant circuit given below, the current through the capacitor and inductor are _________(a) IC = 10.892 mA; IL = 12.732 mA(b) IC = 12.732 mA; IL = 10.892 mA(c) IC = 10.892 mA; IL = 10.892 mA(d) IC = 12.732 mA; IL = 12.732 mAThis question was posed to me during an interview.I want to ask this question from Problems of Parallel Resonance Involving Quality Factor in section Resonance & Magnetically Coupled Circuit of Network Theory
Answer» Right answer is (d) IC = 12.732 mA; IL = 12.732 mA 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. IC = \(\frac{V_A}{X_C} \) = \(\frac{10}{785.417} \) = 12.732 mA IL = \(\frac{V_A}{X_L} \) = \(\frac{10}{785.394} \) = 12.732 mA.

Discussion

100.	For the parallel resonant circuit, the bandwidth of the circuit is ____________(a) ±6.25 kHz(b) ±8.56 kHz(c) ±10.35 kHz(d) ±6.37 kHzI got this question in exam.I'm obligated to ask this question of Problems of Parallel Resonance Involving Quality Factor in chapter Resonance & Magnetically Coupled Circuit of Network Theory
Answer» RIGHT choice is (a) ±6.25 KHZ Explanation: 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 Q = \(\frac{X_L}{R} = \frac{785.394}{7.85}\) = 100.05 ∴ ∆F = \(\frac{f}{Q} = \frac{1.25 × 10^6}{100.05}\) = 12.5 kHz Hence, bandwidth = \(\frac{∆F}{2}\) = 6.25 kHz.

Discussion

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In the circuit shown, the voltage source supplies power which is _____________(a) Zero(b) 5 W(c) 10 W(d) 100 WI had been asked this question in homework.Origin of the question is Dot Convention in Magnetically Coupled Circuits in division Resonance & Magnetically Coupled Circuit of Network Theory

The time constant for the given circuit under resonance condition will be?(a) \(\frac{1}{9}\) s(b) \(\frac{1}{4}\) s(c) 4 s(d) 9 sThis question was posed to me in an interview.Enquiry is from Advanced Problems on Resonance topic in section Resonance & Magnetically Coupled Circuit of Network Theory