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51.	In the circuit given below, RI = 1 MΩ, RO = 10 Ω, A = 10^6 and VI = 1μV. Then the output voltage, input impedance and output impedance respectively are _________(a) 1 V, ∞ and 10 Ω(b) 1 V, 0 and 10 Ω(c) 1V, 0 and ∞(d) 10 V, ∞ and 10 ΩI have been asked this question during an internship interview.I need to ask this question from Advanced Problems on Network Theory in section Network Theory and Complex Circuit Diagram of Network Theory
Answer» The correct ANSWER is (a) 1 V, ∞ and 10 Ω To explain: VO (OUTPUT voltage) = AVI = 10^6 × 10^-6 = 1 V V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2 Here, I1 = 0 Z11 = \(\frac{V_1}{I_1} = \frac{V_O}{0}\) = ∞ Z22 = \(\frac{V_2}{I_2} = \frac{AV_I}{I_2}\) Or, Z22 = \(\frac{1}{I_2}\) = RO = 10 Ω.

52.	An AC source of RMS voltage 20 V with internal impedance ZS = (1+2j) Ω feeds a load of impedance ZL = (7+4j) Ω in the circuit given below. The reactive power is _________(a) 8 VAR(b) 16 VAR(c) 28 VAR(d) 32 VARThe question was asked in an online interview.Asked question is from Advanced Problems on Network Theory in portion Network Theory and Complex Circuit Diagram of Network Theory
Answer» Right answer is (b) 16 VAR Easy explanation: CURRENT I = \(\frac{V}{Z_L+Z_S} = \frac{20∠0°}{8+6j}\) = \(\frac{20}{\sqrt{8^2+6^2}} = \frac{∠0°}{∠arc tan⁡(\frac{3}{4})}\) = \(\frac{20}{10}\) ∠-arc tan⁡(\(\frac{3}{4}\)) = 2∠-arc tan⁡(\(\frac{3}{4}\)) POWER consumed by load = \|I\|^2ZL = 4(7+4j) = 28 + j16 ∴ The REACTIVE power = 16 VAR.

53.	For the circuit given below, the current I in the circuit is ________(a) –j1 A(b) J1 A(c) Zero(d) 20 AThis question was posed to me in an online quiz.The origin of the question is Advanced Problems on Network Theory in chapter Network Theory and Complex Circuit Diagram of Network Theory
Answer» Correct answer is (a) –J1 A The best I can explain: XEQ = sL + \(\frac{R×1/sC}{R+1/sC} = sL + \frac{R}{1+sRC}\) IO = \(\frac{V}{X_{EQ}}\) ∴ I = \(\frac{X_C}{X_C+R}\) IO = \(\frac{1/sC}{\frac{1}{sC}+R} × \frac{V}{\frac{sL(1+sRC)+R}{(1+sRC)}}\) = \(\frac{1}{1+sRC} × \frac{V}{\frac{sL(1+sRC)+R}{(1+sRC)}}\) = \(\frac{V}{sL(1+sRC)+R}\) = \(\frac{V}{j×10^3×20×10^{-3} (1+j×10^3×50×10^{-6}+1)}\) = \(\frac{V}{20j(1+j50×10^{-3})+1}\) = \(\frac{V}{20j-1+1} = \frac{20}{20j}\) = -j1 A.

54.	For the two circuits shown below, the relation between IA and IB is ________(a) IB = IA + 6(b) IB = IA + 2(c) IB = 1.5IA(d) IB = IAThe question was asked in a job interview.My question comes from Advanced Problems on Network Theory in section Network Theory and Complex Circuit Diagram of Network Theory
Answer» Correct option is (c) IB = 1.5IA Explanation: In the circuit of FIGURE (IB), TRANSFORMING 3A SOURCE into 18 V source, all SOURCES are 1.5 times of that in circuit (IA). Hence, IB = 1.5IA.

55.	A 1 μF capacitor is connected to 12 V batteries. The energy stored in the capacitor is _____________(a) 12 x 10^-6 J(b) 24 x 10^-6 J(c) 60 x 10^-6 J(d) 72 x 10^-6 JThis question was posed to me in class test.I want to ask this question from Advanced Problems on Network Theory topic in section Network Theory and Complex Circuit Diagram of Network Theory
Answer» Right ANSWER is (d) 72 x 10^-6 J The explanation: We know that, Energy, E = 0.5 CV^2 = 0.5 X 1 X 10^-6 X 144 = 72 x 10^-6 J.

56.	A two branch circuit has a coil of resistance R1, inductance L1 in one branch and capacitance C2 in the second branch. If R is increased, the dynamic resistance is going to ___________(a) Increase(b) Decrease(c) Remains constant(d) May increase or decreaseI had been asked this question in an online interview.The doubt is from Advanced Problems on Network Theory in portion Network Theory and Complex Circuit Diagram of Network Theory
Answer» CORRECT option is (b) Decrease For explanation: We know that, DYNAMIC resistance = \(\frac{L_1}{R_1 C_2}\) So, if R1 is increased, KEEPING Inductance and CAPACITANCE same, so The Dynamic resistance will decrease, as the denomination is INCREASING.

57.	D is the distance between the plates of a parallel plate capacitor. The dielectric constants are ∈1 and ∈2 respectively. The total capacitance is proportional to ____________(a) \(\frac{∈_1 ∈_2}{∈_1+∈_2}\)(b) ∈1 – ∈2(c) \(\frac{∈_1}{∈_2}\)(d) ∈1 ∈2This question was posed to me in unit test.This intriguing question comes from Advanced Problems on Network Theory topic in portion Network Theory and Complex Circuit Diagram of Network Theory
Answer» RIGHT CHOICE is (a) \(\frac{∈_1 ∈_2}{∈_1+∈_2}\) Easiest explanation: The combination is equal to two capacitors in series. So, C = \(\frac{\BIG[∈_0 ∈_1 \LEFT(\frac{A}{0.5d}\right)\Big]\Big[∈_0 ∈_2 \left(\frac{A}{0.5d}\right)\Big]}{∈_0 ∈_1\frac{A}{0.5d} + ∈_0 ∈_1\frac{A}{0.5d}}\) Hence, C is proportional to \(\frac{∈_1 ∈_2}{∈_1+∈_2}\).

58.	A capacitor, used for power factor correction in a single phase circuit decreases which of the following?(a) Power factor(b) Line current(c) Both Line current and Power factor(d) Neither Line current nor Power factorThe question was posed to me by my school principal while I was bunking the class.The doubt is from Advanced Problems on Network Theory topic in chapter Network Theory and Complex Circuit Diagram of Network Theory
Answer» RIGHT answer is (B) Line current The EXPLANATION: We KNOW that a capacitor is used to INCREASE the Power factor. However, with decrease in line current the power factor is increased. Hence line current decreases.