Explore topic-wise InterviewSolutions in .

The correct OPTION is (C) 27.6 RPM

The best I can EXPLAIN: Meter constant = \(\frac{Number \,of \,revolution}{Energy} = \frac{600 × 230 × 15 × 0.8}{1000}\) = 1656

∴ Speed in rpm = \(\frac{1656}{60}\) = 27.6 rpm.

Right answer is (a) Q = \(\frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\)

The BEST I can EXPLAIN: 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 = \(\frac{X_S}{R_S}= \frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\).

The correct choice is (b) 4.8 μH/rad

Explanation: At FULL SCALE POSITION, \(\frac{1}{2} I^2 \frac{dL}{dθ}\) = TC

 \(\frac{1}{2} 10^2 \frac{dL}{dθ}\) = 240 × 10^-6

∴ \(\frac{dL}{dθ}\) = 4.8 μH/rad.

Correct option is (c) 3 A ± 1.7 %

The explanation: I = [2 + \(\SQRT{2}\) sin (314t +30°) + 2\(\sqrt{2}\) cos (952t + 45°)]

Thermocouple MEASURE the rms value of current.

Irms = \(\Big[2^2 + \left(\frac{\sqrt{2}}{\sqrt{2}}\right)^2 + \left(\frac{2\sqrt{2}}{\sqrt{2}}\right)^2\Big]^{1/2} = \sqrt{9}\) = 3 A ± 1.7%.

The correct answer is (b) \(\FRAC{1.11V_A}{V_B}\)

For explanation I would SAY: 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 = \(\frac{V_A}{V_B/1.11} = \frac{1.11V_A}{V_B}\).

The correct ANSWER is (b) -28.57

To explain I would SAY: Im = 350/1 =350 A

Ip = \(((nI_s^2)^2 + (I_m^2)^2)^{0.5}\) = 490.05

Or, n = \(\frac{350}{7}\) = 50

∴ R = \(\frac{I_P}{I_S}= \frac{490.05}{7}\) = 70

∴ Percentage RATIO error = \(\frac{50-70}{70}\) × 100 = -28.57%.

The CORRECT choice is (c) 1.2

To explain: At equilibrium,

Kθ = \(\frac{1}{2} I^2\frac{DL}{dθ}\)

(25 × 10^-6) θ = \(\frac{1}{2} I^2 (3 – \frac{θ}{2}) × 10^{-6}\)

∴ 2 θ + \(\frac{θ}{2}\) = 3

Or, θ = 1.2.

Correct option is (d) \(\frac{2RC}{3}\)

The best I can EXPLAIN: The simplified circuit is:

Resistance FACED by C with the source shorted,

REQ = \(\frac{R × 2R}{3R} = \frac{2R}{3}\)

TIME constant of the circuit, τ = REQ × C

= \(\frac{2R}{3} × C = \frac{2}{3}\) RC.

The correct choice is (c) 11 H

The best explanation: Effective INDUCTANCE across AB terminals

= L1 + L2 + L3 – 2M12 – 2M13 + 2M23

= 4 + 5 + 6 – 2(1) – 2(3) + 2(2)

= 15 + 4 – 2 – 6 = 11 H.

Correct ANSWER is (C) Q is doubled

Best EXPLANATION: ω2L = 2ω1L

∴ Q2 = \(\FRAC{2ω_1 L}{R}\) = 2Q1

∴ Q is doubled.

Correct choice is (d) P DECREASES 4 times

Explanation: I1 = \(\FRAC{V}{\sqrt{R^2+ω_1^2 L^2}} = \frac{V}{ω_1 L}\)

For a high current coil, ωL >> R

I2 = \(\frac{V_1}{2ω_1 L} = \frac{I_1}{2}\)

∴ P2 = R (\(\frac{I_1}{2}\))^2 = \(\frac{P_1}{4}\)

Therefore, P decreases 4 times.

Right choice is (d) \(\frac{1}{2π\sqrt{6}}\) Hz

To explain: f = \(\frac{1}{2π\sqrt{L_{EQ} C}} \)

Here, LEQ = 2 + 2 + 2 × 1 = 6

So, C = 1 F

FR = \(\frac{1}{2π\sqrt{6 × 1}} = \frac{1}{2π\sqrt{6}}\) Hz.

Correct CHOICE is (c) ±15.8 Ω

The BEST explanation: Given, R1 = 100 ± 5 Ω

R2 = 150 ± 15 Ω

Now, R = R1 + R2

The probable errors in this case, R = ± (\(R_1^2+R_2^2)^{0.5}\) = ± 15.8 Ω.

The CORRECT option is (d) No such value exists

Easiest explanation: H (ω) = \(\frac{V_O}{V_i}\)

 = \(\frac{1}{1+jωRC}\)

∴ GAIN = \(\frac{1}{\sqrt{1 + ω^2 RC}}\)

For any value of ω, R and C gain is ≤ 1.

Right choice is (d) 400∠-90° V

The EXPLANATION is: Q = \(\FRAC{|X_L|}{R} = \frac{|X_C|}{R}\)

So, Q = \(\frac{20}{10}\) = 2

Rms voltage across capacitor, VCrms = QV∠-90°

Or, VCrms = 2 X 200∠-90°

Or, VCrms = 400∠-90° V.

Correct option is (a) 50 Ω

To explain I WOULD say: At MIDBAND frequency, Z = R, Y = \(\FRAC{1}{R}\)

Or, R = \(\frac{1}{20 × 10^{-3}}\) = 50 Ω.

Correct answer is (c) 8 cos 2t V

Best explanation: V2 = \(\FRAC{dI_2}{dt} + \frac{dI_1}{dt}\)

Or, V2 = \(\frac{dI_1}{dt}\)

∴ V2 = 8 cos2t V.

The correct option is (B) 16 cos2t V

Explanation: V1 = \(\frac{2dI_1}{DT} + \frac{1dI_2}{dt}\)

∴ V1 = \(\frac{2dI_1}{dt}\) = 16 cos2t V.

Correct choice is (b) \(\frac{1}{4π\sqrt{3}}\) HZ

The explanation is: LEFF = L1 + L2 + 2M

= 2 + 2 + 2 × 1

Or, LEFF = 6 H

At resonance, ωL = \(\frac{1}{ωC}\)

Or, ω = \(\frac{1}{\sqrt{L_{EFF} C}} = \frac{1}{\sqrt{12}}\)

Or, I = \(\frac{1}{4π\sqrt{3}}\) Hz.

The correct answer is (d) Inderminate due to insufficient DATA

To elaborate: When 2 COILS are connected in series, then effective inductance,

LEFF = L1 + L2 ± 2M

For this case, LEFF = L1 + L2 – 2M

However, L1 + L2 + 2M or L1 + L2 – 2M cannot be determined.

Hence, data is insufficient to calculate the VALUE of coupling coefficient (k).

The correct option is (a) 14 mH

The BEST I can explain: When 2 coils are connected in series, then EFFECTIVE inductance,

LEFF = L1 + L2 ± 2M

For this CASE, LEFF = L1 + L2 – 2M

= 5 + 10 – 2 × 0.5

= 15 – 1

= 14 mH.

The CORRECT CHOICE is (d) MLALB ≤ \(\sqrt{L_A L_B}\)

The best explanation: M = K \(\sqrt{L_A L_B}\)

Or, K = \(\frac{M}{\sqrt{L_A L_B}}\)

∴ K≤1

∴ \(\frac{M}{\sqrt{L_A L_B}}\) ≤ 1

Or, M ≤ \(\frac{M}{\sqrt{L_A L_B}}\).

The correct ANSWER is (c) 11 H

The best explanation: LEFF across AB = L1 + L2 + L3 – 2M12 – 2M13 + 2M23

= (4 + 5 + 6 – 2 × 1 -2 × 3 + 2 × 2)

= (15 – 2 – 6 + 4)

= (15 – 8 + 4)

= (15 – 4)

= 11.

The correct choice is (a) 375 Ω and 75 mH

To ELABORATE: Applying the usual balance condition relation,

Z1 Z4 = Z2 Z3

We have,(R1 + jL1 ω) \(\frac{R_4/jωC_4}{R_4+1/jωC_4}\) = R2 R3

Or, R1 R4 + jL1 ωR4 = R2 R3 + J R2 R3 R4 C4 ω

∴R1 = 2000 × \(\frac{750}{4000}\) = 375 Ω

∴ L1 = 2000 × 750 × 0.5 × 10^-6 = 75 mH.

The correct OPTION is (c) 242 W

Explanation: Watt-meter reading = CURRENT through CC × Voltage across PC × cos (phase angle).

IBR = ICC = \(\frac{220∠120°}{100°}\) = 2.2∠120°

VYB = VPC = 220∠-120°

w = 2.2∠120° × 220∠-120° × cos 240° = – 242 W.

The correct choice is (c) 27.6 RPM

To elaborate: METER constant = \(\FRAC{Number\, of\, revolution}{ENERGY} = \frac{600 × 230 × 15 × 0.8}{1000}\) = 1656

∴ Speed in rpm = \(\frac{1656}{60}\) = 27.6 rpm.

Right choice is (d) Bridge cannot be BALANCED with the given configuration

Best explanation: For Bridge to be balanced, the product of impedances of the OPPOSITE arm should be equal in MAGNITUDE as well as phase angle. Here Z3 Z2 ≠ Z1 Z4 for whatever CHOSEN VALUE. Therefore the Bridge cannot be balanced.

Right answer is (C) -1.0

To explain I would SAY: Turn compensation only alters RATIO error n=400

Ratio error = -0.5% = – \(\frac{0.5}{100}\) × 400 = -2

So, Actual ratio = R = n+1 = 401

Nominal Ratio KN = \(\frac{400}{1}\) = 400

Now, if the number of turns are reduced by one, n = 399, R = 400

Ratio error = \(\frac{K_N-R}{R} = \frac{200-200}{200}\) = 0.

Correct choice is (c) 8.33%

To elaborate: Here, R1 and R2 are in parallel.

Then, \(\frac{1}{R} = \frac{1}{R_1}+ \frac{1}{R_2}\)

Or, R = \(\frac{50}{15}\) kΩ

∴\(\frac{△R}{R} = \frac{△R_1}{R_1^2} + \frac{△R_2}{R_2^2}\)

And △R1 = 0.5 × 10^3,△R2 = 0.5 × 10^3

∴ \(\frac{△R}{R} = \frac{10 × 10^3}{3 × 10 × 10^3} × \frac{0.5 × 10^3}{10 × 10^3} + \frac{10}{3} × \frac{10^3}{5 × 10^3} × \frac{0.5 × 10^3}{5 × 10^3}\)

= \(\frac{0.5}{30} + \frac{1}{15}= \frac{2.5}{30}\) = 8.33%.

Right option is (b) ZERO

For explanation I would SAY: In WHEATSTONE bridge, balance condition is

R1R3 = R2R4

Here, R1 = 5, R2 = 10, R3 = 16, R4 = 8

And when the Wheatstone bridge is BALANCED then, at Vout voltage will be Zero.

Correct choice is (b) Q SIN (2t + 15)

To EXPLAIN: \(\FRAC{f_y}{f_x} = \frac{x-peak}{y-peak}\)

Here, x-peak = 1 and y-peak = 2

∴ y(t) = Q sin (2t + 15).

The correct choice is (c) 14.95 A

The explanation: Voltage DROP per unit LENGTH = \(\frac{1.53}{42}\) = 0.036 V/cm

Voltage drop across 83 cm length = 0.036 × 83 = 2.99 V

∴ CURRENT through resistor, I = \(\frac{2.99}{0.2}\) = 14.95 A.

Correct option is (c) 40 Ω

The explanation is: Z = 20 + j20

V = VR = j (VL – VC)

Given, VL = 2 VC

Or, ZL = 2 ZC

Or, ZL – ZC = 20

Or, 2 ZC – ZC = 20

Or, ZC = 20 Ω

Or, ZL = 2ZC = 40 Ω.

The correct choice is (d) 90∠-32.44°

For explanation I WOULD say: 3VP IP cosθ = 1500

Or, 3\((\frac{V_L}{\SQRT{3}}) (\frac{V_L}{\sqrt{3} Z_L})\) cos θ = 1500

Or, ZL = \(\frac{V_L^2}{1500} = \frac{400^2 (0.844)}{1500}\) = 90 Ω

And θ = ∠-arc cos⁡(0.844)

= ∠-32.44°.

The correct answer is (a) 4

The best explanation: IX = \(\frac{48}{12}\) = 4 A

IY = \(\frac{22}{12}\) = 1.8 A

IZ = \(\frac{14.4}{12}\) = 1.2 A

CURRENT REQUIRED to below the fuse = 20 A

∴ Additional BULBS must draw current = 20 – (4 + 18 + 1.2)

= 20 – 7 = 13

∴ Number of additional bulbs required = \(\frac{13}{3}\) = 4.33

So, 4 additional bulbs are required.

Right ANSWER is (a) -65 V

Easy explanation: Going from 10 V to 0 V, we get,

50 + 10 + E + 5 = 0

Or, E + 65 = 0

Hence, E = -65 V.

Right choice is (C) 4.358

Easiest EXPLANATION: QIN = \(\sqrt{\frac{L}{CR^2} – 1}\)

= \(\sqrt{\frac{1}{2 × 5^2× 10^{-3} – 1}}\)

= \(\sqrt{\frac{1}{50 × 10^{-3} – 1}}\)

= \(\sqrt{19}\) = 4.358.

Correct answer is (C) \(\frac{1}{4π}\) Hz

To explain I would say: IEQ = L1 + L2 + 2M

LEQ = 1 + 2 + 2 × \(\frac{1}{2}\) = 4 H

∴ FO = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{2π\sqrt{4 × 1}} \)

= \(\frac{1}{4π}\) Hz.

The correct option is (b) 3.33 A

Explanation: As the capacitors are in PARALLEL, then the voltage V is given by,

V =\(\frac{1}{C_2} \int I_2\,dt \)

∴ I2 = C2 \(\frac{DV}{dt}\)

That is, \(\frac{I_1}{I_2}= \frac{C_1}{C_2}= \frac{50}{100} = \frac{1}{2}\) ………………….. (1)

Also, I1 + I2 = 5 A ………………………….. (2)

Solving (1) and (2), we GET, I2 = 3.33 A.

Correct ANSWER is (c) 1.67 A

Explanation: As the CAPACITORS are in parallel, then the voltage V is given by,

V = \(\frac{1}{C_1}\int I_1\,dt \)

∴ I1 = C1 \(\frac{dV}{dt}\)

That is, \(\frac{I_1}{I_2}= \frac{C_1}{C_2}= \frac{50}{100} = \frac{1}{2}\) ………………….. (1)

Also, I1 + I2 = 5 A ………………………….. (2)

Solving (1) and (2), we get, I1 = 1.67 A.