InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 351. |
The frame of electric motor is connected to three plates having earthing resistance of 10 Ω, 20 Ω and 30 Ω respectively. The materials of the three plates are copper, aluminium and iron respectively. The percentage of energy dissipated by 10 Ω earthing plate |
| Answer» The parallel combination of 30 Ω and 20 Ω is 12 Ω. Since 12 Ω and 10 Ω are in parallel, the 10 ohm plate draws more than 50% current and dissipates more than 50% energy. | |
| 352. |
In the circuit of figure, the power consumed in resistance R is measured when one source is acting at one time. These values are 18 W, 50 W and 98 W. When all source act together the maximum and minimum power can be |
| Answer» Let R = 1 Ω. Then I1 = 32 A, I2 = 52 A and I3 = 72 A. (I1 + I2 + I3)2 R = (152)2 R = 450 Ω. | |
| 353. |
An RL series circuit has an impedance of 20 Ω when frequency is 25 Hz. At = 50 Hz, the impedance will be |
| Answer» Z = R2 + XL2. When frequency is doubled, XL becomes twice but R remains the same. | |
| 354. |
In figure, the voltmeter is ideal. The transformer has two identical windings with perfect coupling. The reading of voltmeter is |
| Answer» The secondary voltage is equal to primary voltage and the two voltages oppose each other. | |
| 355. |
The resistance of the circuit shown is figure is |
| Answer» The parallel combination of 3, 5 and 4 ohm is short-circuited. Therefore their net resistance is zero. | |
| 356. |
In an R-C circuit, the impedance is 40 Ω at a frequency of 100 Hz. At 200 Hz the impedance should be |
| Answer» Since frequency is doubled, becomes one-fourth but R2 remains the same. | |
| 357. |
An RLC series circuit is fed form 100 V ac supply. Inductance is 1 H and Q = 7.5. At resonance, the voltage across inductance is |
| Answer» Voltage across inductance at resonance = QV = 7.5 X 100 V. | |
| 358. |
When cells having different emfs are connected in parallel, a circulating current flows. |
| Answer» To avoid circulating current cells in parallel should be identical. | |
| 359. |
The self inductances of two coils are 27 H and 3 H. If the winding or coils is such that 50% of flux of one links the other, the mutual inductance is |
| Answer» M = 0.5(27 x 3) | |
| 360. |
In a lead acid cell, the positive plate has spongy lead. |
| Answer» The positive plate has lead peroxide. | |
| 361. |
The rms voltage measured across each of the circuit element in figure is as shown, when the circuit is excited by a sinusoidal voltage E, the rms value of the source voltage is |
| Answer» Voltage across C = -10jv Voltage across L = + 14 jv source voltage Es = VR + jVL - jVC = 3 + j. 14 - j. 10 3 + j4 rms of Es = VR2(VL - VC)2 = 32 + 42 5 volt. | |
| 362. |
For H() to be positive real, the condition Re[H(ω)] ≥ 0 for 0 ≤ ω ≤ ∞ is |
| Answer» H(s) must possess two more properties if it is positive real. | |
| 363. |
In the circuit, shown the current I marked is given by |
| Answer» When 9 V source is connected, 2V source is short circuited [Superposition Theorem]. I' = 9/4.5 = 2 A, I1 = 1.5 A when 2 volt source is connected, then Req = 4.5 I2 = 0.22 A I = I1 + I2 = 0.66 A. | |
| 364. |
In a specimen of cast iron, a field strength of 400 AT/m cause a flux density of 0.6 T. In a specimen of silicon steel, the same value of H would cause B to be |
| Answer» Since μr for silicon steel is more than that of cast iron, flux density will be more than 0.6 T. | |
| 365. |
Volt is equivalent to |
| Answer» | |
| 366. |
A two branch parallel tuned circuit has a capacitance C in one of the two branches and resistance R in second branchs. As C is decreased the dynamic resistance |
| Answer» Dynamic resistance = and increases as C is decreased. | |
| 367. |
A system function the resonant frequency in rad/sec and bandwidth in rad/sec is given by |
| Answer» ωn = 10, ωr 10 B = a 4. | |
| 368. |
The reading of the voltmeter in figure will be __________ volt. |
| Answer» Good lamp will short the circuit. | |
| 369. |
An RLC series circuit has R = 8 Ω, X = 8 Ω and X = 8 Ω Its impedance is |
| Answer» Z = 8 + j(8 - 8). | |
| 370. |
Two similar mutually coupled coils have a total inductance of 900 mH and coefficient of coupling 0.5. The self inductance of each coils is |
| Answer» M = kL1L2, L1 = L2 = L, M = kL. Total inductance = L1 + L2 + 2M = 2L + 2kL = 900 mH or L(2 + 2 x 0.5) = 900 or L = 300 mH. | |
| 371. |
Flux linkages ψ and voltages are related as |
| Answer» | |
| 372. |
Find R for maximum power transfer |
| Answer» Using y - Δ transformation Req = (9 || 8) 11 (9 || 18) 3Ω. | |
| 373. |
A low-pass filter frequency responce H (ω) = A (ω) (ω) does not produce any phase distortion, if |
| Answer» For distortion less transmission line . | |
| 374. |
In a tungsten filament lamp |
| Answer» Resistance at the time of switching on is minimum and increases due to increase of temperature. | |
| 375. |
In figure, capacitor is uncharged. The switch is closed at = 0 at = 0 / is |
| Answer» RC = 10-3, = 0.1 e-1000t. Therefore at t = 0+ equals -100 A/s. | |
| 376. |
In a series RLC circuit, R = 2 kΩ, L = 1H and The resonant frequency is |
| Answer» . | |
| 377. |
The function of a choke in a fluorescent lamp is |
| Answer» At the time of starting the choke provides a high voltage. During working there is voltage drop across choke so that lamp gets proper voltage. | |
| 378. |
A series resonant circuit is fed by a voltage having rms value V. At resonance, the voltage across inductance V and voltage across capacitance V are related as |
| Answer» At resonance XL = XC. Therefore IXL = IXC or VL = VC. | |
| 379. |
Calculate () for ≥ 0, assuming the switch has been in position A for a long time at = 0, the switch is moved to position B. |
| Answer» When switch is in A position, C will be open, t < 0, i(0-) = = 1.11 mA, VC(0-) = 3 k x 3.33 volt at t > 0 i(0+) = 1.11 mA, i(t) = A + B e-at where 5 x 103 at t = 0, A + B = 1.11 t = ∞, A = 0 i(t) = 1.11 e-5 x 103(t) . | |
| 380. |
The function Z() has a pole at infinity if |
| Answer» If numerator is one degree higher than denominator z(s) tends to infinity as s tends to infinity. | |
| 381. |
An input voltage V() = 10 cos ( + 10°) + 10 cos (2 + 10°) is applied to a series combination of resistance R = 1 kΩ inductance L = 1H. The resulting steady-state current () in ampere is |
| Answer» The steady state current i(t) = i1(t) + i2(t) Hence ω1, ω2 = 2. | |
| 382. |
The switch in figure is initially closed. At = 0 the switch is opened. At = 0, the current through inductance is |
| Answer» At t = ∞ inductance behaves as open circuit. | |
| 383. |
A constant low pass T section filter has a characteristic impedance of 600 at zero frequency. At = the characteristic impedance is |
| Answer» Z0T = R0(1 - f2 / fc2)0.5. At f = f0. Z0T = 0. | |
| 384. |
For a network having 1 Ω resistor and 1 F capacitor in series Z() = |
| Answer» | |
| 385. |
In given figure the switch was closed for a long time before opening at = 0. The voltage V at = 0 is |
| Answer» at t = 0, L will act as short circuit hence I = 2.5 A. Vx = 2.5 x 20 50v, but in opposite to current source. Hence Vx = - 50v. | |
| 386. |
The ac bridge shown in the figure is balanced if and Z is equal to |
| Answer» = 375 ∠ - 40 - 30. | |
| 387. |
S is open for a long time and steady state is reached S is closed at = 0. Let the current in the inductor At = 0, is given by |
| Answer» If S is open, for long time then C will be short circuited, voltage across 2Ω resistor = 2 volt (V0) | |
| 388. |
In an air cored coil, a current of 0.7 A causes a flux of 5 x 10 Wb. If the core is made of cast iron and the coil is similar, the same current can set up a flux of about |
| Answer» The permeability of cast iron is about 100. | |
| 389. |
The resonant frequency of the circuit shown in figure is |
| Answer» Leq = L1 + L2 + 2m = 1 + 2 + 2 x = 4 . | |
| 390. |
For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is V() = sin 10 . The output voltage V() is equal to |
| Answer» V0(t) = i(t) . jωc | |
| 391. |
Bus bars are always meant for very heavy currents. |
| Answer» Bus bars are used for heavy as well as light currents. | |
| 392. |
A thermal power plant has an efficiency of 0.4. If 0.6 kg of coal is burnt to generate 1 kWh, the calorific value of coal is |
| Answer» Therefore, calorific value = | |
| 393. |
An RC series circuit has R = 20 Ω and X = 20 Ω. Then Z = |
| Answer» Z = 202 + 202 = 28.28 ohm. | |
| 394. |
An inductance having X = 5 Ω and a capacitance having X = 5 Ω are connected in parallel across 100 V, 50 Hz supply. The current drawn from source is |
| Answer» The total current is | |
| 395. |
The circuit shown in figure, will act as an ideal current source with respect to terminals A and B, when frequency is |
| Answer» to find the frequency. Put XC + XL = 0. | |
| 396. |
The realization of a term in the impedance function will give |
| Answer» Impedance of capacitance | |
| 397. |
An RL admittance function can also be realized as |
| Answer» RL admittance function and RC impedance function have similar forms. | |
| 398. |
Current having waveform in figure flows through 10 Ω resistance. Average power is |
| Answer» RMS current = = 5.773 A, P = I2R = 333.3 W. | |
| 399. |
S is closed for a long time and steady state is reached. S is opened at (0). The voltage marked V is V at = 0 and V at = ∞. Then value of V and V are respectively |
| Answer» For Thevenin Equivalent Resistance, v = 0, I = 0 . | |
| 400. |
Two voltages are = 100 sin (ω + 15°) and = 60 cosω, then |
| Answer» v1 = 100 sin (ωt + 15°) and v2 = 60 cos ωt = 60 sin (ωt + 90°). Hence, v2 is leading v1 by (90 - 15) = 75°. | |