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This section includes 785 Mcqs, each offering curated multiple-choice questions to sharpen your Electronic Devices Circuits knowledge and support exam preparation. Choose a topic below to get started.
1. |
Q = 2000 VAR, pf = 0.9 (leading). The power in complex form is |
A. | 4129.8 j2000 VA |
B. | 2000 j4129.8 VA |
C. | 2000 j41.29.8 VA |
D. | 4129.8 j2000 VA |
Answer» E. | |
2. |
P = 269 W, Q = 150 VAR (capacitive). The power in the complex form is |
A. | 150 – j269 VA |
B. | 150 + j269 VA |
C. | 269 – j150 VA |
D. | 269 + j150 VA |
E. | . The power in the complex form isa) 150 – j269 VAb) 150 + j269 VAc) 269 – j150 VAd) 269 + j150 VA |
Answer» D. 269 + j150 VA | |
3. |
In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66 sin (1000t) V, i(t) = 2.3sin (1000t + 68.3) 3 A. The nature of the elements would be |
A. | R C |
B. | L C |
C. | R L |
D. | R R |
Answer» B. L C | |
4. |
In the bridge shown, Z1 = 300 ohm, Z2 = 400 – j300 ohm, Z3 = 200 + j100 ohm. The Z4 at balance isa) 400 + j300 ohmb) 400 – j300 ohmc) j100 ohmd) -j900 ohm Circuit for Q.6 and Q.76. i1(t) = ? |
A. | 400 + j300 ohmb) 400 – j300 ohmc) j100 ohmd) -j900 ohm Circuit for Q.6 and Q.76. i1(t) = ?a) 2.36 cos (4t 41.07) A |
B. | 400 – j300 ohmc) j100 ohmd) -j900 ohm Circuit for Q.6 and Q.76. i1(t) = ?a) 2.36 cos (4t 41.07) Ab) 2.36 cos (4t 41.07) A |
C. | j100 ohmd) -j900 ohm Circuit for Q.6 and Q.76. i1(t) = ?a) 2.36 cos (4t 41.07) Ab) 2.36 cos (4t 41.07) Ac) 1.37 cos (4t 41.07) A |
D. | -j900 ohm Circuit for Q.6 and Q.76. i1(t) = ?a) 2.36 cos (4t 41.07) Ab) 2.36 cos (4t 41.07) Ac) 1.37 cos (4t 41.07) Ad) 2.36 cos (4t 41.07) AView Answer |
Answer» C. j100 ohmd) -j900 ohm Circuit for Q.6 and Q.76. i1(t) = ?a) 2.36 cos (4t 41.07) Ab) 2.36 cos (4t 41.07) Ac) 1.37 cos (4t 41.07) A | |
5. |
Vc(t) = ? |
A. | 0.89 cos (1000t – 63.43) V |
B. | 0.89 cos (1000t + 63.43) V |
C. | 0.45 cos (1000t + 26.57) V |
D. | 0.45 cos (1000t – 26.57) V |
Answer» B. 0.89 cos (1000t + 63.43) V | |
6. |
i(t) = ? |
A. | 20 cos (300t + 68.2) A |
B. | 20 cos(300t – 68.2) A |
C. | 2.48 cos(300t + 68.2) A |
D. | 2.48 cos(300t – 68.2) A |
Answer» E. | |
7. |
Given V1 = 25∠0° V and V2 = 10∠36.87° V are connected in series. Find the resultant voltage Vs = V1 + V2.(Take cos 36.87° = 0.8, and sin 36.87° = 0.6) |
A. | (17 + j31) V |
B. | (6 + j33) V |
C. | (33 + j6) V |
D. | (33 + j19) V |
Answer» D. (33 + j19) V | |
8. |
In the circuit shown in the figure, the value of node voltage |
A. | 22 + j2 V |
B. | 2 + j 22 V |
C. | 22 – j 2 V |
D. | 2 – j 22 V |
Answer» E. | |
9. |
In an RLC series circuit, if the inductive reactance voltage exceeds the capacitive reactance voltage, the reactive component of power will: |
A. | be in-phase with the current |
B. | lead the current by 90° |
C. | lag the current by 90° |
D. | lead the voltage by 90° |
Answer» D. lead the voltage by 90° | |
10. |
In a purely inductive circuit the current ______ the voltage by _______. |
A. | lags, 0° |
B. | leads, 90° |
C. | lags, 90° |
D. | lags, 45° |
Answer» D. lags, 45° | |
11. |
In a series RL circuit, the two resistors of R = 5 Ω and 10 Ω, are connected in series with the inductor of L H. The supply voltage is 50 V. If the power consumed by the 5 Ω resistor is 10 W, then power factor of the circuit is: |
A. | 0.828 |
B. | 0.424 |
C. | 0.567 |
D. | 0 |
Answer» C. 0.567 | |
12. |
In the given network \({V_1} = 100\angle 0^\circ \;V,\;{V_2} = 100\angle - 120^\circ \;V,\;{V_3} = 100\angle + 120^\circ \;V\). the phasor current \(i\) (in Ampere) is |
A. | \(173.2\angle - 60^\circ\) |
B. | \(173.2\angle - 120^\circ\) |
C. | \(100.0\angle - 60^\circ\) |
D. | \(100.0\angle - 120^\circ\) |
Answer» B. \(173.2\angle - 120^\circ\) | |
13. |
Consider the circuit shown belowThe power factor of the above circuit as seen by the source is |
A. | 0.988 LAG |
B. | 0.988 LEAD |
C. | 0.235 LAG |
D. | 0.235 LEAD |
Answer» C. 0.235 LAG | |
14. |
A two-terminal black box contains one of the R.L.C. elements. The black box is connected to a 220 V AC supply. The current through the source is I. When a capacitance of 0.1 F is inserted in series between the source and the box, the current through the source is 2I. The element is: |
A. | resistance |
B. | an inductance |
C. | a capacitor of 0.5 F |
D. | it is not possible to determine the element |
Answer» C. a capacitor of 0.5 F | |
15. |
If the true power is 120 W and the power factor is 0.68, then what will be the apparent power (in VA)? |
A. | 81.6 |
B. | 17.6 |
C. | 176.47 |
D. | 103.2 |
Answer» D. 103.2 | |
16. |
Identify the type / status of circuit the below diagram represents. |
A. | Inductive |
B. | Capacitive |
C. | At resonance |
D. | Resistive |
Answer» C. At resonance | |
17. |
A voltage wave is given by V = 200 sin 314t. What is the maximum value? |
A. | 1600 V |
B. | 800 V |
C. | 400 V |
D. | 200 V |
Answer» E. | |
18. |
If \(V_1\) and \(V_3\) are the rms values of the fundamental and third harmonics of an alternating quantity, then the rms value of |
A. | \(\sqrt{V_1^2 + V_3^2}\) |
B. | \(V_1/V_3\) |
C. | \(V_1+V_3\) |
D. | \(V_1-V_3\) |
Answer» B. \(V_1/V_3\) | |
19. |
If admittance Y = a + jb, then a = ? |
A. | Conductance |
B. | Susceptance |
C. | Resistance |
D. | Reactance |
Answer» B. Susceptance | |
20. |
______ is defined as the no. of the cycle completed by an alternating quantity in one second. |
A. | Frequency |
B. | Cycle |
C. | Amplitude |
D. | Instantaneous value |
Answer» B. Cycle | |
21. |
Let the voltage and current in an element be \(V = V_m \angle \theta_V\) and \(I = I_m \angle \theta_I\). The complex power delivered to the element is defined as: |
A. | \(S = \frac{V_mI_m}{2} \angle(\theta_V - \theta_I)\) |
B. | \(S = V_m I_m \angle (\theta_V - \theta_I)\) |
C. | \(S = V_m I_m^*\) |
D. | \(S = \frac{V_mI_m}{2} \ \angle (\theta_I-\theta_V)\) |
Answer» C. \(S = V_m I_m^*\) | |
22. |
For a series RL circuit, i(t) = √2 sin (ωt – 45°). If ωL = 1 Ω, the value of R is |
A. | 1 Ω |
B. | 3 Ω |
C. | √3 Ω |
D. | 3√3 Ω |
Answer» B. 3 Ω | |
23. |
Crest or amplitude factor of an alternating quantity is the ratio of: |
A. | (rms value)/(maximum value) |
B. | (maximum value)/(average value) |
C. | (rms value)/(average value) |
D. | (maximum value)/(rms value) |
Answer» E. | |
24. |
A 230 V RMS source supplies power to two loads connected is parallel. The final load draws 10 kW at 0.8 leading power factor & the second one draws 10 kVA at 0.8 lagging power factor. The complex power delivered by the source is |
A. | (18 + j1.5) kVA |
B. | (18 – j1.5) kVA |
C. | (20 + j1.5) kVA |
D. | (20 – j1.5) kVA |
Answer» C. (20 + j1.5) kVA | |
25. |
In a given network (shown in Figure), a steady state is reached with switch k open. At t = 0, which is closed. Determine the values of I1, I2 and I3 at t = 0+ |
A. | 1 A, 1/3 A, 0A |
B. | 1/3 A, 1/3 A, 1/3 A |
C. | 1 A, 0A, 0A |
D. | 1 A, 1 A, 0 A |
Answer» B. 1/3 A, 1/3 A, 1/3 A | |
26. |
A phasor1. may be a scalar or a vector2. is a time-dependent quantity3. is a complex quantityWhich of the above statements are correct? |
A. | 1, 2 and 3 |
B. | 1 and 2 only |
C. | 1 and 3 only |
D. | 2 and 3 only |
Answer» E. | |
27. |
In a tank, where V is the initial voltage to which the capacitor is charged, C is the capacitance of the tank circuit and L is the inductance of the circuit, the peak value of the circulating current is given by which of the following expressions? |
A. | \(I = V\sqrt {L/C}\) |
B. | \(I = V\sqrt {LC}\) |
C. | \(I= \frac{V}{{\sqrt {LC} }}\) |
D. | \(I = V\sqrt {\frac{C}{L}}\) |
Answer» E. | |
28. |
A supply of 250 V, 50 Hz is applied to a series RC circuit. If the power absorbed by the resistor be 400 W at 160 V, the value of the capacitor C will be nearly |
A. | 30.5 μF |
B. | 41.5 μF |
C. | 64.0 μF |
D. | 76.8 μF |
Answer» C. 64.0 μF | |
29. |
A 50 kW load at a power factor of 0.8 lagging is supplied from a single-phase AC supply. Reactive power drawn from the source is: |
A. | 37.5 kVAr lagging |
B. | 625 kVAr Leading |
C. | 37.5 kVAr leading |
D. | 625 kVAr lagging |
Answer» B. 625 kVAr Leading | |
30. |
In purely resistive circuits, the ________ and the applied ________ are ________ phase with each other |
A. | Current, voltage, out of |
B. | Frequency, voltage, in |
C. | Frequency, voltage, out of |
D. | Current, voltage, in |
Answer» E. | |
31. |
If the rms value and average value of half-wave-rectified alternating current are \(\frac{{{I_m}}}{2}\) and \(\frac{{{I_m}}}{\pi }\) respectively, then the form factor of the half-wave rectified current will be given as: |
A. | \(\frac{2}{\pi }\) |
B. | 2π |
C. | \(\frac{\pi }{2}\) |
D. | π2 |
Answer» D. π2 | |
32. |
A voltage is said to be alternating when it changes in ___________. |
A. | neither magnitude nor direction |
B. | magnitude only |
C. | direction only |
D. | both magnitude and direction |
Answer» E. | |
33. |
For a parallel RLC circuit, the critically damped condition for \(\alpha = \frac{1}{{2RC}}\) damping factor and \({\omega _0} = \frac{1}{{\sqrt {LC} }}\) resonant frequency is |
A. | α = 1/ω0 |
B. | α > ω0 |
C. | α < ω0 |
D. | α = ω0 |
Answer» E. | |
34. |
In the waveform shown, RMS value of voltage is |
A. | \(\frac{{{200}}}{\pi }\) |
B. | \(\frac{{{100}}}{\pi }\) |
C. | 200 V |
D. | 100 V |
Answer» E. | |
35. |
If an AC circuit is supplied with an active power of 500 W and reactive power 500 VAR, then the load power factor of the circuit is: |
A. | 0.5 |
B. | 1 |
C. | 1/√2 |
D. | 0 |
Answer» D. 0 | |
36. |
A resistance and a coil are connected in series and supplied from a single phase, 100 V, 50 Hz ac source as shown in the figure below. The rms values of possible voltages across the resistance and coil respectively, in volts, are |
A. | 65, 35 |
B. | 50, 50 |
C. | 60, 90 |
D. | 60, 80 |
Answer» E. | |
37. |
A series RLC circuit consisting of R = 10 Ω, XL = 20 Ω and XC = 20 Ω, is connected across an AC supply of 100 V (rms). The magnitude and pulse angle (with respect to supply voltage) of the voltage across the induction coil are respectively. |
A. | 100 V; 90° |
B. | 100 V; -90° |
C. | 200 V; -90° |
D. | 200 V; 90° |
Answer» E. | |
38. |
Find the average value of sinusoidal alternating current having peak value of 50 A. |
A. | 63.66 A |
B. | 31.83 A |
C. | 12.50 A |
D. | 25.00 A |
Answer» C. 12.50 A | |
39. |
If the voltage and current in a circuit are given by:ν = 10 sin (ωt – π/6) andi = 10 sin (ωt + π/6),The power consumed is given as: |
A. | 100 watts |
B. | 50 watts |
C. | 86.6 watts |
D. | 25 watts |
Answer» E. | |
40. |
Determine the value of phase current (in A) for a balanced delta connected system, when the value of the current is 8.7 A |
A. | 8 |
B. | 7 |
C. | 6 |
D. | 5 |
Answer» E. | |
41. |
An alternating current of 50 Hz frequency and 100 A maximum value is given by _________. |
A. | I = 200 sin 628 t |
B. | I = 100 sin 314 t |
C. | I = 100√2 sin 157 t |
D. | I = 100√2 sin 314 t |
Answer» C. I = 100√2 sin 157 t | |
42. |
For the circuit shown, if the power consumed by 5 Ω resistor is 10 W, then :1.|I| = \(\sqrt2 \) A2. Total impedance = 5 Ω 3. Power factor = 0.866Which of the above are correct? |
A. | 1 and 3 only |
B. | 1 and 2 only |
C. | 2 and 3 only |
D. | 1, 2, 3 |
Answer» B. 1 and 2 only | |
43. |
A current impulse, 5 δ(t) units, is forced through an ideal capacitor C. The voltage VC(t) across the capacitor is given by |
A. | 5t |
B. | 5u(t) – C |
C. | \(\frac{5u\left( t \right)}{C}\) |
D. | \(\frac{5t}{C}\) |
Answer» D. \(\frac{5t}{C}\) | |
44. |
In an AC network, the load connected is (10 + j10). The phase relation between the voltage applied and the current through the load is: |
A. | voltage and current are in phase with each other |
B. | voltage leads current by 45° |
C. | voltage lags current by 30° |
D. | voltage lags current by 45° |
Answer» C. voltage lags current by 30° | |
45. |
In the circuit show in the following figure, the current through the 1 Ω resistor is |
A. | (1 + 5 cos 2t) A |
B. | (5 + cos 2t) A |
C. | (1 – 5 cos 2t) A |
D. | 6 A |
Answer» B. (5 + cos 2t) A | |
46. |
In parallel LC circuit, what will be the value of current (in A) at resonance? |
A. | 0 |
B. | 10 |
C. | 100 |
D. | ∞ |
Answer» B. 10 | |
47. |
Determine the current in the given circuit, if the source voltage is vs = 12 cos (1000t + 15°). |
A. | 0.24 cos (1000t + 15° - tan-1 3/4) A |
B. | 0.24 cos (1000t + 15° - tan-1 4/3) A |
C. | 0.24 cos (1000t + 15° + tan-1 3/4) A |
D. | 0.24 cos (1000t + 15° + tan-1 4/3) A |
Answer» C. 0.24 cos (1000t + 15° + tan-1 3/4) A | |
48. |
Inductive reactance X is a function of inductance L and frequency f. The value of X increases when |
A. | both L and f increase |
B. | L increases and f decreases |
C. | both L and f decrease |
D. | L decreases and f increases |
Answer» B. L increases and f decreases | |
49. |
In a series R-C circuit, the AC current is: |
A. | Lagging behind the applied voltage |
B. | Leading the applied voltage |
C. | Out of phase with applied voltage |
D. | In phase with the applied voltage |
Answer» C. Out of phase with applied voltage | |
50. |
Consider the driving point impedance functions which are to be realized using passive elements1.\(\frac{s+3}{s^{2} (s+5)}\)2.\(\frac{s^{2} + 3}{s^{2} (s+5)}\)which of the above are realizable? |
A. | 1 only |
B. | 2 only |
C. | Both 1 and 2 |
D. | neither 1 nor 2 |
Answer» E. | |