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This section includes 1136 Mcqs, each offering curated multiple-choice questions to sharpen your Electronics & Communication Engineering knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
A transistor with a = 0.9 and ICBO = 10 μA is biased so that IBQ = 90 μA. Then IEQ will be |
| A. | 5.15 mA |
| B. | 5.2 mA |
| C. | 5 mA |
| D. | 990 mA |
| Answer» E. | |
| 52. |
The transistor of following figure in Si diode with a base current of 40 μA and ICBO = 0, if VBB = 6V, RE = 2 kΩ and β = 90, IBQ = 20 μA then IEQ = |
| A. | 1800 μA |
| B. | 180 μA |
| C. | 2000 μA |
| D. | 3000 μA |
| Answer» B. 180 μA | |
| 53. |
A transistor with a = 0.9 and ICBO = 10 μA is biased so that IBQ = 90 μA. Then ICQ will be |
| A. | 910 mA |
| B. | 5 mA |
| C. | 5.25 mA |
| D. | 4.95 mA |
| Answer» B. 5 mA | |
| 54. |
For a BJT if β = 50, ICEO = 3 μA and IC = 1.2 mA then IE |
| A. | 1 mA |
| B. | 1.2 mA |
| C. | 1.3 mA |
| D. | 1.4 mA |
| Answer» C. 1.3 mA | |
| 55. |
In the following figure C = 0.02 μF, and Vth is known to be of frequency ω = 107 rad/sec and rd = 2.5 Ω and ZTh = RTh = 10 Ω Then phase angle between id and Vd |
| A. | 26.5° |
| B. | 25° |
| C. | 30° |
| D. | 0° |
| Answer» B. 25° | |
| 56. |
For a BJT if β = 50, ICEO = 3 μA and IC = 1.2 mA then IB |
| A. | 24 μA |
| B. | 23 μA |
| C. | 20 μA |
| D. | 10 μA |
| Answer» B. 23 μA | |
| 57. |
A Ge diode operated at a junction temperature of 27°C. For a forward current of 10 mA, VD is found to be 0.3 V. If VD = 0.4 V then forward current will be |
| A. | 400 mA |
| B. | 475 mA |
| C. | 477 mA |
| D. | 480 mA |
| Answer» D. 480 mA | |
| 58. |
Values of VT at 20°C for an ideal P-N diode |
| A. | 25.27 mV |
| B. | 30.3 mV |
| C. | 23 mV |
| D. | 50 mV |
| Answer» B. 30.3 mV | |
| 59. |
Percentage increase in the reverse saturation current of a diode if the temperature is increased from 25°C to 50°C |
| A. | 50% |
| B. | 500% |
| C. | 565.7% |
| D. | 575.6% |
| Answer» D. 575.6% | |
| 60. |
Voltage VL in the circuit when Vs < 0 where D is an ideal diode. (Take R1 = Rs = RL = 1 Ω) |
| A. | 0 |
| B. | 1 |
| C. | 1.5 |
| D. | cannot find |
| Answer» B. 1 | |
| 61. |
A distorted sinusoidal has the amplitudes A1, A2, A3 ... of the fundamental, second, harmonic third harmonic, ... respectively. The total harmonic distortion is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |
| 62. |
The common emitter short circuit current gain β of a transistor |
| A. | is a monotonically increasing function of the collector current Ic |
| B. | is a monotonically decreasing function of Ic |
| C. | increasing with Ic, for low Ic, reaches a maximum and then decreases with further increase in Ic |
| D. | is not a function of Ic |
| Answer» D. is not a function of Ic | |
| 63. |
An npn transistor has a Beta cutoff frequency fβ of 1 MHz, and a common emitter short circuit low frequency current gain β0 of 200. It unity gain frequency fT and the alpha cut off frequency fa2 respectively are |
| A. | 200 MHz, 201 MHz |
| B. | 200 MHz, 199 MHz |
| C. | 199 MHz, 200 MHz |
| D. | 201 MHz, 200 MHz |
| Answer» B. 200 MHz, 199 MHz | |
| 64. |
In the circuit of the given figure, assume that the diodes are ideal and the meter is an average indicating ammeter, the ammeter will read |
| A. | 0.42 A |
| B. | 0.4 A |
| C. | [C]. |
| D. | 0.4/p mA |
| Answer» E. | |
| 65. |
For the circuit shown in the figure, the capacitor C is initially uncharged. At t = 0, the switch S is closed. The voltage Vc across the capacitor at t = 1 millisecond is (In the figure shown above, the op-amp is supplied with ± 15 V and the ground has been shown by the symbol) |
| A. | 0 V |
| B. | 6.3 V |
| C. | 9.45 volts |
| D. | 10 volts |
| Answer» E. | |
| 66. |
A silicon PN junction diode under reverse bias has depletion region of width 10 μm. The relative permittivity of silicon, εr = 11.7 and the permittivity of free space ε0 = 8.85 x 10-12 F/m. The depletion capacitance of the diode per square meter is |
| A. | 100 μF |
| B. | 10 μF |
| C. | 1 μF |
| D. | 20 μF |
| Answer» C. 1 μF | |
| 67. |
In the op-amp circuit given in the figure, the load current iL is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» B. [B]. | |
| 68. |
Assuming VCE sat = 0.2 V and β = 50, the minimum base current (IB) required to drive the transistor in the given figure to saturation is |
| A. | 56 μA |
| B. | 140 μA |
| C. | 60 μA |
| D. | 3 μA |
| Answer» B. 140 μA | |
| 69. |
In the figure, assume the op-amp is to be ideal. The output Vo if the circuit is |
| A. | 10 cos (100t) |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» B. [B]. | |
| 70. |
Volt equivalent temperature in a PN Junction diode is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | 11600 T |
| Answer» B. [B]. | |
| 71. |
An inverting amplifier has R1 = 10 kΩ, and Rf = 150 kΩ then the O/P voltage, if input voltage Vi = 1 volt. |
| A. | -15 V |
| B. | -10 V |
| C. | 15 V |
| D. | -14 V |
| Answer» B. -10 V | |
| 72. |
An R-C coupled amplifier has mid-frequency gain of 200 and a frequency response from 100 Hz to 20 kHz. A negative feedback network with β = 0.2 is incorporated into the amplifier circuit, the Bandwidth will be |
| A. | 50 kHz |
| B. | 20 kHz |
| C. | 100 kHz |
| D. | infinite |
| Answer» D. infinite | |
| 73. |
In a negative feedback amplifier A = 100, β = 0.04 and Vs = 50 mV, then feedback will be |
| A. | 4 |
| B. | 8 |
| C. | 10 |
| D. | 2 |
| Answer» B. 8 | |
| 74. |
A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltage |
| A. | 220 V |
| B. | 200 V |
| C. | 198 V |
| D. | 0 volts |
| Answer» D. 0 volts | |
| 75. |
In a half wave rectifier, Vrms is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | None of these |
| Answer» B. [B]. | |
| 76. |
For a BJT, β and a are related as |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» E. | |
| 77. |
Calculate the conductivity of pure silicon at room temperature when the concentration of carriers is 1.6 x 1010 per cm3. Take μe = 1500 cm2/V-sec, μh = 500 cm2/V-sec at room temperature, |
| A. | 5.12 x 10-16 |
| B. | 5 x 10+16 |
| C. | 1016 |
| D. | 5 x 1020 |
| Answer» B. 5 x 10+16 | |
| 78. |
The turn ratio of a transformer is 20:1, if a load of 10Ω is connected across the secondary, what will be the effective resistance seen looking into the Primary? |
| A. | 2 kΩ |
| B. | 4 kΩ |
| C. | 8 kΩ |
| D. | 32 kΩ |
| Answer» C. 8 kΩ | |
| 79. |
Calculate the O/P voltage of a non-inverting constant gain multiplier with R1 = 100 kΩ, Rf= 600 kΩ, V1 = 2 volt, then O/P voltage Vo of non inverting op-amp is given by |
| A. | 16 V |
| B. | 14 V |
| C. | 20 V |
| D. | 15 V |
| Answer» C. 20 V | |
| 80. |
Oscillation frequency in Wein bridge oscillator is |
| A. | [A]. |
| B. | [B]. |
| C. | f = 2p CR |
| D. | [D]. |
| Answer» B. [B]. | |
| 81. |
The RC network of wein bridge oscillator consists of resistors and capacitors of values R = 220 kΩ and C = 250 PF Determine the frequency of oscillation |
| A. | 2.98 kHz |
| B. | 3 kHz |
| C. | 2 kHz |
| D. | 10 kHz |
| Answer» B. 3 kHz | |
| 82. |
Oscillation frequency in colpitt's oscillator is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | none of these |
| Answer» B. [B]. | |
| 83. |
The current through R1 is(If β = 99, VBE = 0.74 V) |
| A. | 0.1 mA |
| B. | 0.35 mA |
| C. | 0.2 mA |
| D. | none |
| Answer» B. 0.35 mA | |
| 84. |
Gain of the amplifier is 'A'. Then the I/P impedance and O/P impedance of the closed loop amplifier shown below would be |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |
| 85. |
In a circuit of figure, Vs = 10 cos(ωt) power drawn by the 2Ω resistor is 4 watts. The power factor is |
| A. | 0.3 |
| B. | 0.4 |
| C. | 0.6 |
| D. | none |
| Answer» D. none | |
| 86. |
If an amplifier with gain of - 100 and feedback of β = - 0.1 has a gain change of 20% due to temperature the change in gain of the feedback amplifier will be |
| A. | 0.02% |
| B. | 2% |
| C. | 0.1% |
| D. | 0.01% |
| Answer» C. 0.1% | |
| 87. |
In the circuit shown in figure, RC = 10k, RE = 150Ω, β = 100, I = 1 mA. The value of , will be |
| A. | 50 |
| B. | 100 |
| C. | 192 |
| D. | 400 |
| Answer» D. 400 | |
| 88. |
A voltage gain of a CE amplifier connected in collector to base configuration is 50. Collector to base resistor RF = 1 kΩ, RC = 4Ω, the O/P resistance R0 =? |
| A. | 800 Ω |
| B. | 813 Ω |
| C. | 817 Ω |
| D. | 819 Ω |
| Answer» C. 817 Ω | |
| 89. |
In a closed loop non-inverting amplifier, constructed using single break-frequency op-amp, bandwidth with feedback can be written as. (f0 - break frequency, β - gain of the feedback circuit, A - open loop voltage gain, Af - closed loop voltage gain) |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» B. [B]. | |
| 90. |
Regulation of the DC power supply of 12 V, 100 mA is the effective resistance of power supply is 20 Ω |
| A. | 22% |
| B. | 131% |
| C. | 30% |
| D. | 20% |
| Answer» E. | |
| 91. |
For the amplifier in figure, β = 800. The mid-band voltage gain V0/Vi is |
| A. | 0 |
| B. | < 1 |
| C. | ≈ 1 |
| D. | 800 |
| Answer» D. 800 | |
| 92. |
For transistor 2 N 338 the manufacturer specifies Pmax = 100 mW at 250°C free air temperature and maximum junction temperature of 125°C. Its thermal resistance is |
| A. | 10° C/W |
| B. | 100° C/W |
| C. | 1000° C/W |
| D. | 5000° C/W |
| Answer» D. 5000° C/W | |
| 93. |
An amplifier has open loop gain of 100, input impedance 1 kΩ and output impedance 100 Ω. If negative feedback with β = 0.99 is used, the new input and output impedances are |
| A. | 10 Ω and 1 Ω |
| B. | 10 Ω and 10 kΩ |
| C. | 100 kΩ and 1 Ω |
| D. | 100 kΩ and 10 Ω |
| Answer» D. 100 kΩ and 10 Ω | |
| 94. |
A class C amplifier has input frequency of 200 kHz. If width of collector pulses is 0.1 μs, duty cycle is |
| A. | 1% |
| B. | 2% |
| C. | 10% |
| D. | 20% |
| Answer» C. 10% | |
| 95. |
The ripple factor of a power supply is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |
| 96. |
For the amplifier circuit of figure, the h parameters of transistor are hib = 25 Ω, hfb = 0.999, hob = 10-6 Ω. The voltage gain is |
| A. | 0.999 |
| B. | 1.98 |
| C. | 20 |
| D. | 400 |
| Answer» E. | |
| 97. |
In an amplifier with a gain of - 1000 and feedback factor β = - 0.1, the change in gain is 20% due to temperature. The change in gain for feedback amplifier will be |
| A. | 10% |
| B. | 5% |
| C. | 0.2% |
| D. | 0.01% |
| Answer» D. 0.01% | |
| 98. |
In a transistor IE = 10 mA, ICO = 0.5 μA and a = 0.995. Then ICEO = |
| A. | 100 μA |
| B. | 25 μA |
| C. | 10.1 mA |
| D. | 10.5 mA |
| Answer» B. 25 μA | |
| 99. |
For the circuit of figure the critical frequency is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |
| 100. |
The critical frequency of the circuit of figure is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |