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Multiple choice question for engineering

Set 1

1. An amplifier operating from ±3V provide a 2.2V peak sine wave across a 100 ohm load when provided with a 0.2V peak sine wave as an input from which 1.0mA current is drawn. The average current in each supply is measured to be 20mA. What is the amplifier efficiency?
a) 20.2%
b) 25.2%
c) 30.2%
d) 35.2%

2. In order to prevent distortion in the output signal after amplification, the input signal must be
a) Higher than the positive saturation level of the amplifier
b) Lower than the negative saturation level of the amplifier
c) Must lie with the negative and the positive saturation level of the amplifier
d) Both higher than the positive saturation level of the amplifier and lower than the negative saturation level of the amplifier

Answer: c [Reason:] Higher than the positive saturation and lower than the negative saturation level of the amplifier are the desired characteristics in order to prevent distortion.

3. The voltage gain of the amplifier is 8 and the current gain is 7. The power gain of the amplifier is
a) 56 db
b) 17.481 db
c) 34.963 db
d) 1 db

Answer: b [Reason:] The power gain is given by 10 log (7 X 8) db.

4. Statement 1: Voltage gain of -5 means that the output voltage has been attenuated.
Statement 2: Voltage gain of -5db means that the output voltage has been attenuated.
a) Statement 1 and Statement 2 are true
b) Statement 1 and Statement 2 are false
c) Only Statement 1 is true
d) Only Statement 2 is true

Answer: d [Reason:] A negative voltage gain means that a phase difference of 1800 has been introduced in the output waveform when compared to the input waveform. A voltage gain of -5db means that the signal has been attenuated.

5. Which of the following isn’t true?
a) Both transformer and amplifier can provide voltage gain
b) Both transformer and amplifier can provide current gain
c) Both transformer and amplifier can provide power gain
d) None of the mentioned

Answer: c [Reason:] For an ideal transformer the power input is always equal to the power output. In real conditions there is slight loss of power when transferring the power from an input source to an output source. Amplifiers only provide power gain.

6. Symmetrically saturated amplifiers operating in clipping mode can be used to convert a sine wave to a
a) Square wave
b) Pseudo Square wave
c) Sawtooth wave
d) Triangular wave

Answer: b [Reason:] Clipping circuits with low peak values of the output signals are used to generate pseudo square waves if the input signal is very large as compared to the output signal.

7. What is meant by stability of the an amplified signal?
a) The amplified signal must have a finite amplitude
b) The amplified signal should not have self oscillation
c) The input and the output signal must be proportional
d) The ratio of the input and the output signal must be finite

Answer: b [Reason:] The ability of the amplifier to prevent self oscillation is a measure of its stability.

8. If Av, Ai and Ap represents the voltage gain, current gain and power gain ratio of an amplifier which of the below is not the correct expression for the corresponding values in decibel?
a) Current gain: 20 log Ai db
b) Voltage gain: 20 log Av db
c) Power gain: 20 log Ap db
d) Power gain: 10 log Ap

Answer: c [Reason:] Power gain is given by 10 log Ap db.

9. An amplifier has a voltage gain of 100 V/V and a current gain of 1000A/A. the value of the power gain decibel is
a) 30 db
b) 40 db
c) 50 db
d) 60 db

Answer: c [Reason:] Power gain in db is given by 10 log (100 X 1000) db.

10. The units of voltage gain is
a) It has no units, it is a ratio
b) Decibels (db)
c) All of the mentioned
d) None of the mentioned

Answer: a [Reason:] Voltage gain (Vo) = output voltage/input voltage (Vi). It is also expresses as 20 log (Vo/Vi) db.

Set 2

(Q.1-Q.3) An AM signal is represented by x(t) = (20 + 4sin(500πt)) cos(2πt x 105)V.

1. The modulation index is
a) 20
b) 4
c) 0.2
d) 10

Answer: c [Reason:] 20 + 4sin(500πt) = 20(1 + 0.2sin(500πt)), hence the modulation index is 0.2.

2. Total signal power is
a) 208 W
b) 204 W
c) 408 W
d) 416 W

Answer: b Explantion: Pc = 20×20/2 or 200 W. Pt = Pc(1 + 0.2×0.2/4) or 204 W.

3. Total sideband power is
a) 4 W
b) 8 W
c) 16 W
d) 2 W

Answer: a [Reason:] Pt – Pc = 204 – 200 = 4W.

4. An AM broadcast station operates at its maximum allowed total output of 50 kW with 80% modulation. The power in the intelligence part is
a) 12.12 kW
b) 31.12 kW
c) 6.42 kW
d) none of the mentioned

Answer: a [Reason:] Pi = Pt – Pc = 50 – 37.88 kW or 12.12 kW.

5. The aerial current of an AM transmitter is 18 A when unmodulated but increases to 20 A when modulated.The modulation index is
a) 0.68
b) 0.73
c) 0.89
d) 0.95

Answer: a [Reason:] 400/326 = 1 + (α2)/2, therefore α is 0.68.

6. A modulating signal is amplified by a 80% efficiency amplifier before being combined with a 20 kW carrier to generate an AM signal. The required DC input power to the amplifier, for the system to operate at 100% modulation, would be
a) 5 kW
b) 8.46 kW
c) 12.5 kW
d) 6.25 kW

Answer: c [Reason:] Pi = Pt -Pc = 30 – 20 = 10 kW. DC input = 10/0.8 or 12.5 kW.

7. A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal is
a) 2.23 kW
b) 2.36 Kw
c) 1.18 kW
d) 1.26 kW

Answer: a [Reason:] Pt = Pc (1 + 0.49/2).

8. A carrier is simultaneously modulated by two sine waves with modulation indices of 0.4 and 0.3. The resultant modulation index will be
a) 1.0
b) 0.7
c) 0.5
d) 0.35

Answer: c [Reason:] α2 = 0.32 + 0.42 = 0.52 or α = 0.5.

9. In a DSB-SC system with 100% modulation, the power saving is
a) 100%
b) 55%
c) 75%
d) 100%

Answer: b [Reason:] This is so because the power is suppressed by two thirds of the total. hence the power saving is 66%.

10. A 10 kW carrier is sinusoidally modulated by two carriers corresponding to a modulation index of 30% and 40% respectively. The total radiated power is
a) 11.25 kW
b) 12.5 kW
c) 15 kW
d) 17 kW

Answer: a [Reason:] The required answer is 1 (1 + 0.42 + 0.32) or 11.25 kW.

Set 3

(Q1-Q.6) An amplifier is measured to have an internal resistance of 10 kΩ, voltage gain of 100V/V and output resistance of 100 Ω. Also, when a load resistance of 1 kΩ is connected between the output resistance if found to decrease to 8 kΩ. If the amplifier is fed with the signal source having an internal resistance of 2 kΩ, then
1. Find Gm.
a) 1 A/V
b) 10 A/V
c) 100 A/V
d) 1000 A/V

Answer: a [Reason:] Gm = (voltage gain in V/V) / (output resistance) or 100/100 A/V.

2. Find Av.
a) 9.09 V/V
b) 10 V/V
c) 90.9 V/V
d) 100 V/V

Answer: c [Reason:] Av = Avo X Rl/Ro+Rl or 100 X 1000/1000+100 or 90.9 V/V.

3. Find Gvo.
a) 53.3 V/V
b) 63.3 V/V
c) 73.3 V/V
d) 83.3 V/V

Answer: d [Reason:] Gvo = (Avo X input resistance) / (input resistance + signal resistance).

4. Find Gv.
a) 53.4 V/V
b) 72.7 V/V
c) 83.3 V/V
d) 90.9 V/V

Answer: b [Reason:] Gv = (Gvo X Av) / Avo or 83.3 X 90.9 / 100 V/V.

5. Find R out.
a) 146 Ω
b) 292 Ω
c) 584 Ω
d) 1168 Ω

Answer: a [Reason:] Rout = Rl (1 – Gvo/Gv). Put in the respective values and solve.

6. Find Ai.
a) 182 A/A
b) 364 A/A
c) 546 A/A
d) 728 A/A

Answer: d [Reason:] (Vo X R in) / (Vi X Rl) gives the required value of Ai.

(Q.7-Q.10) The circuit shown below is a small sine wave signal with average zero and transistor ß =

7. Find the value of R(E) to establish a dc emitter current of about 0.5 mA.
a) 28.57 kΩ
b) 57.04 kΩ
c) 114.08 kΩ
d) 228.16 kΩ

8. Find R(C) to establish a dc collector voltage of about +5V.
a) 5 kΩ
b) 10 kΩ
c) 15 kΩ
d) 20 kΩ

Answer: d [Reason:] Vc = 15 – Rc.Ic 5 = 15 – Rc * 0.99 * 0.5m Rc = 20.2kΩ = 20kΩ.

9. For R (L) = 10 kΩ and transistor Ro = 200 kΩ, determine the overall voltage gain.
a) -21 V/V
b) -42 V/V
c) -86 V/V
d) -123 V/V

Set 4

1. A solid copper sphere, 10 cm in diameter is deprived of 1020 electrons by a charging scheme. The charge on the sphere is
a) 160.2 C
b) -160.2 C
c) 16.02 C
d) -16.02 C

Answer: c [Reason:] n 1020, Q = ne = e 1020 = 16.02 C. Charge on sphere will be positive.

2. A lightning bolt carrying 15,000 A lasts for 100 s. If the lightning strikes an airplane flying at 2 km, the charge deposited on the plane is
a) 13.33 C
b) 75 C
c) 1500 C
d) 1.5 C

Answer: d [Reason:] dQ = i dt = 15000 x 100µ = 1.5 C.

3. If 120 C of charge passes through an electric conductor in 60 sec, the current in the conductor is
a) 0.5 A
b) 2 A
c) 3.33 mA
d) 0.3 mA

Answer: b [Reason:] i = dQ/dt = 120/60 = 2A.

4. The energy required to move 120 coulomb through
3 V is
a) 25 mJ
b) 360 J
c) 40 J
d) 2.78 mJ

Answer: b [Reason:] W = Qv = 360 J.

5. Consider the circuit graph shown in figure below. Each branch of circuit graph represent a circuit element. The value of voltage V1 is

a) 30 V
b) 25 V
c) 20 V
d) 15 V

Answer: d [Reason:] 100 = 65 + V2 => V2 = 35 V V3 – 30 = V2 => V3 = 65 V 105 – V3 + V4 – 65 = 0 => V4 = 25 V V4 + 15 – 55 + V1 = 0 => V1 = 15 V.

6. Req = ?

a) 11.86 ohm
b) 10 ohm
c) 25 ohm
d) 11.18 ohm

Answer: d [Reason:] Req – 5 = 10(Req + 5)/(10 + 5 +Req). Solving for Req we have Req = 11.18 ohm.

7. In the circuit the dependent source

a) supplies 16 W
b) absorbs 16 W
c) supplies 32 W
d) absorbs 32 W

Answer: d [Reason:] P = VIx = 2Ix Ix = 2 x 16 or 32 watt (absorb).

8. Twelve 6 resistor are used as edge to form a cube. The resistance between two diagonally opposite corner of the cube is (in ohm)
a) 5/6
b) 6/5
c) 5
d) 6

9. The energy required to charge a 10 µF capacitor to 100 V is
a) 0.1 J
b) 0.05 J
c) 5 x 10(-9) J
d) 10 x 10(-9) J

Answer: b [Reason:] Energy provided is equal to 0.5 CVxV.

10. A capacitor is charged by a constant current of 2 mA and results in a voltage increase of 12 V in a 10 sec
interval. The value of capacitance is
a) 0.75 mF
b) 1.33 mF
c) 0.6 mF
d) 1.67 mF

Answer: d [Reason:] Voltage gain = 1/C x i(t2 – t1), where t2 – t1 = 10s. Hence C = 1.67 mF.

Set 5

1. In which of the following configuration does a MOSFET works as an amplifier?
a) Common Source (CS)
b) Common Gate (CG)
c) Common drain (CD)
d) All of the mentioned

Answer: d [Reason:] There are three basic configurations for connecting the MOSFET as an amplifier. Each of these configurations is obtained by connecting one of the three MOSFET terminals to ground, thus creating a two-port network with the grounded terminal being common to the input and output ports.

2. The MOSFET in the following circuit is in which configuration?

a) Common Source (CS)
b) Common Gate (CG)
c) Common Drain (CD)
d) None of the mentioned

Answer: b [Reason:] It is the circuit for Common gate configuration.

3. The MOSFET in the following circuit is in which configuration?

a) Common Source (CS)
b) Common Gate (CG)
c) Common Drain (CD)
d) None of the mentioned

Answer: c [Reason:] It is the circuit for Common drain configuration.

4. The MOSFET in the following circuit is in which configuration?

a) Common Source (CS)
b) Common Gate (CG)
c) Common Drain (CD)
d) None of the mentioned

Answer: a [Reason:] It is the circuit for Common source configuration.

(Q.5-Q.10) Reference circuit for Q.5-Q.10 The circuit below is the characterization for the amplifier as a functional block.

5. If the value of Rin for the common source configuration is R1 and that for common source with a source resistance configuration is R2 ideally. The ratio of R1/R2 will be
a) R1/R2 = 1
b) 0 < R1/R2 < 1
c) R1/R2 > 1
d) R1/R2 = 0

Answer: a [Reason:] Ideally both must have infinite resistance.

6. Which is true for the value of Avo for common source (Represented by A1) and common source with a source resistance (represented by A2).
a) A1 = A2
b) A1 > 2
c) A1 < A2
d) |A1| < |A2|

Answer: c [Reason:] A1 = -gmRD and A2 = -gmRD/1+gmRS Reference circuit for Common source configuration Reference circuit for common source with source resistance RS

7. Which of the following is true for the voltage gain (AV) for the common source configuration (represented by A1) and the common gate configuration (represented by A2)?
a) A1 = A2
b) |A1| = |A2| and A1 ≠ A2
c) |A1| > |A2|
d) |A1| < |A2|

Answer: b [Reason:] A1 = -gm(RL||RD) and A1 = gm(RL||RD) Reference figure for common source configuration Reference figure for common gate configuration

8. The value of the voltage gain (Av) for the common source with source resistance (represented by A1) and common gate configuration (represented by A2) are related to each other by
a) A1 > A2
b) |A1| > |A2|
c) A1 < A2
d) A1 > A2 and |A1| > |A2|

Answer: c [Reason:] A1 = – gm(RL||RD)/ 1 + gmRS and A2 = gm(RL||RD) Reference figure for common source with source resistance configuration Reference figure for common gate configuration

9. In which of the following configuration is the input resistance (Ri) not equal to zero ideally?
a) Common source configuration
b) Common source configuration with source resistance
c) Common gate configuration
d) Source follower configuration