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# Multiple choice question for engineering

## Set 1

1. Efficiency of a centre tapped full wave rectifier is _________
a) 50%
b) 46%
c) 70%
d) 81.2%

Answer: d [Reason:] Efficiency of a rectifier is the effectiveness to convert AC to DC. It’s obtained by taking ratio of DC power output to maximum AC power delivered to load. It’s usually expressed in percentage. For centre tapped full wave rectifier, it’s 81.2%.

2. A full wave rectifier supplies a load of 1KΩ. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?
a) 0.562V
b) 0.785V
c) 0.954V
d) 0.344V

Answer: c [Reason:] The ripple voltage is (Vϒ)RMS=ϒVDC /100. VDC=0.636*VRMS* √2=0.636*220* √2=198V and ripple factor ϒ for full wave rectifier is 0.482. Hence, (Vϒ)RMS=0.482*198 /100=0.954V.

3. A full wave rectifier delivers 50W to a load of 200Ω. If the ripple factor is 2%, calculate the AC ripple across the load.
a) 2V
b) 5V
c) 4V
d) 1V

Answer: a [Reason:] We know that, PDC=VDC2/RL. So, VDC=(PDC*RL)1/2=100001/2=100V. Here, ϒ=0.02 ϒ=VAC/VDC=VAC/100.So, VAC=0.02*100=2V.

4. A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.
a) 368.98mW
b) 275.2mW
c) 145.76mW
d) 456.78mW

Answer: b [Reason:] The AC power input PIN=IRMS2(RF+Rr). IRMS=Im/√2=Vm/(Rf+RL)√2=30/(1500+10)*1.414=13.5mA So, PIN=(13.5*10-3)2*(1500+10)=275.2mW.

5. In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________

a) 54%
b) 48%
c) 26%
d) 81%

Answer: b [Reason:] The ripple factor ϒ= [(IRMS/IAVG)2 – 1]1/2. IRMS =Im /√2=Vm/(Rf+RL)√2=200/1.01=198. (Secondary line to line voltage is 800/2=400. Due to centre tap Vm=400/2=200) IRMS=198/√2=140mA, IAVG=2*198/π=126mA. ϒ=[(140/126)2-1]1/2=0.48. So, ϒ=48%.

6. If input frequency is 50Hz for a full wave rectifier, the ripple frequency of it would be _________
a) 100Hz
b) 50Hz
c) 25Hz
d) 500Hz

Answer: a [Reason:] In the output of the centre tapped rectifier, one of the half cycle is repeated. The frequency will be twice as that of input frequency. So, it’s 100Hz.

7. Transformer utilization factor of a centre tapped full wave rectifier is_________
a) 0.623
b) 0.678
c) 0.693
d) 0.625

Answer: c [Reason:] Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.693.

8. In the circuits given below, the correct full wave rectifier is _________
a)
b)
c)
d)

Answer: c [Reason:] When the input is applied, a full wave rectifier should have a current flow. The flow should be in the same direction for both positive and negative half cycles. Only the third circuit satisfies the above condition.

9. If the peak voltage on a centre tapped full wave rectifier circuit is 5V and diode cut in voltage is 0.7. The peak inverse voltage on diode is_________
a) 4.3V
b) 9.3V
c) 5.7V
d) 10.7V

Answer: b [Reason:] PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, if PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=2Vm-Vd = 10-0.7 = 9.3V.

10. In a centre tapped full wave rectifier, the input sine wave is 250sin100t. The output ripple frequency will be _________
a) 50Hz
b) 100Hz
c) 25Hz
d) 200Hz

Answer: b [Reason:] The equation of sine wave is in the form Vmsinωt. So, by comparing we get ω=100. Frequency, f =ω/2=50Hz. The output of centre tapped full wave rectifier has double the frequency of inpu. Hence, fout = 100Hz.

## Set 2

1. The diode in a half wave rectifier has a forward resistance RF. The voltage is Vmsinωt and the load resistance is RL. The DC current is given by _________
a) Vm/√2RL
b) Vm/(RF+RL
c) 2Vm/√π
d) Vm/RL

Answer: b [Reason:] For a half wave rectifier, the IDC=IAVG=Im/π I= Vmsinωt/(RF+RL)=Imsinωt Im =Vm/ RF+RL So, IDC=Im/π=Vm/(RF+RL).

2. The below figure arrives to a conclusion that _________

a) for Vi > 0, V0=-(R2/R1)Vi
b) for Vi > 0, V0=0
c) Vi < 0, V0=-(R2/R1)Vi
d) Vi < 0, V0=0

Answer: b [Reason:] The given op-amp is in inverting mode and this makes the output voltage to have a phase shift of 180°. The output voltage is now negative. So, the diode 1 is reverse biased and diode 2 is forward biased. Then output is clearly zero.

3. What is the output as a function of the input voltage (for positive values) for the given figure. Assume it’s an ideal op-amp with zero forward drop (Di=0)

a) 0
b) -Vi
c) Vi
d) 2Vi

Answer: c [Reason:] When the input of the inverted mode op-amp is positive, the output is negative. The diode is reverse biased. The input appears at the output.

4. In a half wave rectifier, the sine wave input is 50sin50t. If the load resistance is of 1K, then average DC power output will be?
a) 3.99V
b) 2.5V
c) 5.97V
d) 6.77V

Answer: b [Reason:] The standard form of a sine wave is Vmsinωt. BY comparing the given information with this equation, Vm =50. Power=Vm2/RL=50*50/1000=2.5V.

5. In a half wave rectifier, the sine wave input is 200sin300t. The average value of output voltage is?
a) 57.876V
b) 67.453V
c) 63.694V
d) 76.987V

Answer: c [Reason:] Comparing with the standard equation, Vm=200V. Average value is given by, Vavg=Vm/π. So, 200/π=63.694.

6. Efficiency of a half wave rectifier is
a) 50%
b) 60%
c) 40.6%
d) 46%

Answer: c [Reason:] Efficiency of a rectifier is the effectiveness to convert AC to DC. For half wave it’s 40.6%. It’s given by, Vout/Vin*100.

7. If peak voltage for a half wave rectifier circuit is 5V and diode cut in voltage is 0.7, then peak inverse voltage on diode will be?
a) 5V
b) 4.9V
c) 4.3V
d) 6.7V

Answer: c [Reason:] PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, If the PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=Vm-Vd=5-0.7=4.3V.

8. Transformer utilisation factor of a half wave rectifier is _________
a) 0.234
b) 0.279
c) 0.287
d) 0.453

Answer: c [Reason:] Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.287.

9. If the input frequency of a half wave rectifier is 100Hz, then the ripple frequency will be_________
a) 150Hz
b) 200Hz
c) 100Hz
d) 300Hz

Answer: c [Reason:] The ripple frequency of the output and input is same. This is because, one half cycle of input is passed and other half cycle is seized. So, effectively the frequency is the same.

10. Ripple factor of a half wave rectifier is_________(Im is the peak current and RL is load resistance)
a) 1.414
b) 1.21
c) 1.4
d) 0.48

Answer: b [Reason:] The ripple factor of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 1.21.

## Set 3

1. In the Hall Effect, the directions of electric field and magnetic field are parallel to each other.
The above statement is
a) True
b) False

Answer: b [Reason:] To make Lorentz force into the effect, the electric field and magnetic field should be perpendicular to each other.

2. Which of the following parameters can’t be found with Hall Effect?
a) Polarity
b) Conductivity
c) Carrier concentration
d) Area of the device

Answer: d [Reason:] The Hall Effect is used for finding the whether the semiconductor is of n-type or p-type, mobility, conductivity and the carrier concentration.

3. In the Hall Effect, the electric field is in x direction and the velocity is in y direction. What is the direction of the magnetic field?
a) X
b) Y
c) Z
d) XY plane

Answer: c [Reason:] The Hall Effect satisfies the Lorentz’s Force which is E=vxB So, the direction of the velocity, electric field and magnetic field should be perpendicular to each other.

4. What is the velocity when the electric field is 5V/m and the magnetic field is 5A/m?
a) 1m/s
b) 25m/s
c) 0.2m/s
d) 0.125m/s

Answer: a [Reason:] E=vxB v=E/B =5/5 =1m/s.

5. Calculate the hall voltage when the Electric Field is 5V/m and height of the semiconductor is 2cm.
a) 10V
b) 1V
c) 0.1V
d) 0.01V

Answer: c [Reason:] Vh=E*d =5*2/100 =0.1V.

6. Which of the following formulae doesn’t account for correct expression for J?
a) ρv
b) I/wd
c) σE
d) µH

Answer: d [Reason:] B=µH So, option d is correct option.

7. Calculate the Hall voltage when B=5A/m, I=2A, w=5cm and n=1020.
a) 3.125V
b) 0.3125V
c) 0.02V
d) 0.002V

Answer: d [Reason:] Vh=BI/wρ =5=2/ (5*10-2*105*1.6*10-19) =0.002V.

8. Calculate the Hall Effect coefficient when number of electrons in a semiconductor is 1020.
a) 0.625
b) 0.0625
c) 6.25
d) 62.5

Answer: b [Reason:] R=1/ρ =1/(1.6*10-19*1020) =0.0625.

9. What is the conductivity when the Hall Effect coefficient is 5 and mobility is 5cm2 /s.
a) 100 S/m
b) 10 S/m
c) 0.0001S/m
d) 0.01 S/m

Answer: c [Reason:] µ=σR σ =µ/R =5*10-4/5 =0.0001 S/m.

10. In Hall Effect, the electric field applied is perpendicular to both current and magnetic field?
a) True
b) False

Answer: a [Reason:] In Hall Effect, the electric field is perpendicular to both current and magnetic field so that the force due to magnetic field can be balanced by the electric field or vice versa.

## Set 4

1. For an ideal diode which of the following is true?
a) It allows the passage of current in the forward bias with zero potential drop across the diode
b) It does not allow the flow of current in reverse bias
c) All of the mentioned
d) None of the mentioned

Answer: c [Reason:] Both of the facts mentioned hold true for an ideal operational amplifier.

2. Consider an AC sine wave voltage signal being used to connect a diode and a resistor as shown in the figure. The variation of the voltage across the diode (Vd) with respect to time (t) is given by

a)
b)
c)
d) None of the mentioned

Answer: c [Reason:] Vd = Vi – Vo.

3. The figure below shows a circuit for charging a 12-V battery. If Vs is a sinusoid with 24-V peak amplitude, the fraction of each cycle during which the diode conducts is

a) One quarter of a cycle
b) One-third of a cycle
c) One half of the cycle
d) Three quarters of a cycle

Answer: b [Reason:] For one half of the cycle the diode conducts when 24 cos(Θ) = 12 or Θ is 600. Hence for a complete cycle the diode does not conducts for 2Θ or 1200. Hence the diode does not conducts for a third of a cycle.

4. Diodes can be used in the making of
a) Rectifiers
b) LED lamps
c) Logic gates
d) All of the mentioned

Answer: d [Reason:] Diodes are used to make rectifiers (full wave or half wave rectifiers are the most common examples), LED lamps uses diodes (diodes are generally doped for their use in this purpose) and logic gates can also be made using diodes using the fact that diodes are conducting only in the forward biased configuration.

5. For the connections shown below, the equivalent logic gate is

a) OR gate
b) AND gate
c) XOR gate
d) NAND gate

Answer: a [Reason:] The following circuit behaves as a OR logic gate.

6. For the connections shown below, the equivalent logic gate is

a) OR gate
b) AND gate
c) XOR gate
d) NAND gate

Answer: b [Reason:] The circuit shown behaves as an AND logic gate.

7. The figure below shows a circuit for an AC voltmeter. It utilizes a moving-coil meter that gives a full-scale reading when the average current flowing through it is 1 mA. The moving-coil meter has a 50-Ω resistance. The value of R that results in the meter indicating a full-scale reading when the input sine-wave voltage VI is 20 V peak-to-peak is

a) 3.183 kΩ
b) 3.133 kΩ
c) 3.183 Ω
d) 56.183 Ω

Answer: b [Reason:] Average value of input voltage= 20/π or 6.366 V. Since there is diode half of the time there is no voltage across the voltmeter hence the average value of the input voltage is 10/π or 3.183V. The total resistance required to generate 1mA current is 3.183 kΩ out of which 50Ω is provided by the moving coil. Therefore, a resistance of 3.133 kΩ is required.

8. The value of I and V for the circuit shown below are

a) 2A and 5V respectively
b) -2A and 5V respectively
c) -2A and -5V respectively
d) 2A and 0V respectively

Answer: d [Reason:] I = V/R or 5/2.5 or 2A For an ideal diode when it is conducting the potential drop across its terminal is zero.

9. The units frequently used to measure the forward bias and reverse bias current of a diode are
a) µA and µA respectively
b) µA and mA respectively
c) mA and µA respectively
d) mA and mA respectively

Answer: c [Reason:] The currents in forward bias is generally in MA and in the reverse bias it is very small and is generally measured in µA.

10. A diode
a) Is the simplest of the semiconductor devices
b) Has a characteristic that closely follows that of a switch
c) Is two terminal device
d) All of the mentioned

Answer: d [Reason:] Both the statements are true for a diode.

## Set 5

1. What is the minimum number of terminals required in an IC package containing four operational amplifiers (quad op amps)?
a) 12
b) 13
c) 14
d) 15

Answer: c [Reason:] The minimum no of pins required by dual-op-amp is 8. Each op-amp has 2 input terminals(4 pins) and one output terminal(2 pins). Another 2 pins are required for power. Similarly, The minimum no of pins required by dual-op-amp is 14: 4*2 + 4*1 + 2 = 14.

2. Which of the following is not a property of an ideal operational amplifier?
a) Zero input impedance
b) Infinite bandwidth
c) Infinite open loop gain
d) Zero common-mode gain or conversely infinite common mode-rejection.

Answer: a [Reason:] An ideal operational amplifier does not has a zero input impedance.

3. In an ideal op amp the open-loop gain is 103. The op amp is used in a feedback circuit, and the voltages appearing at two of its three signal terminals are measured as v2 = 0V and v3 = 2V where it is assumed that v1 and v2 are input terminals and v3 is the output terminal. The value of the differential (vd) and common-mode (vcm)signal is
a) Vd = 2 mV and vcm = 1 mv
b) Vd = 2 mV and vcm = -1 mV
c) Vd = 2 mV and vcm = 2mV
d) Vd = 2 mV and vcm = -2mV

Answer: b [Reason:] Vc = 0.5(V1 + V2) and Vd = V2 – V1.

4. Consider the figure given below. Known that vo = 4V and vi = 2V, determine the gain for the op amp assuming that it is ideal except for the fact that it has finite gain

a) 1001
b) 2002
c) 3003
d) 4004

Answer: b [Reason:] The Voltage at the positive input has to be -3.000v, vi = -3.020v A = vo / vi – vr = -2 / -3.020 -(-3) = 100.

5. Which of the following is not a terminal for the operational amplifier?
a) Inverting terminal
b) Non-inverting terminal
c) Output terminal
d) None of the mentioned

Answer: d [Reason:] There are three terminals for the operational amplifier.

6. Operational amplifiers are
a) Differential input and single-ended output type amplifier
b) Single-ended input and single-ended output type amplifier
c) Single-ended input and differential output type amplifier
d) Differential input and differential output type amplifier

Answer: a [Reason:] It is another way to refer to op amps based on its terminal characteristics.

7. Express the input voltages v1 and v2 in terms of differential input (vd) and common-mode input(vc). Given v2 > v2.
a) Vd = V1 – V2, Vc = 0.5(V1 + V2)
b) Vd = V2 – V1, Vc = V1 + V2
c) Vd = V1 – V2, Vc = V1 + V2
d) Vd = V2 – V1, Vc = 0.5(V1 + V2)

Answer: d [Reason:] This is the correct mathematical representation.

8. What is the minimum number of pins for a dual operational amplifier IC package?
a) 4
b) 6
c) 8
d) 10

Answer: c [Reason:] The minimum no of pins required by dual-op-amp is 8. Each op-amp has 2 input terminals(4 pins) and one output terminal(2 pins). Another 2 pins are required for power.

9. For an ideal operational amplifier (except for the fact that it has finite gain) one set of the value for the input voltages (v2 is the positive terminal v1 is the negative terminal) and output voltage (v0) as determined experimentally is v1= 2.01V, v2=2.00V and v0= -0.99V. Experiment was carried with different values of input and output voltages. Which of the following is not possible considering experimental error?
a) v1= 1.99V, v2= 2.00V, v0 = 1.00V
b) v1= 1.00V, v2= 1.00V, v0 = 0V
c) v1= 1.00V, v2= 1.10V, v0 = 10.1V
d) v1= 0.99V, v2= 2.00V, v0 = 1.00V