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# Multiple choice question for engineering

## Set 1

1. In a 3-phase full wave converter, if V is the maximum value of line voltage at the input, then each SCR is subjected to a peak negative voltage of
a) V
b) 3V
c) √3V
d) V/2

Answer: c [Reason:] PIV = √3 Vml in case of a semi-converter.

2. In a 3-phase semi-converter, firing angle is less than 60°, as such each SCR and diode conduct respectively for __________ (in degrees)
a) 60, 60
b) 90, 30
c) 120, 120
d) 180, 180

Answer: c [Reason:] Each will conduct for 120° per cycle whatever the firing angle is.

3. The effect of source inductance on the performance of a 3-phase controlled converter is to
a) increase the average load voltage
b) reduce the average load voltage
c) make the load current continuous
d) remove ripples from the load current

Answer: b [Reason:] It reduces the average value of the output voltage by introducing a overlap delay μ.

4. Which of the following converter circuits would require a neutral point?
a) 3-phase semi-converter
b) 3-phase full converter
c) 3-phase full converter with freewheeling diode
d) 3-phase half wave converter

Answer: d [Reason:] Half wave converter would require delta-star transformer, the secondary winding star connection requires a neutral point.

5. The range of firing angle for a 3-phase, 3-pulse converter feeding a resistive load is __________ (in degrees).
a) 0 to 180
b) 0 to 150
c) 30 to 150
d) 30 to 180

Answer: b [Reason:] Firing angle for a 3-phase, 3-pulse converter feeding a resistive load is 0 to 150 degrees.

6. A 3-phase bridge converter is given a three-phase supply in the phase sequence R-Y-B. Let the neutral to R phase voltage be Vm sinωt. The first SCR (connected to R phase) is fired at an angle of 15°. What is the maximum value at the output terminals at this instant?
a) Vm/√2
b) Vm
c) 1.5 Vm
d) 3 Vm

Answer: c [Reason:] In case of a 3 phase bridge converter, the maximum value of voltage at the output terminal is always 1.5 Vm.

7. The PIV experienced by each SCRs in M-3 converter is __________ times that in a 3-phase full converter having the same output voltage.
a) 0.5
b) 1
c) 2
d) 1.5

Answer: c [Reason:] In case of M-3 type of connection, the devices have to handle more peak inverse voltage (PIV) than the 3-phase full converter which has 6 SCRs.

8. Each SCR of a 3-phase 6-pulse converter conducts for
a) 120 degrees
b) 60 degrees
c) 180 degrees
d) 360 degrees

Answer: a [Reason:] A 3-phase 6-pulse converter is nothing but the 3-pulse full controlled converter using 6 devices each conducting for 120°.

9. A 3-phase full converter has an average output voltage of 365 V for zero degree firing and resistive load. For a firing angle of 90 degree, the output voltage would be
a) 125 V
b) 569 V
c) 365 V
d) zero

Answer: d [Reason:] Cos 90 = 0.

10. Semi-converters are
a) single quadrant converters
b) double quadrant converters
c) three quadrant converters
d) none of the mentioned

Answer: a [Reason:] Semi-converters are single quadrant converters, because the voltage and current can only be both positive due to the diodes connected.

## Set 2

1. Which of the following is a defined quantity?
a) Pressure
b) Polarity
c) Money
d) Length

Answer: d [Reason:] The SI units are based on 7 defined quantities namely length, mass, time, electric current, thermodynamic temperature, amount of substance and also on the luminous intensity.

2. The basic unit for luminous intensity is
a) Ampere
b) Candela
c) Coulomb

Answer: b [Reason:] The unit for luminous intensity is candela and its symbol is cd.

3. Which of the following is a correct relation?
a) Giga>Mega>Tera
b) Mega>Tera>Giga
c) Tera>Mega>Giga
d) Tera>Giga>Mega

Answer: d [Reason:] Tera – 1012 Giga – 109 Mega – 106.

4. Charge is
a) Unipolar
b) Bipolar
c) Tripolar
d) Non – Polar in nature

Answer: b [Reason:] Charge is bipolar since it can be expressed in terms of positive and negative.

5. Separation of charge creates
a) Current
b) Voltage
c) Resistance
d) Friction

Answer: b [Reason:] An electric force called voltage is created by separation of charge where as an electric fluid called current is created by motion of charge.

6. The energy per unit charge is
a) Voltage
b) Power
c) Current
d) Work

Answer: a [Reason:] Voltage v=dw/dq and its SI unit is volt.

7. If charge q = 3t2 + 2 then current is given by
a) 6t + 2
b) 3t2
c) 6t
d) 3t2 + 2

Answer: c [Reason:] I = dq/dt = d/dt(3t2 + 2) = 6t.

8. If energy w = 200t2+ 99 et + 2 then calculate at 0.1 sec
a) 148.52
b) 149.41
c) 149.95
d) 148.39

Answer: b [Reason:] p=dw/dt =400t +99et =400(0.1) + 99e(0.1) =149.41

9. For the following circuit power is given by a) –Vi
b) Vi
c) 0
d) 1

Answer: a [Reason:] P = vi. The algebraic sign of power depends on movement of charge through the drop and rise of voltage.

10. ‘Positive Power’ meaning ___________
a) Power is being delivered to circuit
b) Power is being extracted from circuit
c) No power supply
d) Input and output powers are equal

Answer: a [Reason:] P > 0 means +ve Power Being delivered P < 0 means –ve Power Being extracted.

## Set 3

1. Find the voltage across 24Ω resistor by using Thevenin’s theorem. a) 8V
b) 9V
c) 1V
d) 6V

Answer: a [Reason:] 1. Remove 24Ω resistor and calculate the voltage across the open circuit. 2. Calculate the thevenin’s resistance and by using it, the thevenin’s current. 3. V24Ω=I*R (can also verify by using Nodal analysis).

2. Calculate Thevenin’s voltage for the network shown below where the voltage source is 4V. a) 6V
b) 4.71V
c) 5V
d) 1V

Answer: c [Reason:] In the circuit given, thevenin’s voltage is nothing but the open circuit voltage which is Vx. Applying KVL, it is obtained.

3. Find the Thevenin’s resistance for the network given. a) 6.75Ω
b) 5.85Ω
c) 4.79Ω
d) 1.675Ω

Answer: a [Reason:] Remove all the voltage/current sources and calculate the equivalent resistance.

4. Find the current through (5+j4) Ω resistor. a) 0.9-j0.2 A
b) 0.78-j0.1 A
c) 2.7-j0.5 A
d) 1A

Answer: a [Reason:] 1. Remove the 5+j4 Ω branch and calculate thevenin’s voltage. (V= v across 6Ω resistor- v across 8Ω resistor) 2. Calculate Zth. (10//6 and 8//8) 3. Current= (Vth/ (Zth+Z).

5. The voltage across 6Ω resistor is __________ a) 7.5V
b) 6.78V
c) 20V
d) 8.5V

Answer: d [Reason:] Remove the resistor across which voltage is to be calculated and short circuit it. By using short circuit current and resistance calculate the current across 6Ω resistor and thereby voltage. (In this 10Ω resistor is also short-circuited since 10//0).

6. Find the Norton’s current for the circuit given below. a) 5A
b) 3.33A
c) 4A
d) 1.66A

Answer: c [Reason:] IN= (20/10) + (10/5).

7. Calculate IN for the given network. a) 0A
b) 1A
c) 4.37A
d) 0.37A

Answer: a [Reason:] Using nodal analysis Vx is calculated. IN =Vx/4.

8. Calculate RTh for the network given. a) 8Ω
b) 7Ω
c) 2Ω
d) 1Ω

Answer: b [Reason:] 5//20 and then in series with 3Ω resistor.

9. Thevenin’s equivalent circuit consists of a ____________
a) Voltage source in series with a resistor
b) Current source in parallel with a resistor
c) Voltage source in parallel with a resistor
d) Current source in series with a resistor

Answer: a [Reason:] Thevenin’s equivalent circuit contains a Voltage source in series with a resistor.

10. Norton’s equivalent circuit consists of a _____________
a) Voltage source in series with a resistor
b) Current source in parallel with a resistor
c) Both voltage and current sources
d) Current source in series with a resistor

Answer: b [Reason:] Norton’s equivalent circuit consists of a Current source in parallel with a resistor.

11. Thevenin’s voltage is equal to ____________
a) Short circuit voltage
b) Open circuit current
c) Open circuit voltage
d) Short circuit current

Answer: c [Reason:] Thevenin’s voltage is equal to open circuit voltage.

12. Norton’s current is equal to ____________
a) Short circuit voltage
b) Open circuit current
c) Open circuit voltage
d) Short circuit current

Answer: d [Reason:] Norton’s current is equal to Short circuit current.

13. Thevenin’s resistance RTh = ___________
a) VTh/ISC
b) VSC/ITh
c) VTh/ITh
d) VSC /ISC

Answer: a [Reason:] Thevenin’s resistance is defined as the ratio of open circuit voltage to the short circuit current across the terminals of the original circuit.

14. What is the expression forthe thevenin’s current if there is an external resistance in series with the RTh?
a) VTh/ITh
b) VTh/ (RTh-R)
c) VTh/ (RTh+R)
d) VTh/RTh

Answer: c [Reason:] ITh= VTh/ (RTh+R).

15. One can find the thevenin’s resistance simply by removing all voltage/current sources and calculating equivalent resistance.
a) False
b) True

Answer: b [Reason:] Yes. One can find the thevenin’s resistance simply by removing all voltage/current sources and calculating equivalent resistance.

## Set 4

1. The Wheatstone Bridge is mainly used to measure ______________
a) Currents
b) Voltages
c) Node potentials
d) Resistances

Answer: d [Reason:] Resistances can be measured by various methods. Wheatstone bridge is one such method. In this method resistances in the range of 1Ω to 1 MΩ can be measured.

2. The relation between the resistances in the given Wheatstone bridge circuit is _____________ a) P/S = R/Q
b) PR = QS
c) P/Q = R/S
d) PQ = RS

Answer: c [Reason:] The relation is P/Q=R/S or PS=QR.

3. Find the unknown resistance value in given circuit. a) 10.2Ω
b) 11.7Ω
c) 10.5Ω
d) 11.5Ω

Answer: a [Reason:] A/B=C/D. Using this D= 10.2Ω.

4. Lower resistances are difficult to measure using Wheatstone bridge circuit because of ____________
a) Leakage currents
b) I2R effects
c) Power dissipation
d) Thermal breakdown

Answer: b [Reason:] A standard Wheatstone bridge couldn’t measure lower resistances because of thermoelectric voltages which are generated at the junctions of the dissimilar metals and also because of thermal heating effects- that is, i2R effects.

5. If P/Q=1, unknown resistance S=1000Ω and R could be varied from 0 to 100Ω then the bridge could be ___________
a) A balanced circuit
b) A rectified circuit
c) An unbalanced circuit
d) An identical circuit

Answer: c [Reason:] P/Q=R/S. If P/Q=1 then according to given range of R and S, the bridge circuit could never be a balanced one.

6. The other name for Delta connection is ___________
a) Star connection
b) Pi connection
c) T connection
d) Y connection

Answer: b [Reason:] Delta connection is also known as Pi connection because the ∆ can be shaped into π without disturbing the electrical equivalence of both the structures.

7. Star connection can also be called as Y (or) T connection.
a) True
b) False

Answer: a [Reason:] Star connection can also be called as Y (or) T connection because the star can be shaped into Y or T without disturbing the electrical equivalence of both the structures.

8. If R2 = RC RA / (RA +RB +RC ) then R3 equals?
a) RA RB / (RA +RB +RC )
b) RC RA / (RA +RB +RC )
c) RB RC / (RA +RB +RC )
d) RX RA / (RA +RB +RC )

Answer: a [Reason:] R3 = RA RB / (RA +RB +RC ).

9. Convert the given Delta circuit to star circuit and give the Ra , Rb and Rc values. a) Ra=5Ω, Rb = 4.5Ω, Rc=4.67Ω
b) Ra=4Ω, Rb=4.30Ω, Rc=4.66Ω
c) Ra=3Ω, Rb=4Ω, Rc =5Ω
d) Ra=5.2Ω, Rb=4.2Ω, Rc =4.89Ω

Answer: b [Reason:] By using the standard formulae the delta circuit can be converted into star circuit.

10. Find VAB if iAB = 5A. a) 32.76V
b) 35.56V
c) 36.12V
d) 34.21V

Answer: d [Reason:] By converting the star circuits into the delta and then measuring the equivalent resistance, voltage value can be calculated using this resistance and the given current value.

11. Convert the given star network into Pi network and calculate the sum of all the resistances in the obtained Pi network. a) 125.5Ω
b) 122.5Ω
c) 127.8Ω
d) 129.8Ω

Answer: b [Reason:] Conversion of given network into delta gives the resistances. After that sum of the resistances equals 122.5Ω.

12. The star and delta networks would be electrically equal if resistances measured between any pair of terminals __________
a) Is different
b) Greater in star
c) Greater in delta
d) Is equal

Answer: d [Reason:] The star and delta networks would be electrically equal if a resistance measured between any pair of terminals is same.

13. A Wheatstone bridge is balanced when the galvanometer shows __________ reading.
a) 0A
b) 1A
c) Infinity
d) -1A

Answer: a [Reason:] A Wheatstone bridge is balanced when the galvanometer shows 0A reading when resistors obey P/Q=R/S.

14. __________ are difficult to measure using Wheatstone bridge.
a) Higher resistances
b) Currents
c) Lower resistances
d) Voltages

Answer: c [Reason:] Specifically Kelvin Bridge is used for measuring lower resistances.

15. What will be the resistance between B and C when the network given below is converted into delta? a) 13Ω
b) 8.66Ω
c) 6.5Ω
d) 7.33Ω

Answer: b [Reason:] Resistance between B and C = 2+4+ ((2*4)/3).

## Set 5

1.Dependent sources are _____________ types.
a) 3
b) 2
c) 4
d) 1

Answer: c [Reason:] Dependent sources are 4 types. Voltage controlled voltage/current source and current controlled current/voltage source.

2. In case of a dependent voltage/current source, the value of this voltage/current source depends on _________
a) Voltage/current sources of an external circuit
b) Voltage/current source present somewhere in the circuit
c) Only on voltage sources
d) Only on current sources

Answer: b [Reason:] The name dependent itself tells us that they are dependent on some other source. A dependent voltage/current source depends on the value of the voltage/current source present somewhere in the circuit itself.

3. Find i0 and v0 in the above circuit..
a) 26A, 260v
b) 28A, 280v
c) 27A, 275v
d) 29A, 285v

Answer: b [Reason:] Applying KVL in loop1: 300= 3i + 10i0 ———- (1) and i0= i+3i =4i (1) → 300= 43i ,on solving i =6.976A, i0=27.90A, v0=279V.

4.The value of the voltage controlled current source ia=βva given β=0.3 and va=9.5mV.
a) 2.5 mA
b) 2.85 mA
c) 1.75 mA
d) 1.2 mA

Answer: b [Reason:] ia= 0.3*9.5*10-3=2.85mA.

5.Find I0 in the following circuit, R1= 1.1 kilo ohms, R2=3.25 kilo ohms, V= 3.7 v. a) 1.5 mA
b) 2 mA
c) 0.5 mA
d) 1.2 mA

Answer: d [Reason:] By using the fact that the current is same in series connection resistors and voltage is same parallel, the above problem can be solved. I0 is divided into αIx and Ix. So, calculation these two gives the required current value.

6. The value of the current controlled voltage source ,given β=0.8 and ia=9.5mA, is ___________
a) 8mV
b) 7.6mV
c) 0.0011mV
d) 0.0051mV

Answer: b [Reason:] va=βia

7. In a VCIS which is the controlled source and which one is the dependent source?
a) V-contorller, I-dependent
b) V-dependent, I-controller
c) Both V and I are controllers
d) Both V and I are dependent

Answer: a [Reason:] VCIS: Voltage-controlled current source. ia=βva, current value depends on the voltage value so voltage source is the controller and current is the dependent source.

8. In an ICVS which is the controlled source and which one is the dependent source?
a) V-controller, I-dependent
b) V-dependent, I-controller
c) Both V and I are controllers
d) Both V and I are dependent

Answer: b [Reason:] ICVS :Current-controlled voltage source. va=βia , voltage value depends on the current value so current source is the controller and voltage is the dependent source.

9. What is the other name for Dependent sources?
a) Uncontrolled sources
b) Time response elements
c) Steady state elements
d) Controlled sources