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# Multiple choice question for engineering

## Set 1

1. What is the coupling coefficient when all the flux of coil 1 links with coil 2?
a) 0
b) 100
c) 1
d) Insufficient information provided

Answer: c [Reason:] When all the flux of coil 1 links with coil 2 it is known as an ideal coupling where the coupling coefficient is 1.

2. What is the coupling coefficient when there is ideal coupling?
a) 0
b) 100
c) 1
d) Insufficient information provided

Answer: c [Reason:] When all the flux of coil 1 links with coil 2 it is known as an ideal coupling where the coupling coefficient is 1.

3. Can the coupling coefficient practically ever be equal to 1?
a) Yes
b) No
c) Depends on current in coil 1
d) Depends on current in coil 2

Answer: b [Reason:] Coupling coefficient can never be equal to 1 because all the flux of coil 1 can never link with coil 2. There are bound to be losses.

4. Mutual inductance between to coupled coils depends on?
a) Amount of flux linkage
b) Rate of change of flux linkage
c) Rate of change of current
d) Flux density

Answer: b [Reason:] Faraday’s law of induction states that, the magnitude of the induced EMF is the product of the number of turns of the coil and the rate of change of flux linkage in it. Hence, the mutual inductance depends on the rate of change of flux linkage.

5. Which, among the following, is the correct formula to fing coupling coefficient?
a) k=M/sqrt(L1L2)
b) k=M/sqrt(L12)
c) k=M/sqrt(L22)
d) k=M/(L1L2)

Answer: a [Reason:] The correct formula for coupling coefficient is k=M/sqrt(L1L2). Where, L1 and L2 are the inductance values of the first and second coil respectively and M is the mutual inductance.

6. What happens to coupling coefficient when the flux linkage of coil 1 and coil 2 increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Answer: a [Reason:] When the flux linkage of coil 1 and coil 2 increases, its mutual inductance increases. The coupling coefficient is directly proportional to the mutual inductance hence as mutual inductance increases, the coupling coefficient increases.

7. What is the SI unit of coupling coefficient?
a) H
b) H-1
c) No unit
d) H2

Answer: c [Reason:] The expression to find mutual inductance is k=M/sqrt(L1L2)= H/sqrt(H*H)= 1. Therefore it does not have any unit.

8. Find the coupling coefficient if the Mutual inductance is 20H, the inductance of coil 1 is 2H and the inductance of coil 2 is 8H.
a) 5
b) 20
c) 2
d) 8

Answer: a [Reason:] we know that: k=M/sqrt(L1L2) Substituting the values from the question, we get k=5.

9. Find the value of x if the Mutual inductance is x H, the inductance of coil 1 is 2H and the inductance of coil 2 is 8H. The coupling coefficient is 5.
a) 10H
b) 20H
c) 16H
d) 15H

Answer: b [Reason:] we know that: k=M/sqrt(L1L2) Substituting the values from the question, we get M=20H.

10. Find the value of x if the Mutual inductance is 20H, the inductance of coil 1 is xH and the inductance of coil 2 is 8H. The coupling coefficient is 5.
a) 2H
b) 4H
c) 6H
d) 8H

Answer: a [Reason:] we know that: k=M/sqrt(L1L2) Substituting the values from the question, we get L1=2H.

## Set 2

1. In a series RLC circuit, the phase difference between the current in the capacitor and the current in the resistor is?
a) 0 degrees
b) 90 degrees
c) 180 degrees
d) 360 degrees

Answer: a [Reason:] In a series RLC circuit, the phase difference between the current in the capacitor and the current in the resistor is 0 degrees because the same current flows in the capacitor as well as the resistor.

2. In a series RLC circuit, the phase difference between the current in the inductor and the current in the resistor is?
a) 0 degrees
b) 90 degrees
c) 180 degrees
d) 360 degrees

Answer: a [Reason:] In a series RLC circuit, the phase difference between the current in the inductor and the current in the resistor is 0 degrees because the same current flows in the inductor as well as the resistor.

3. In a series RLC circuit, the phase difference between the current in the capacitor and the current in the inductor is?
a) 0 degrees
b) 90 degrees
c) 180 degrees
d) 360 degrees

Answer: a [Reason:] In a series RLC circuit, the phase difference between the current in the inductor and the current in the capacitor is 0 degrees because the same current flows in the inductor as well as the capacitor.

4. In a series RLC circuit, the phase difference between the current in the circuit and the voltage across the resistor is?
a) 0 degrees
b) 90 degrees
c) 180 degrees
d) 360 degrees

Answer: a [Reason:] In a series RLC circuit, the phase difference between the voltage across the resistor and the current in the circuit is 0 degrees because they are in phase.

5. In a series RLC circuit, the phase difference between the current in the circuit and the voltage across the capacitor is?
a) 0 degrees
b) 90 degrees
c) 180 degrees
d) 360 degrees

Answer: b [Reason:] In a series RLC circuit, the phase difference between the voltage across the capacitor and the current in the circuit is 90 degrees.

6. In a series RLC circuit, the phase difference between the current in the circuit and the voltage across the resistor is?
a) 0 degrees
b) 90 degrees
c) 180 degrees
d) 360 degrees

Answer: b [Reason:] In a series RLC circuit, the phase difference between the voltage across the inductor and the current in the circuit is 90 degrees.

7. _________ the resonant frequency, the current in the inductor lags the voltage in a series RLC circuit.
a) Above
b) Below
c) Equal to
d) Depends on the circuit

Answer: a [Reason:] The current in the inductor lags the voltage in a series RLC circuit above the resonant frequency.

8. _________ the resonant frequency, the current in the capacitor leads the voltage in a series RLC circuit.
a) Above
b) Below
c) Equal to
d) Depends on the circuit

Answer: b [Reason:] The current in the capacitor leads the voltage in a series RLC circuit below the resonant frequency.

9. The current in the inductor ___________ the voltage in a series RLC circuit above the resonant frequency.
b) Lags
c) Equal to
d) Depends on the circuit

Answer: a [Reason:] The current in the inductor lags the voltage in a series RLC circuit above the resonant frequency.

10. The current in the capacitor ___________ the voltage in a series RLC circuit below the resonant frequency.
b) Lags
c) Equal to
d) Depends on the circuit

Answer: b [Reason:] The current in the capacitor leads the voltage in a series RLC circuit above the resonant frequency.

## Set 3

1. The value of the 3 resistances when connected in star connection is_________ a) 2.32ohm,1.22ohm, 4.54ohm
b) 3.55ohm, 4.33ohm, 5.67ohm
c) 2.78ohm, 1.67ohm, 0.83ohm
d) 4.53ohm, 6.66ohm, 1.23ohm

Answer: d [Reason:] Following the delta to star conversion: R1=10*5/(10+5+3) R2=10*3/(10+5+3) R3=5*3/(10+5+3).

2. Which, among the following is the right expression for converting from delta to star?
a) R1=Ra*Rb/(Ra+Rb+Rc), R2=Rb*Rc/(Ra+Rb+Rc), R3=Rc*Ra/(Ra+Rb+Rc)
b) R1=Ra/(Ra+Rb+Rc), R2=Rb/(Ra+Rb+Rc), Rc=/(Ra+Rb+Rc)
c) R1=Ra*Rb*Rc/(Ra+Rb+Rc), R2=Ra*Rb/(Ra+Rb+Rc), R3=Ra/(Ra+Rb+Rc)
d) R1=Ra*Rb*Rc/(Ra+Rb+Rc), R2=Ra*Rb*Rc/(Ra+Rb+Rc), R3=Ra*Rb*Rc/(Ra+Rb+Rc)

Answer: a [Reason:] After converting to star, each star connected resistance is equal to the product of the resistances it is connected to and the total sum of the resistances. Hence R1=Ra*Rb/(Ra+Rb+Rc), R2=Rb*Rc/(Ra+Rb+Rc), R3=Rc*Ra/(Ra+Rb+Rc).

3. Find the equivalent star network. a) 2.3ohm, 2.3ohm, 2.3ohm
b) 1.2ohm, 1.2ohm, 1.2ohm
c) 3.3ohm, 3.3ohm, 3.3ohm
d) 4.5ohm, 4.5ohm, 4.5ohm

Answer: b [Reason:] The 6 ohm and 9 ohm resistances are connected in parallel. Their equivalent resistances are: 6*9/(9+6)=3.6 ohm. The 3 3.6 ohm resistors are connected in delta. Converting to star: R1=R2=R3= 3.6*3.6/(3.6+3.6+3.6)=1.2 ohm.

4. Star connection is also known as__________
a) Y-connection
b) Mesh connection
c) Either Y-connection or mesh connection
d) Neither Y-connection nor mesh connection

Answer: a [Reason:] The star connection is also known as the Y-connection because its formation is like the letter Y.

5. Rab is the resistance between the terminals A and B, Rbc between B and C and Rca between C and A. These 3 resistors are connected in delta connection. After transforming to star, the resistance at A will be?
a) Rab*Rac/(Rab+Rbc+Rca)
b) Rab/(Rab+Rbc+Rca)
c) Rbc*Rac/(Rab+Rbc+Rca)
d) Rac/(Rab+Rbc+Rca)

Answer: a [Reason:] When converting from delta to star, the resistances in star connection is equal to the product of the resistances it is connected to, divided by the total sum of the resistance. Hence Rab*Rac/(Rab+Rbc+Rca).

6. Rab is the resistance between the terminals A and B, Rbc between B and C and Rca between C and A. These 3 resistors are connected in delta connection. After transforming to star, the resistance at B will be?
a) Rac/(Rab+Rbc+Rca)
b) Rab/(Rab+Rbc+Rca)
c) Rbc*Rab/(Rab+Rbc+Rca)
d) Rab/(Rab+Rbc+Rca)

Answer: c [Reason:] When converting from delta to star, the resistances in star connection is equal to the product of the resistances it is connected to, divided by the total sum of the resistance. Hence Rab*Rbc/(Rab+Rbc+Rca).

7. Rab is the resistance between the terminals A and B, Rbc between B and C and Rca between C and A. These 3 resistors are connected in delta connection. After transforming to star, the resistance at C will be?
a) Rac/(Rab+Rbc+Rca)
b) Rab/(Rab+Rbc+Rca)
c) Rbc*Rac/(Rab+Rbc+Rca)
d) Rab/(Rab+Rbc+Rca)

Answer: c [Reason:] When converting from delta to star, the resistances in star connection is equal to the product of the resistances it is connected to, divided by the total sum of the resistance. Hence Rac*Rbc/(Rab+Rbc+Rca).

8. Find the current in the circuit. a) 0.54A
b) 0.65A
c) 0.67A
d) 0.87A

Answer: a [Reason:] The 3 5 ohm resistors are connected in delta. Changing it to star: R1=R2=R3= 1.67 ohm. One of the 1.67 ohm resistors are connected in series with the 2 ohm resistor and another 1.67 ohm resistor is connected in series to the 3 ohm resistor. The resulting network has a 1.67 ohm resistor connected in series with the parallel connection of the 3.67 and 4.67 resistors. The equivalent resistance is: 3.725A. I=2/3.725= 0.54A.

9. If a 6 ohm, 2ohm and 4ohm resistor is connected in delta, find the equivalent star connection.
a) 1ohm, 2ohm, 3ohm
b) 2ohm, 4ohm, 7ohm
c) 5ohm, 4ohm, 2ohm
d) 1ohm, 2ohm, 32/3ohm

Answer: d [Reason:] Using the delta to star conversion formula: R1=2*6/(2+6+4) R2=2*4/(2+6+4) R3=4*6/(2+6+4).

10. If a 4ohm, 3ohm and 2ohm resistor is connected in delta, find the equivalent star connection.
a) 8/9ohm, 4/3ohm, 2/3ohm
b) 8/9ohm, 4/3ohm, 7/3ohm
c) 7/9ohm, 4/3ohm, 2/3ohm
d) 8/9ohm, 5/3ohm, 2/3ohm

Answer: a [Reason:] Using the delta-star conversion formula: R1=4*3/(2+3+4) R2=2*3/(2+3+4) R3=2*4/(2+3+4).

## Set 4

1. The unit for dielectric strength is ____________
a) V/m2
b) MV/m2
c) MV/m
d) Vm

Answer: b [Reason:] Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and length, its unit is MV/m.

2. If the Voltage increases, what happens to dielectric strength?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Answer: a [Reason:] Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and length. Hence as potential increases, dielectric strength also increases.

3. The electric fields of dielectrics having the same cross sectional area in series is related to their relative permittivities in which way?
a) Directly proportional
b) Inversely proportional
c) Equal
d) Not related

Answer: b [Reason:] Let us consider two plates having fields E1 and E2 and relative permittivities e1 and e2. Then, E1=Q/(e0*e1*A) and E2=Q/(e0*e2*A), where e0=absolute permittivity and A=area of cross section. From the given expression, we see that E1/E2=e2/e1, hence the electric field is inversely proportional to the relative permittivities.

4. If the potential difference in a material is 4MV and the thickness of the material is 2m, calculate the dielectric strength.
a) 2MV/m
b) 4MV/m
c) 6MV/m
d) 8MV/m

Answer: a [Reason:] Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness. Dielectric strength= V/t= 4/2= 2MV/m.

5. If the dielectric strength of a material is 4MV/m and its potential difference is 28MV, calculate the thickness of the material.
a) 4m
b) 7m
c) 5m
d) 11m

Answer: b [Reason:] Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness. V/dielectric strength= t= 28/4=7m.

6. If the thickness of the material increases, what happens to the dielectric strength?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Answer: b [Reason:] Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness. Hence as thickness increases, dielectric strength decreases.

7. The thickness of a material having dielectric strength 10MV/m is 5m, calculate the potential difference.
a) 2MV
b) 10MV
c) 50MV
d) 100MV

Answer: c [Reason:] Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness. V=t*dielectric strength= 5*10=50MV.

8. What happens to the potential drop between the two plates of a capacitor when a dielectric is introduced between the plates?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Answer: b [Reason:] When a dielectric is introduced between the two plates of a parallel plate capacitor, the potential difference decreases because the potential difference of the dielectric is subtracted from it.

9. A dielectric is basically a ____________
a) Capacitor
b) Conductor
c) Insulator
d) Semiconductor

Answer: c [Reason:] A dielectric is basically an insulator because it has all the properties of an insulator.

10. What happens to the potential difference between the plates of a capacitor as the thickness of the dielectric slab increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Answer: b [Reason:] When a dielectric is introduced between the plates of a capacitor, its potential difference decreases. New potential difference= potential difference without dielectric-potential difference of dielectric. Hence as the thickness of the dielectric slab increases, a larger value is subtracted fro the original potential difference.

## Set 5

1. According to Faraday’s laws of electromagnetic inductance, an emf is induced in a conductor whenever?
a) The conductor is perpendicular to the magnetic field
b) Lies in the magnetic field
c) Cuts magnetic lines of flux
d) Moves parallel to the magnetic field

Answer: c [Reason:] An emf is induced, according to Faraday’s laws of electromagnetic inductance, whenever the conductor in a magnetic field cuts the magnetic lies of flux, varying the flux per unit area, and hence the magnetic field which induces an emf.

2. Direction of induced emf is determined by __________
a) Fleming’s left hand rule
b) Fleming’s right hand rule
d) Right hand thumb rule

Answer: b [Reason:] Fleming’s left hand rule stated that if the index finger points towards magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

3. “The direction of an induced e.m.f. is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that e.m.f.”, this is the statement for?
a) Fleming’s left hand rule
b) Fleming’s right hand rule
d) Lenz’s law

Answer: d [Reason:] The above statement is that of Lenz’s law. It is used to determine the direction of the induced emf.

4. According to Fleming’s right hand rule, the thumb points towards?
a) Current
b) E.M.F.
c) Motion of the conductor
d) Magnetic flux

Answer: c [Reason:] Fleming’s left hand rule stated that if the index finger points towards magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

5. According to Fleming’s right hand rule, the index finger points towards?
a) Current
b) E.M.F.
c) Motion of the conductor
d) Magnetic flux

Answer: d [Reason:] Fleming’s left hand rule stated that if the index finger points towards magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

6. According to Fleming’s right hand rule, the middle finger points towards?
a) Current
b) E.M.F.
c) Motion of the conductor
d) Magnetic flux

Answer: b [Reason:] Fleming’s left hand rule stated that if the index finger points towards magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

7. The relation between direction of induced emf and direction of motion of the conductor is?
a) Parallel
b) Equal
c) Not related
d) Perpendicular

Answer: d [Reason:] According to Fleming’s right hand rule, the induced emf, the motion of the conductor and the magnetic flux are mutually perpendicular.

8. The relation between direction of induced emf and direction of magnetic flux is _______
a) Parallel
b) Equal
c) Not related
d) Perpendicular

Answer: d [Reason:] According to Fleming’s right hand rule, the induced emf, the motion of the conductor and the magnetic flux are mutually perpendicular.

9. The relation between direction of magnetic flux and direction of motion of the conductor is _______
a) Parallel
b) Equal
c) Not related
d) Perpendicular