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# Multiple choice question for engineering

## Set 1

1. Which among the following is a unit for electrical energy?
a) V(volt)
b) Wh(kilowatt-hour)
c) Ohm
d) C(coloumb)

Answer: b [Reason:] Watt is a unit of power and hour is a unit of time. Energy is the product of power and time, hence the unit for power is kWh.

2. A bulb has a power of 200W. What is the energy dissipated by it in 5 minutes?
a) 60J
b) 1000J
c) 60kJ
d) 1kJ

Answer: c [Reason:] Here, Power = 200w and time = 5min. E=Pt => E= 200*5= 1000Wmin=60000Ws= 60000J= 60kJ.

3. Out of the following, which one is not a source of electrical energy?
a) Solar cell
b) Battery
c) Potentiometer
d) Generator

Answer: c [Reason:] A potentiometer is an instrument used for measuring voltage hence it is not a source for electrical energy.

4. Calculate the energy dissipated by the circuit in 50 seconds.

a) 50kJ
b) 50J
c) 100j
d) 100kJ

Answer: a [Reason:] Here V = 100 and R = 10. Power in the circuit= V2/R= 1002/10= 1000W. Energy= Pt= 1000*50= 50000J= 50kJ.

5. Which among the following is an expression for energy?
a) V2It
b) V2Rt
c) V2t/R
d) V2t2/R

Answer: c [Reason:] Expression for power= VI, substituting I from ohm’s law we can write, P=V2/R. Energy is the product of power and time, hence E=Pt= V2t/R.

6. Calculate the energy in the 10 ohm resistance in 10 seconds.

a) 400J
b) 40kJ
c) 4000J
d) 4kJ

Answer: b [Reason:] Since the resistors are connected in parallel, the voltage across both the resistors are the same, hence we can use the expression P=V2/R. P=2002/10= 4000W. E=Pt= 4000*10=40000Ws= 40000J= 40kJ.

7. A battery converts___________
a) Electrical energy to chemical energy
b) Chemical energy to electrical energy
c) Mechanical energy to electrical energy
d) Chemical energy to mechanical energy

Answer: b [Reason:] A battery is a device in which the chemical elements within the battery react with each other to produce electrical energy.

8. A current of 2A flows in a wire offering a resistance of 10ohm. Calculate the energy dissipated by the wire in 0.5 hours.
a) 72Wh
b) 72kJ
c) 7200J
d) 72kJh

Answer b [Reason:] Here I (current) = 2A and Resistance(R) = 10ohm. Power= I2R= 22*10=40. Energy=Pt= 40*0.5*60*60= 72000J=72kJ.

9. Calculate the energy in the 5 ohm resistor in 20 seconds.

a) 21.5kJ
b) 2.15kJ
c) 2.15J
d) 21.5kJ

Answer: a [Reason:] The current in the circuit is equal to the current in the 5 ohm resistor since it a series connected circuit, hence I=220/(5+10)=14.67A. P=I2R= 14.672*5=1075.8W. E=Pt= 1075.8*20= 21516J=21.5kJ.

10. Practically, if 10kJ of energy is supplied to a device, how much energy will the device give back?
a) Equal to10kJ
b) Less than 10kJ
c) More than 10kJ
d) Zero

Answer: b [Reason:] Practically, if 10kJ of energy is supplied to a system, it returns less than the supplied energy because, some of the energy is lost as heat energy, sound energy etc.

## Set 2

1. Work done in charging a capacitor is ____________
a) QV
b) 1/2QV
c) 2QV
d) QV2

Answer: b [Reason:] We know that work done= Q2/2C. Substituting C as Q/V, we get work done= Q/2V.

2. Energy stored in 2000mF capacitor charged to a potential difference of 10V is?
a) 0.1J
b) 0.2J
c) 0.3J
d) 0.4J

Answer: a [Reason:] From the expression: WD= CV2/2= 0.1J.

3. When do we get maximum energy from a set of capacitors?
a) When they are connected in parallel
b) When they are connected in series
d) Both in series and parallel
d) Insufficient information provided

Answer: a [Reason:] We get maximum energy when capacitors are connected in parallel because the equivalent capacitance is larger than the largest individual capacitance when connected in parallel. The relation between capacitance and energy is: Energy=CV2/2, hence as the capacitance increases, the energy stored in it also increases.

4. If the charge stored in a capacitor is 4C and the value of capacitance is 2F, calculate the energy stored in it.
a) 2J
b) 4J
c) 8J
d) 16J

Answer: b [Reason:] The expression for finding the value of energy is: U=Q2/2C= 4*4/(2*2)= 4J.

5. If the charge in a capacitor is 4C and the energy stored in it is 4J, find the value of capacitance.
a) 2F
b) 4F
c) 8F
d) 16F

Answer: a [Reason:] The expression for finding the value of energy is: U=Q2/2C. Substituting the values of U and Q, we get C=2F.

6. If the charge in a capacitor is 4C and the energy stored in it is 4J, calculate the voltage across its plates.
a) 2V
b) 4V
c) 8V
d) 16V

Answer: a [Reason:] The expression for finding the value of energy is: U=Q2/2C. Substituting the values of U and Q, we get C=2F. V=Q/C, hence V=4/2=2V.

7. Calculate the energy in the 2F capacitor.

a) 8.6kJ
b) 64kJ
c) 64J
d) 6.4kJ

Answer: d [Reason:] From the expression: WD= CV2/2= 2*802/2=6400J=6.4kJ.

8. Calculate the energy in the 4F capacitor.

a) 128kJ
b) 1.28kJ
c) 12.8kJ
d) 1280J

Answer: c [Reason:] From the expression: WD= CV2/2= 4*802/2=12800J=12.8kJ.

9. Calculate the energy stored in the combination of the capacitors.

a) 192kJ
b) 1.92kJ
c) 19.2kJ
d) 1920J

Answer: c [Reason:] The equivalent capacitance is: Ceq=4+2=6F. From the expression: WD= CV2/2= 6*802/2=19200J=19.2kJ.

10. What is the energy in a capacitor if the voltage is 5V and the charge is10C?
a) 6.25J
b) 2.35J
c) 6.54J
d) 4.55J

Answer: a [Reason:] We know that Q/V=C. Hence the value of capacitance is 2F. From the expression: U=Q2/2C, we get U= 6.25J.

## Set 3

1. If the current in a coil having a constant inductance of L henrys grows at a uniform rate, what is the value of the average current?
a) I
b) I/2
c) I/4
d) 0

Answer: b [Reason:] The average current is the average of the current which flows in the inductor. Hence it is I/2.

2. What is the power in the magnetic field if the current in a coil has a constant inductance of L henrys grows at a uniform rate?
a) LI/2t
b) LI2/2t
c) L/2It
d) L/2I2t

Answer: b [Reason:] The average current is I/2. The e.m.f. induced in the coil is LI/t V. Power= VI, hence the average power= I/2*LI/t.

3. What is the energy stored in the magnetic field if the current in a coil has a constant inductance of L henrys grows at a uniform rate?
a) LI/2
b) LI2/2
c) L/2I
d) L/2I2

Answer: b [Reason:] The average current is I/2. The e.m.f. induced in the coil is LI/t V. Power= VI, hence the average power= I/2*LI/t. The total energy stored= power*t= LI2/2.

4. Find the average current in an inductor if the total current in the inductor is 26A.
a) 10A
b) 26A
c) 13A
d) 5A

Answer: c [Reason:] Average current= I/2. Substituting the value of I from the equation, average current= 13A.

5. Calculate the power in an inductive circuit if the inductance is 10H, the current flowing in the inductor is 2A in 4s.
a) 50W
b) 4W
c) 5W
d) 10W

Answer: c [Reason:] The expression for power in an inductive circuit is: P= LI2/2 Substituting the values from the given question, we get P=5W.

6. Calculate the value of stored energy in an inductor if the value of inductance is 20H and 4A of current flows through it.
a) 220J
b) 150J
c) 190J
d) 160J

Answer: d [Reason:] The expression for energy in an inductor is: W= LI2/2t Substituting the values from the given question, we get W= 160J.

7. Calculate the emf induced in an inductor if the inductance is 10H and the current is 2A in 4s.
a) 2.5V
b) 1.5V
c) 3.5V
d) 5V

Answer: a [Reason:] The expression for emf in an inductive circuit is: emf= LI/2t Substituting the values from the given question, we get emf= 2.5V.

8. Calculate the value of emf in an inductor if the value of inductance is 15H and an average current of 5A flows through it in 10s.
a) 15V
b) 7.5V
c) 10V
d) 5.5V

Answer: b [Reason:] The expression for emf in an inductive circuit is: emf= LI/2t, where I/2 is the average current. Substituting the values from the given question, we get emf= 2.5V.

9. Calculate the current in an inductor if the energy stored is 160J and the inductance is 20H.
a) 1A
b) 2A
c) 3A
d) 4A

Answer: d [Reason:] The expression for energy in an inductor is: W= LI2/2t Substituting the values from the given question, we get I=4A.

10. Find the time taken for the current in an inductor to change to 2A from 0A if the power in the inductor is 5W. The value of inductance is 10H.
a) 1s
b) 2s
c) 3s
d) 4s

Answer: d [Reason:] The expression for power in an inductive circuit is: P= LI2/2 Substituting the values from the given question, we get t=4s.

## Set 4

1. RMS stands for ____________
a) Root Mean Square
b) Root Mean Sum
c) Root Maximum sum
d) Root Minimum Sum

Answer: a [Reason:] RMS stands for Root Mean Square. This value of current is obtained by squaring all the current values, finding the average and then finding the square root.

2. If Im is the maximum value of a sinusoidal voltage, what is the instantaneous value?
a) i=Im/2
b) i=Imsinθ
c) i=Imcosθ
d) i=Im

Answer: b [Reason:] The instantaneous value of a sinusoidal varying current is i=Imsinθ , where theta is the angle from instant of zero current.

3. In a sinusoidal wave, average current is always _______ rms current.
a) Greater than
b) Less than
c) Equal to
d) Not related

Answer: b [Reason:] The average value of current is the sum of all the currents divided by the number of currents whereas RMS current is obtained by squaring all the current values, finding the average and then finding the square root. Hence RMS current is greater than average current.

4. Average value of current over a half cycle is?
a) 0.67Im
b) 0.33Im
c) 6.7Im
d) 3.3Im

Answer: a [Reason:] Integrating idθ in the interval from 0 to pi ( since we are finding the average current over one half cycle) , we get i=0.61Im.

5. Find the average value of current when the current that are equidistant are 4A, 5A and 6A.
a) 5A
b) 6A
c) 15A
d) 10A

Answer: a [Reason:] The average value of current is the sum of all the currents divided by the number of currents. Therefore average current= (5+4+6)/3=5A.

6. What is the correct expression for the rms value of current?
a) Irms=Im/2
b) Irms=Im/√2
c) Irms=Im/4
d) Irms=Im

Answer: b The correct expression for the rms value of current is Irms= Im/√2, where Im is the maximum or peak value of the current.

7. __________ current is found by dividing the area enclosed by the half cycle by the length of the base of the half cycle.
a) RMS current
b) Average current
c) Instantaneous current
d) Total current

Answer: b [Reason:] The average value of current is the sum of all the currents divided by the number of currents. Hence it can also be found by dividing the area enclosed by the half cycle by the length of the base of the half cycle.

8. Peak value divided by the rms value gives us?
a) Peak factor
b) Crest factor
c) Both peak and crest factor
d) Neither peak nor crest factor

Answer: c [Reason:] Peak and crest factor both mean the same thing. Hence the peak value divided by the rms value gives us the peak or crest factor.

9. What is the effective value of current?
a) RMS current
b) Average current
c) Instantaneous current
d) Total current

Answer: a [Reason:] Effective current is also known as the effective current. RMS stands for Root Mean Square. This value of current is obtained by squaring all the current values, finding the average and then finding the square root.

10. What is the type of current obtained by finding the square of the currents and then finding their average and then fining the square root?
a) RMS current
b) Average current
c) Instantaneous current
d) Total current

Answer: a [Reason:] RMS stands for Root Mean Square. This value of current is obtained by squaring all the current values, finding the average and then finding the square root.

## Set 5

1. Which among the following is the correct expression for force between the plates of a parallel plate capacitor?
a) F=epsilon*(V/x)2/2
b) F=epsilon*(V*x)2/2
c) F=epsilon/(V/x)2/2
d) F=epsilon*(V/x)2/3

Answer: a [Reason:] The force is proportional to the square of the potential gradient per metre and the area. Hence the force F=epsilon*(V/x)2/2.

2. When the area of cross section of the plate increases, what happens to the force between the plates?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Answer: a [Reason:] The force of attraction between the two plates of the capacitor is directly proportional to the area of cross section of the plates, hence as area of cross section increases, the force of attraction also increases.

3. When the potential gradient increases, what happens to the force between the plates?
a) Increases
b) Decreases
c) Remains the same
d) becomes zero

Answer: a [Reason:] The force of attraction between the two plates of the capacitor is directly proportional to the potential gradient, hence as potential gradient, the force of attraction also increases.

4. In which of the following mediums, will the force of attraction between the plates of a capacitor be greater?
a) Air
b) Water
c) Does not depend on medium
d) Cannot be determined

Answer: b [Reason:] The absolute permittivity(epsilon) of water is greater than that of air. The expression relating F and epsilon is: F=epsilon*(V/x)2/2. From this expression we can see that as epsilon increases, the force of attraction also increases.

5. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Calculate the force on each plate if the potential difference between the plates is 1kV.
a) 350N
b) 0.035kN
c) 0.035N
d) 3.35kN

Answer: c [Reason:] From the given data: A=pi*d2/4=0.007854m2 Potential gradient= V/x= 10^6V/m F=epsilon*(V/x)2/2 Therefore, F=0.035N.

6. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is ‘a’ mm. The capacitor is placed in air. Force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.
a) 1m
b) 1cm
c) 10cm
d) 1mm

Answer: d [Reason:] From the given data: A=pi*d2/4=0.007854m2 Potential gradient= V/x= 1000/a F=epsilon*(V/x)2/2 Substituting the given values, we find a=1mm.

7. A metal parallel plate capacitor has ‘a’mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.
a) 10mm
b) 100mm
c) 1000m
d) 1000cm

Answer: b [Reason:] From the given data: A=pi*d2/4=pi*a2/4 Potential gradient= V/x= 106V/m F=epsilon*(V/x)2/2 Substituting the given values, we get d=100mm.

8. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Calculate the potential difference between the plates if the force on each plate is 0.035N.
a) 1kV
b) 1V
c) 2kV
d) 2V

Answer: a [Reason:] From the given data: A=pi*d2/4=0.007854m2 Potential gradient= V/x= 1000*V F=epsilon*(V/x)2/2 Substituting the given values in the above expression, we get V=1kV.

9. What happens to the force of attraction between the capacitors when the potential difference between the plates decreases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero