# 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)

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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

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3. Out of the following, which one is not a source of electrical energy?

a) Solar cell

b) Battery

c) Potentiometer

d) Generator

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

a) 50kJ

b) 50J

c) 100j

d) 100kJ

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^{2}/R= 100

^{2}/10= 1000W. Energy= Pt= 1000*50= 50000J= 50kJ.

5. Which among the following is an expression for energy?

a) V^{2}It

b) V^{2}Rt

c) V^{2}t/R

d) V^{2}t^{2}/R

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^{2}/R. Energy is the product of power and time, hence E=Pt= V

^{2}t/R.

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

a) 400J

b) 40kJ

c) 4000J

d) 4kJ

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^{2}/R. P=200

^{2}/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

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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

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^{2}R= 2

^{2}*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

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^{2}R= 14.67

^{2}*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

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## Set 2

1. Work done in charging a capacitor is ____________

a) QV

b) 1/2QV

c) 2QV

d) QV^{2}

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^{2}/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

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^{2}/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

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^{2}/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

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^{2}/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

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^{2}/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

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^{2}/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

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^{2}/2= 2*80

^{2}/2=6400J=6.4kJ.

8. Calculate the energy in the 4F capacitor.

a) 128kJ

b) 1.28kJ

c) 12.8kJ

d) 1280J

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^{2}/2= 4*80

^{2}/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

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^{2}/2= 6*80

^{2}/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

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^{2}/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

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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) LI^{2}/2t

c) L/2It

d) L/2I^{2}t

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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) LI^{2}/2

c) L/2I

d) L/2I^{2}

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^{2}/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

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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

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^{2}/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

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^{2}/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

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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

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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

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^{2}/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

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^{2}/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

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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

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3. In a sinusoidal wave, average current is always _______ rms current.

a) Greater than

b) Less than

c) Equal to

d) Not related

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4. Average value of current over a half cycle is?

a) 0.67Im

b) 0.33Im

c) 6.7Im

d) 3.3Im

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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

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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

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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

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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

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9. What is the effective value of current?

a) RMS current

b) Average current

c) Instantaneous current

d) Total current

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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

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## 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

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^{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

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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

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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

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^{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

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^{2}/4=0.007854m

^{2}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

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^{2}/4=0.007854m

^{2}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

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^{2}/4=pi*a

^{2}/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

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^{2}/4=0.007854m

^{2}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

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10. What happens to the force of attraction between the capacitors when the distance of separation between the plates increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero