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This mock test of Test: Power Energy for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
This contains 20 Multiple Choice Questions for Electrical Engineering (EE) Test: Power Energy (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Which of the following is not an expression power?

Solution:

Power is the product of voltage and current. Writing I in terms of V, we get P=V^{2}/R and writing V in terms of I, we get P=I^{2}r.

QUESTION: 2

Which of the following statements are true?

Solution:

Power is proportional to both voltage and current. Hence both the options are right.

QUESTION: 3

A 250V bulb passes a current of 0.3A. Calculate the power in the lamp.

Solution:

Here, V = 250v and I = 0.3A. P=VI. Which implies that, P=250*0.3=75W.

QUESTION: 4

Kilowatt-hour(kWh) is a unit of?

Solution:

Power is the energy per unit time. That is, P=E/t. If the unit of power is kW and the unit of time is hour, then the unit of energy=unit of power*unit of time=kWh.

QUESTION: 5

Calculate the power in the 20 ohm resistance.

Solution:

Here V = 200v and Resistance( R) = 20ohm. P=V^{2}/R= 200^{2}/20=2000W=2kW

QUESTION: 6

A current of 5A flows in a resistor of 2 ohm. Calculate the power in the resistor and the energy dissipated in 300 seconds.

Solution:

P = I^{2}R = 5^{2}*2 = 50W.

E = Pt = 50*300 = 15000J = 15kJ.

QUESTION: 7

Calculate the power across each 20 ohm resistance.

Solution:

This is a series connected circuit hence the current across each resistance is the same. To find current: I=V/R=200/40=5A.

To find power: P=I^{2}R=(5)^{2}*20 = 500W. Since both the resistors have a resistance of 20 ohm, the power across both is the same.

QUESTION: 8

Calculate the power across each 10 ohm resistance.

Solution:

This parallel connected circuit, hence the voltage across each of the resistors are the same. P=V2/R= 1002/10=1000W=1kW. Since both the resistors receive the same amount of voltage, the Power in both are the same.

QUESTION: 9

Calculate the work done in a resistor of 20 ohm carrying 5A of current in 3 hours.

Solution:

To find power: P=I^{2}R=5^{2}*20=500W=0.5kW.

To find Work done: W=Pt=0.5*3=1.5kWh.

QUESTION: 10

The SI unit of power is?

Solution:

Power = energy/time =J/s(joules per second). Hence the Si unit of Power is J/s.

QUESTION: 11

Which among the following is a unit for electrical energy?

Solution:

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.

QUESTION: 12

A bulb has a power of 200W. What is the energy dissipated by it in 5 minutes?

Solution:

Here, Power = 200w and time = 5min. E=Pt => E= 200*5= 1000Wmin=60000Ws= 60000J= 60kJ.

QUESTION: 13

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

Solution:

A potentiometer is an instrument used for measuring voltage hence it is not a source for electrical energy.

QUESTION: 14

Calculate the energy dissipated by the circuit in 50 seconds.

Solution:

Here V = 100 and R = 10. Power in the circuit= V^{2}/R= 100^{2}/10= 1000W.

Energy= Pt= 1000*50= 50000J= 50kJ.

QUESTION: 15

Which among the following is an expression for energy?

Solution:

Expression for power= VI, substituting I from ohm’s law we can write, P=V^{2}/R. Energy is the product of power and time, hence E=Pt= V^{2}t/R.

QUESTION: 16

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

Solution:

Since the resistors are connected in parallel, the voltage across both the resistors are the same, hence we can use the expression P=V^{2}/R. P=200^{2}/10= 4000W. E=Pt= 4000*10=40000Ws= 40000J= 40kJ.

QUESTION: 17

A battery converts___________

Solution:

A battery is a device in which the chemical elements within the battery react with each other to produce electrical energy.

QUESTION: 18

A current of 2A flows in a wire offering a resistance of 10ohm. Calculate the energy dissipated by the wire in 0.5 hours.

Solution:

Here I (current) = 2A and Resistance(R) = 10ohm. Power= I^{2}R= 2^{2}*10=40. Energy=Pt= 40*0.5*60*60= 72000J=72kJ.

QUESTION: 19

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

Solution:

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=I^{2}R= 14.67^{2}*5=1075.8W. E=Pt= 1075.8*20= 21516J=21.5kJ.

QUESTION: 20

Practically, if 10kJ of energy is supplied to a device, how much energy will the device give back?

Solution:

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.

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