Objective Questions Electrical Engineering

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1) Resistivity of a wire depends on

A. material

B. length

C. cross section area

D. all of above

 A  B  C  D

Resistivity is a basic property of any materials. It is defined as the resistance offered by a cube of unit volume of the materials. Hence resistivity of a wire depends on its material.

2) When n numbers resistances of each value r are connected in parallel, then the resultant resistance is x. When these n resistances are connected in series, total resistance is

A. nx.

B. n2x.

C. x/n.

D. rnx.

 A  B  C  D

n numbers resistances of each value r are connected in parallel, then the resultant resistance is x, that means r/n = x ⇒ r = nx.When these n resistances are connected in series, total resistance is nr = n.(nx) = n2x [Since r = nx alraedy proved]

3) Resistance of a wire is r ohms. The wire is stretched to double its length, then its resistance will be

A. r/2

B. 4r

C. 2r

D. r/4

 A  B  C  D

Resistance r = ρl/a. When the wire of length l is stretched to 2l, then the cross - sectional area of the wire becomes a/2. Now new value of resistance, r' = ρ2l/(a/2) = 4ρl/a = 4r.

4) Kirchhoff's second law is based on law of conservation of

A. charge

B. energy

C. momentum

D. mass

 A  B  C  D

Kirchhoff's voltage law (KVL) is also called Kirchhoff's second law. The principle of conservation of energy implies that the directed sum of the electrical potential differences (voltage) around any closed network is zero.

5) One coulomb of electrical charge is contributed by how many electrons ?

A. 0.625 X 1019.

B. 1.6 X 1019.

C. 1019.

D. 1.6 X 1012.

 A  B  C  D

Electrical charge of one electron is 1.6 X 10 - 19 coulomb, hence one coulomb implies 1/(1.6 X 10 - 19) or 0.625 X 1019 numbers of electrons.

6) Tow bulbs marked 200 watts - 250 V, and 100 watts - 250 V are joined in series to 250 V supply. The power consumed by the circuit is

A. 33 watt

B. 200 watt

C. 300 watt

D. 67 watt.

 A  B  C  D

The resistance of first and second bulb are (250)2/200 and (250)2/100 Ω respectively. The total resistance when the bulbs are connected in series will be (250)2/200 + (250)2/100 Ω. The total power consumption when they joined in series to 250 V supply. The power consumed in the circuit will be (250)2/{(250)2(1/200 + 1/100)} = 20000/300 = 67 watt.

7) Ampere second is the unit of

A. conductance

B. power

C. energy

D. charge

 A  B  C  D

Electrical current is transfer electrical charge per second. Therefore Ampere = coulomb/second hence coulomb = ampere X second or ampere second.

8) Which of the following is not the unit of electrical power

A. volt/ampere

B. volt ampere

C. watt

D. joule/second

 A  B  C  D

Unit of electrical power is watt and watt means joule/second. Again electrical power = voltage X current, hence volt ampere may be another expression for unit of power. But impedance = voltage/current, hence volt/ampere may be expression for unit of impedance not power.

9) One kilowatt hour is same as

A. 36 X 105 watt

B. 36 X 105 ergs

C. 36 X 105 joules

D. 36 X 105 BTU


 A  B  C  D

Kilowatt hour is the unit of energy and 1 kilowatt hour = 1000 X 1 watt X 3600 second = 36 X 105 watt second = 36 X 105 joule.

10) An electric current of 6 A is same as

A. 6 joule/second

B. 6 Coulomb/second.

C. 6 watt/second

D. none of the above.

 A  B  C  D

Current is rate of charge transferred per second. A current of 6 ampere implies 6 coulomb charge transferred through a cross section of conductor per second. Therefore 6 Ampere = 6 coulomb/second.

11) A circuit contains two un equal resistor in parallel

A. voltage drops across both are same

B. currents in both are same

C. heat losses in both are same

D. voltage drops are according to their resistive value

 A  B  C  D

Whatever may be the value of resistance the voltage drops, across all the resistors connected in parallel, are always same

12) Conductance of any conductor is expressed as

A. ampere/watt

B. mho

C. volt2/watt

D. watt/ampere2

 A  B  C  D

Conductance is reciprocal of resistance that means conductance = (resistance) - 1 . Hence unit of conductance will be 1/ohm and this is known as mho

13) A copper wire of length l and diameter d has potential difference V applied at its two ends. The drift velocity is vd. If the diameter of the wire is made d/2, then the drift velocity becomes

A. vd.

B. 4vd.

C. vd/4.

D. vd/2.

 A  B  C  D

The drift velocity is a basic property of conductor material and hence it does not depend upon the length or diameter of the conductor.

14) Two resistances R1 and R2 give combined resistances 4.5Ω and 1Ω when they are connected in series and parallel respectively. What would be the values of these resistances ?

A. 3Ω and 6Ω

B. 1.5Ω and 3Ω

C. 3Ω and 9Ω

D. 6Ω and 9Ω

 A  B  C  D

Here, R1 + R2 = 4.5..................(1)
and R1.R2/(R1 + R2) = 1
⇒ R1.R2/4.5 = 1
⇒ R1.R2 = 4.5 ......................(2)
Combining (1) & (2) we get R1 = 1.5 Ω or 3 Ω and R2 = 3 Ω or 1.5 Ω

15) Which of the following may be value of resistivity of copper

A. 1.7 X 10 - 6

B. 1.7 X 10 - 5

C. 1.7 X 10 - 4

D. 1.7 X 10 - 3.

 A  B  C  D

The value of resistivity of copper is 1.7 X 10 - 6

16) Mass of a proton is how many times greater than mass of an electron

A. 184000

B. 18400

C. 1840

D. 184

 A  B  C  D

Mass of a proton is 1840 times greater than mass of an electron.


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