Basic Electrical Objective Questions and Answers

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17) Two equal resistors R connected in series across a voltage source V dissipate power P. What would be the power dissipated in the same resistors when they are connected in parallel across the same voltage source ?

A. 4P

B. P

C. 2P

D. 16P

 A  B  C  D

Say R is the resistance of the identical two resistors. When they are connected in series across a voltage source V, the equivalent resistance of the combination is 2R and then total power dissipated by the resistors will be P = V2/2R. When they are connected in parallel across the same voltage source V, the equivalent resistance of the combination is R/2 then total power dissipated by the resistors will be V2/R/2 = 4V2/2R = 4P.

18) Two identical resistors are first connected in parallel then in series. The ratio of resultant resistance of the first combination to the second will be

A. 4

B. 0.25

C. 2

D. 0.5

 A  B  C  D

Let us consider the value resistance of the equal resistors is R. So equivalent resistance of parallel combination of the resistors is R/2, and equivalent resistance of series combimnation of the resistors is 2R. So ratio of these two combination will be (R/2)/2R = 1/4 = 0.25

19) The ratio of the resistance of a 200W, 230V lamp to that of a 100W, 115V lamp will be

A. 0.5

B. 2

C. 4

D. 0.25

 A  B  C  D

Resistance of the first lamp R1 = 2302/200 Ω
Resistance of the first lamp R2 = 1152/100 Ω
Therefore, R1/R2 = (2302/200)/(1152/100) = 2

20) The resistance of 200W 200V lamp is

A. 100 Ω

B. 200 Ω

C. 400 Ω

D. 800 Ω

 A  B  C  D

Resistance R = V2/W. Here, V = 200 V and W = 200 watts.
Therefore, resistance of 200W 200V lamp is 2002/200 = 200 Ω

21) Tow 1 kΩ 1 W resistors are connected in series. Their combine resistance and wattage will be

A. 2 kΩ, 0.5 W

B. 1 kΩ, 1 W

C. 0.5 kΩ, 2 W

D. 2 kΩ, 1 W

 A  B  C  D

Wattage W = V2/R.......(I)
Here, for each resistor, W = 1 W and R = 1 kΩ and putting these values in equation (I), we get V2 = 1, When two 1 kΩ resistance are connected in series, combined resistance will be 2 kΩ and putting this value and and V2 = 1 in equation (I) we again get, combined wattage W = 1/2 watt.

22) Three 3 Ω resistors are connected to form a triangle. What is the resistance between any two of the corners ?

A. 9 Ω

B. 6 Ω

C. 3 Ω

D. 2 Ω.

 A  B  C  D

Whenever we look at the said triangle from any two of its corners, we will find that it is just a parallel combination of one 3 Ω and one 6 Ω (3 + 3 = 6) resistor. Thus, the resistance aross these two corner points of the triangle will be 3X6/(3+6) = 18/9 = 2 Ω

23) A wire of 0.14 mm diameter and specific resistance 9.6 μΩ - cm is 440 cm long. The resistance of the wire will be

A. 9.6 Ω

B. 11.3 Ω

C. 13.7 Ω

D. 27.4 Ω

 A  B  C  D

Cross - sectional area of the conductor is (π/4)X0.0142 = 0.000154 cm2.
The resistance will be 9.6X440/0.000154 = 27428571 μΩ = 27.4 Ω

24) A 10 Ω resistor is stretched to increase its length double. Its resistance will now be

A. 40 Ω

B. 20 Ω

C. 10 Ω

D. 5 Ω

 A  B  C  D

The resistance of a conductor is directly proportional to its length and inversely proportional to its cross - sectional area. As the wire is stretched to its double length, it's cross - sectional area will become half, hence, the resistance of the stretched wire will become 4 times.

25) Specific resistance is measured in

A. mho

B. ohm

C. ohm - cm

D. ohm/cm

 

 A  B  C  D

The resistance R = ρl/a where R is the resistance of any substance in ohm, ρ is the specific resistance of material of that substance, l and a are length in cm and cross - sectional area in cm2 of that substance respectively. Therefore, ρ = R.a/l and its unit may be ohm X cm2/cm or ohm - cm

26) A wire of resistance R has it length and cross - section both doubled. Its resistance will become

A. 0.5R

B. R.

C. 2R

D. 4R

 A  B  C  D

The resistance of a conductor is directly proportional to its length and inversely proportional to its cross - sectional area. As the length and cross - sectional area both have become double, there will no change in resistance of the wire.

27) 5 X 1016 electrons pass across the section of a conductor in 1 minutes and 20 seconds. The current flowing is

A. 0.1 mA.

B. 1 mA

C. 10 mA

D. 100 mA

 A  B  C  D

The charge of an electron is 1.6 X 10-19 coulomb. Therefore total charge passes across the section of a conductor in 1 minutes and 20 seconds is 5 X 1016 X 1.6 X 10-19 = 8 X 10 - 3. Therefore, charge passes the section in one second, is 8 X 10 - 3 coulomb/80 second = 10 - 4 coulomb/second (or Amp) = 0.1 mA

28) Three element having conductance G1, G2 and G3 are connected in parallel. Their combined conductance will be

A. (G1 + G2 + G3) - 1

B. G1 + G2 + G3

C. 1/G1 + 1/G2 + 1/G3

D. (1/G1 + 1/G2 + 1/G3) - 1

 A  B  C  D

We know that conductance is reciprocal of resistance i.e. resistance = 1 / conductance. Let's resistances of the said conductors are R1, R2 and R3 hence, G1 = 1/R1, G2 = 1/R2 and G3 = 1/R3. The resistance of their parallel combination will be (1/R1 + 1/R2 + 1/R3) - 1 = (G1 + G2 + G3) - 1 . Therefore, conductance of the combination will be
G1 + G2 + G3

29) Which of the following has negative temperature coefficient ?

A. Electrolytes.

B. Brass.

C. Copper.

D. Silver.

 A  B  C  D

Normally non - metallic substance has negative temperature coefficient. Electrolytes is a non metallic substance.

30) Which of the following has positive temperature coefficient ?

A. Germanium.

B. Gold

C. Paper

D. Rubber

 A  B  C  D

Normally metallic substance has positive temperature coefficient. Gold is a metallic substance.

31) A 60 W bulb in series with a room heater is connected across the mains. If the 60 W bulb is replaced by 100 W bulb

A. the heater output will increase

B. the heater output will decrease

C. the heater output will be same

D. the heater output will slightly decrease.

 A  B  C  D

The wattage rating of any electrical component (bulb) is inversely proportional to its resistance. Hence 100 W bulb has smaller resistance than 60 W bulb. So if 60 W bulb is replaced by 100 W bulb in the above case then, the overall resistance of the series combination of heater and bulb is reduced hence current increases accordingly. Therefore the output of the heater is increased as the current through it increases.

32) A cube of material of side 1 cm has a resistance of 0.002 Ω between its opposite faces. If the same volume of the material has a length of 4 cm and a uniform cross - section, the resistance of this length will be

A. 0.128 Ω

B. 0.064 Ω

C. 0.032 Ω

D. 0.016 Ω

 A  B  C  D

Here, the cube of material of side 1 cm has a resistance of 0.002 Ω between its opposite faces that means the resistivity of the material is 0.002 Ω. Now the length of the material has become 4 cm, hence for same volume 1 cm3 the cross - sectional area of the material will be 1/4 or 0.25 cm2. The new resistance will be 0.002X4/0.25 = 0.032 Ω


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