Current Electricity Test

Result

Current Electricity Test - 7

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

If voltage across a bulb rated 220V, 100W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is

A
20%

B
2.5%

C
5%

D
10%

Solution

Power, P = V²/R
As the resistance of the bulb is constant
∴ ΔP/P = 2ΔV / V
% decrease in power = ΔP/P × 100
= 2ΔV/V × 100
= 2 × 2.5% = 5%
Question 2
4 / -1

Four wires of the same diameter are connected, in turn, between two points maintained at a constant potential difference. Their resistivities and lengths are; ρ and L (wire 1), 1.2ρ and 1.2L (wire 2), 0.9ρ and 0.9L (wire 3) and ρ and 1.5L (wire 4). Rank the wires according to the rates at which energy is dissipated as heat, greatest first,

A
4 > 3 > 1 > 2

B
4 > 2 > 1 > 3

C
1 > 2 > 3 > 4

D
3 > 1 > 2 > 4

Solution

Resistance of a wire, R = ρl/A
Rate of energy dissipated as heat is
H = V²/R = V²A/ρl
For a wire 1,

For a wire 2,

=
For a wire 3,

For a wire 4,

=
∴H₃ > H₁ > H₂ > H₄.
Question 3
4 / -1

A heater coil is rated 100 W, 200 V. It is cut into two idential parts. Both parts are connected together in parallel, to the same source of 200 V. The energy liberated per second in the new combination is

A
100 J

B
200 J

C
300 J

D
400 J

Solution

The resistance of heater coil,

= = 400Ω
The resistance of either half part = 200 Ω.
Equivalent resistance when both parts are connected in parallel.

The energy liberated per second when the combination is connected to a source of 200 V,
=
= 400 J
Question 4
4 / -1

In the circuit shown in figure heat developed across 2Ω, 4Ω and 3Ωresistances are in the ratio

A
2 : 4 : 3

B
8 : 4 : 12

C
4 : 8 : 27

D
8 : 4 : 27

Solution

Current through 2Ω,

Hence produced per second, H₁
=
Current through, 4Ω,

Hence produced per second H₂
=
Current through, 3Ω,I
Heat produced, H₃ = I² × 3 = 3I²
=
∴ H₁ : H₂ : H₃ = 8 : 4 : 27
Question 5
4 / -1

Two 2Ω resistances are connected in parallel in circuit X and in series in circuit Y. The batteries in the two circuits are identical and have zero internal resistance. Assume that the energy transferred to resistor A in circuit X within a certain time is W. The energy transferred to resistor B in circuit Y in the same time will be

A
1/4W

B
1/2W

C
2W

D
4W

Solution

In a circuit X, both the resistance are in parallel, Therefore V is the same and, Power (energy transferred in unit time) is
V²/2 = W
In a circuit Y, both resistance are in series.
Therefore, V_B + V_B' = V or V_B = V/2
In a circuit Y, power supplied to
B =