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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
Power, P = V^{2}/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%
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,
Resistance of a wire, R = ρl/A
Rate of energy dissipated as heat is
H = V^{2}/R = V^{2}A/ρl
For a wire 1,
For a wire 2,
=
For a wire 3,
For a wire 4,
=
∴H_{3} > H_{1} > H_{2} > H_{4}.
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
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
In the circuit shown in figure heat developed across 2Ω, 4Ω and 3Ωresistances are in the ratio
Current through 2Ω,
Hence produced per second, H_{1}
=_{ }
Current through, 4Ω,
Hence produced per second H_{2}
=
Current through, 3Ω,I
Heat produced, H_{3} = I^{2} × 3 = 3I^{2}
=
∴ H_{1} : H_{2} : H_{3} = 8 : 4 : 27
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
In a circuit X, both the resistance are in parallel, Therefore V is the same and, Power (energy transferred in unit time) is
V^{2}/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 =
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