The voltage across an electric bulb is decreased by 2%. Assuming that the resistance of the filament remains same, the power consumed will

Decrease by 2%

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increase by 2%

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decrease by 4%

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increase by 4%

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Solution

The correct option is C
decrease by 4%

$P = \frac{V^{2}}{R} ∴ \frac{Δ P}{P} = \frac{2 Δ V}{V} + \frac{2 Δ R}{R}$

$\frac{Δ P}{P} = 2 \times 2 + 0 = 4$

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The voltage across an electric bulb is decreased by 2%. Assuming that the resistance of the filament remains same, the power consumed will

The correct option is C decrease by 4% P=V2R ∴ΔPP=2ΔVV+2ΔRR ΔPP=2×2+0=4

The correct option is C
decrease by 4%

$P = \frac{V^{2}}{R} ∴ \frac{Δ P}{P} = \frac{2 Δ V}{V} + \frac{2 Δ R}{R}$

$\frac{Δ P}{P} = 2 \times 2 + 0 = 4$