NEET-XII-Physics

33: Thermal and Chemical Effects of Electric Current

with Solutions - page 3
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  • #7
    A servo voltage stabiliser restricts the voltage output to 220 V ± 1%. If an electric bulb rated at 220 V, 100 W is connected to it, what will be the minimum and maximum power consumed by it?
    Ans : Output voltage, V = 220 V ± 1% = 220 V ± 2.2 V
    The resistance of a bulb that is operated at voltage V and consumes power P is given by
    `` R=\frac{{V}^{2}}{P}=\frac{(220{)}^{2}}{100}``
    `` \Rightarrow R=\frac{48400}{100}=484\,\mathrm{\,\Omega \,}``
    (a) For minimum power to be consumed, output voltage should be minimum. The minimum output voltage,
    V' = (220 - 2.2) V
    = 217.8 V
    The current through the bulb,
    `` i\text{'}=\frac{V\text{'}}{R}=\frac{217.8}{484}=0.45\,\mathrm{\,A\,}``
    Power consumed by the bulb, P' = i' × V'
    = 0.45 × 217.8 = 98.0 W
    (b) For maximum power to be consumed, output voltage should be maximum. The maximum output voltage,
    V" = (220 + 2.2) V
    = 222.2 V
    The current through the bulb,
    `` i"=\frac{V"}{R}=\frac{222.2}{484}=0.459\,\mathrm{\,A\,}``
    Power consumed by the bulb,
    P" = i" × V"
    = 0.459 × 222.2 = 102 W
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