A typical light dimmer, used to dim the stage lights in a theatre consists of a variable induction for L (where inductance is adjustable between zero and L_max), is connected in series with a light bulb B as shown. The mains electrical supply is 220 V at 50 Hz and the light bulb is rated at 220V, 1100 W. What L_max is required if the rate of energy dissipation in the light bulb is to be varied by a factor of 5 from its upper limit of 1100 W?

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0.28 H

0.69 H

0.56 H

0.38 H

answer is B.

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Detailed Solution

For power to be consumed at the rate of

$\frac{1100}{5} = 220 W,$ we have $P = E_{v} I_{v} cosθ$

$220 = \frac{220 \times 220}{\sqrt{R^{2} + L^{2} ω^{2}}} \times \frac{R}{\sqrt{R^{2} + L^{2} ω^{2}}}$

$where R = \frac{V^{2}}{P} = \frac{220^{2}}{1100} = 44 Ω$

$220 = \frac{(220)^{2} \times 44}{44^{2} + (Lω)^{2}} \Rightarrow 44^{2} + (Lω)^{2} = 220 \times 44$

$(Lω)^{2} = \sqrt{220 \times 44 - 44^{2}}$

$= \sqrt{44 (220) - 44)} = \sqrt{44 \times 176} = 88 Ω$

$L = \frac{88}{2 π \times f} = \frac{88}{2 π \times 50} = \frac{88}{2 \times 22} \times \frac{7}{50} = 0 .28 H$

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