Electro Magnetic Induction And Alternating Currents Question 419
Question: 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 }} $ connected in series with a light bulb B as shown. The mains electrical supply is 220 V a 50 Hz, the light bulb is rated at 220 V, 1100 W. What $ {L_{\max }} $ is required if the rate of energy dissipation in to mains the light bulb is to be varied by a factor of 5 from its upper limit of 1100W?
Options:
A) 0.69 H
B) 0.28 H
C) 0.38 H
D) 0.56 H
Show Answer
Answer:
Correct Answer: B
Solution:
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For power to be consumed at the rate of
$ \frac{1100}{5}=220W, $
we have $ P=E_{v}I_{v}\cos \theta $ $ 220=\frac{220\times 220}{\sqrt{R^{2}+L^{2}}{{\omega }^{2}}}\times \frac{R}{\sqrt{R^{2}+L^{2}{{\omega }^{2}}}} $
Where $ R=\frac{V^{2}}{P}=\frac{220^{2}}{1100}=44\Omega $
$ 220=\frac{{{(220)}^{2}}\times 44}{44^{2}+{{(L\omega )}^{2}}};44^{2}+{{(L\omega )}^{2}}=220 $
$ {{(L\omega )}^{2}}=\sqrt{220\times 44-44^{2}} $
$ =\sqrt{44(220-44)}=\sqrt{44\times 176}=88\Omega $
$ L=\frac{88}{2\pi \times f}=\frac{88}{2\pi \times 50}=\frac{88}{2\times 22}\times \frac{7}{50}=0.28H $
