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A 50 W, 100V lamp is to be connected to ...

A 50 W, 100V lamp is to be connected to an ac mains of 200V, 50 Hz. What capacitor is essential to be put in series with the lamp?

Text Solution

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As resistance of lamp R=`R=(V^(2))/(W)=(100^(2))/(50)=200Omega`
and maximum current `I=(V)/(R)=(100)/(200)=(1)/(2)Amp`
So, when the lamp is put in series with a capacitance and run at 200
`Z=(V)/(I)=(200)/(1//2)=400Omega`
Now`Z,=sqrt(R^(2)+(1)/(omegaC)^(2))`
`C=(100)/(pisqrt12)muF=9.2muF`
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