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  • (i) When an AC source is connected to an ideal capacitor, show that the average power supplied by the source over a complete cycle is zero. (ii) A bulb is connected in series with a variable capacitor and an A.C. source as shown. What happens to

(i) When an AC source is connected to an ideal capacitor, show that the average power supplied by the source over a complete cycle is zero.
(ii) A bulb is connected in series with a variable capacitor and an A.C. source as shown. What happens to the brightness of the bulb when the key is plugged in and capacitance of the capacitor is gradually reduced?

 

 

 

 
 
 
 
 

Answers (1)

(i) Let the voltage of the AC source connected be V=V_{m}\sin \omega t

for an ideal capacitor voltage lags behind the current

\therefore I=I_{m}\sin \left ( \omega t+\frac{\pi }{2} \right )

Average power

^{P}av=\frac{1}{T}\int_{o}^{T}V_{m}I_{m}\sin \omega t\; \sin \left ( \omega t+\frac{\pi }{2} \right )dt

          =\frac{1}{T}\int_{o}^{T}V_{m}I_{m}\sin \omega t\; \cos \omega t \; dt

          =\frac{V_{m}I_{m}}{2T}\int_{o}^{T}\sin 2\; \omega t

\Rightarrow ^{P}av=\frac{V_{m}I_{m}}{2T}\int_{o}^{T}\sin \; \omega t

                =-\frac{V_{m}I_{m}}{2T}\left [ \cos \frac{2\pi }{T}t \right ]_{o}^{T}

               =-\frac{V_{m}I_{m}}{2\pi }\left [ \cos 2\pi -\cos o \right ]

     ^{P}av=0

(ii) When capacitance is reduced, the capacitive reactance X_{c}=\frac{1}{wc} increases. So as capacitance is gradually reduced, the brightness also reduces.

Posted by

Safeer PP

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