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A 220 V A.C. supply is connected betwe...


A 220 V A.C. supply is connected betweeen points A and B (See figure). What will be the potential difference V acorss the capacitor ?

A

200V

B

110 V

C

0 V

D

`220 sqrt(2)` V

Text Solution

Verified by Experts

The correct Answer is:
D
`220 sqrt(2) V`
In the given figure, applying Kirchhoff.s second law in the closed loop containing A.C. source, ideal diode and capacitor, we can write at time t,
`V=V_(D)+V_(C )`
`V_(m)sin(omega t)=IR_(D)+V_(C ) " " …(1)`
Resistance of ideal diode in forward biased condition is `R_(D)=0 ` and so,
`V_(C )=V_(m)sin (omega t)`
`therefore (V_(C ))_("max")=V_(m)`
`=V_("rms")xx sqrt(2)(because V_("rms")=(V_(m))/(sqrt(2)))`
`=220sqrt(2)` volt
Resistance of ideal diode in reverse biased condition is `R_(D)=oo` and so taking I = 0, from equation (1),
`V_(m) sin (omega t)=V_(C )`
`therefore (V_(C ))_("max")=V_(m)`
`=220sqrt(2)` volt
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