When 1 `mu`F capacitor is connected across `V = 100 sqrt2 sin(100t)`Volt, current passing through milliammeter connected in the circuit is ……...

Options

A. 10
B. 20
C. 40
D. 80

Answer

A

Explanation

Here `V=100sqrt2 sin (100t)` …(1)
Comparing with `V=V_m sin (omegat)` …(2)
We get `V_m=100sqrt2` Volt
`omega=100 "rad"/s`
Now, `X_C=1/(omegaC)=1/(100xx1xx10^(-6))=10^4 Omega`
Now, `I=V/Z=V/X_C =(V_m//sqrt2)/X_C=100/10^4=10^(-2)A`
`therefore` Reading obtained in milliammeter
`=10^(-2) xx 10^3` mA
=10 mA