Discussion Forum

An alternating voltage e =200 $\sqrt{2}$ $\sin$ (100 t) volt is connected to 1 $\mu$ F  capacitor through AC ammeter. The reading of ammeter is 

Answer:

Explanation:

 Given, e =200 $\sqrt{2}$ $\sin$ (100 t)   .......(i)

 Capacitance of capacitor  (C) = 1 $\mu$ F= 1 x 10^-6 F

 The standard  equation  of voltage of AC is given by

 V= V₀ $\sin \omega$t  ..........(ii)

 On comparing Eqs.(i) and (ii) , we get

 $V_{0}= 200\sqrt{2}$

 $\omega$ = 100

We know that,

 $l_{rms}= \frac{V_{rms}}{X_{c}}$

 But   $V_{rms}= \frac{V_{0}}{\sqrt{2}}$     $l_{rms}=\frac{V_{0}}{\sqrt{2}X_{c}}$

$l_{rms}=\frac{V_{0}\omega C}{\sqrt{2}}$    $\left(X_{c}=\frac{1}{\omega C}\right)$

$l_{rms}=\frac{200\times\sqrt{2}\times100\times1 \times 10^{-6}}{\sqrt{2}}$

$l_{rms}=20\times 10^{-3}A=20 mA$