Question

An alternating voltage E = 200 √2 sin(100 t) is connected to a 1 microfarad capacitor through an ac ammeter. The reading of the ammeter shall be(a) 10 mA(b) 20 mA(c) 40 mA(d) 80 mA

Answer 1

answer : option (b)

explanation : equation of alternating voltage , E = 200√2sin(100t)

compare this equation with$E_0sin(\omega t)$
so, $E_0=200\sqrt{2}$ volts
$\omega$ = 100 rad/s

now, $E_{rms}=\frac{E_0}{\sqrt{2}}$

= 200√2/√2 = 200 volts .

given, capacitance of capacitor, C = 10^-6F
so, reactance of capacitor, $X_C=\frac{1}{\omega C}$
=$\frac{1}{10^{-6}\times100}$
=10⁴

so, the reading of Ammeter , $I=\frac{E_{rms}}{X_C}$

= 200/10⁴ A = 20 × 10^-3 A = 20mA

hence, option (b) is correct.

Answer 2

Explanation:

i hope it's helpful ans 20mA