Question

A 15.0 uF capacitor is connected to a 220V, 50 Hz source. Find the capacitive reactance and the current (rms
and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?​

Answer 1

Explanation:

using ,

capacitive reactance=1/wc

w=2×3.14×frequecy

c=15×10^( -6)

in this way we can have value of capacitive reactance

if "f" is doubled then capacitive reactance becomes half

now for current,i=£/capacitive react

if "f" is doubled current too becomes double

Answer 2

$\chapter alternating current$

$\C = 15.0 uF = 15 X 10 ^-6 F$

$\Vrms = 220 V$

ц = 50 HZ

Xc = 1/cw

= w = 2πц

= 2 X 3.14 x 50

= 314 rad /s //

Xc = 1/ 15.0 X 10^-6 X 314

= $\1/4710 X10^-6$

=$\212.3 ohm$

Irms = vrms / Xc

$\220v/212.3ohm$

=  1.035 A //

= ц₂ = 100 Hz

= ш₂ = 2πν₂

= 2 x 3.14 X 100

$\628 rad/s$ //

Xc = 1/ 15.0 X 10^-6 X 628

= 106.1 ohm //

Irms = Vrms / Xc

= 220 / 106.1 = 2.0735 A //