Question

A 100 micro farad capacitor in series with a 40 ohm resistance is connected to a 110 v 60 hz supply what is the maximum current

Answer 1

Explanation:

I = v/z

V= 110

Z= √r²+reactance

Reactance= 1 / Omega c

Omega = 2π f

=120

C = 100× 10 ^ - 6

:-) plz do it by urself

Answer 2

Given that,

Capacitance, $P=100\ \mu F=10^{-4}\ F$

Resistance, R = 40 ohms

RMS voltage, $E_v=110\ V$

Frequency, f = 60 Hz

To find,

The maximum current in the circuit.

Solution,

Peak voltage is given by :

$E_o=\sqrt{2} E_v\\\\E_o=\sqrt{2}\times 110\\\\E_o=155.56\ V$

Angular frequency is given by :

$\omega=2\pi f\\\\\omega=2\pi \times 60\\\\\omega=376.99\ rad/s$

The maximum current is given by :

$I=\dfrac{E_o}{\sqrt{R^2+\dfrac{1}{\omega^2 C^2}} }\\\\I=\dfrac{155.56}{\sqrt{(40)^2+\dfrac{1}{(376.99)^2\times (10^{-4})^2}} }\\\\I=3.24\ A$

So, the maximum current in the circuit is 3.24 A.

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