Question

A series RLC circuit consists of a 90.0 Ω resistor, a 0.170 H inductor, and a 45.0 μF capacitor. It is attached to a 120 V/60 Hz power line

What is the peak current I at this frequency?

What is the average power loss?

Answer 1

Solution: From the question we have:

A series RLC circuit consists of a 90.0 Ω resistor, a 0.170 H inductor, and a 45.0 μF capacitor

It is attached to a 120 V/60 Hz power line

So,

ω = 2*3.1416*60 = 377 rad/sec
Xl = ω*L = 377*0.17 = 64.09 ohm (j64.09)
Xc = 10^6/(45*377) = 58.94 ohm (-j58.94)
X = Xl +Xc = j60.09-j58.94 = -j1.14 ohm
Z = √R^2+X^2 = √90^2+1.14^2 = 90.00 ohm

A)What is the peak current I at this frequency?

Irms = 120/90.00 = 1.33 A
Ipeak = Irms*√2 = 1.33* √2 A=1.88 A ANSWER

B)What is the average power loss?

=>P loss = R*Irms^2

= 90 *1.33^2

=159.201 WATT answer