AC circuit help?

Question

In a series L-R-C circuit, the components have the following values: L=20.0 mH C=140 nF, and R=350ohms. The generator has an rms voltage of 120 V and a frequency of 1.25 kHz.

1) Determine the average power supplied by the generator.

2)Determine the average power dissipated in the resistor.

I can't seem to find help on this from the book so can someone please eloborate.?

Answer 1

In a series LCR circuit the impedance Z = √{R² + (XL ~ XC)²}
R is the resistance, XL the inductive reactance & XC the capacitive reactance.
R = 350 ohms
XL = 2πfL= (2 x 22 x 1250 x 20 x 10^ -3) / 7 = 157.14 ohms
XC = 1 / (2πfC)= (1x7) / (2 x 22 x 1250 x 140 x 10^ -9)
.......= (7 x 10^9) / (2 x 22 x 1250 x 140) = 909.1 ohms
Substituting the above values in the formula;
Z = √ {(350)² + (157.14 ~ 909.1)²}
Z = √{(350)² + (751.96)²} = √{122500 + 565444} = 829 ohms
Current (I) in the circuit = V / Z = 120 / 829 = 0.145 Amps
(1) Average Power = V x I = 120 x .145 = 17.4 watt
=======================================
(2) Power dissipated in the resistor = I²R = (0.145)² x350
..............................".........................= 7.36 watts
===========================================

Answer 2

P=VI
Oh you said AC circuit. Sorry...
In AC circuit power would have an imaginary and a real component. It is the real component we are after.

P=VI were I=V/Z
P=V^2/Z
I'm not sure about Z. If the component are in series
Z= R + jLw - j/wc where w=2 pi f
Can you handle the rest?

Answer 3

Lrc Generator

Answer 4

actually for the 1st part answer will be Vrms x Irms X power factor=7.32 W

Answer 5

remember to set your calc to radians