# Average power absorbed by resistor:

1. Dec 30, 2013

### Color_of_Cyan

1. The problem statement, all variables and given/known data

http://imageshack.us/a/img4/5259/wn5b.jpg [Broken]

Find the average power absorbed by the resistor in the circuit
if v1(t) = 10cos(377t + 60°)V
and v2(t) = 20cos(377t + 120°)V

2. Relevant equations

average power = (1/2)Vmax*Imax*cosø

x=rcosθ
y=rsinθ

mag = (r2 + j2)1/2
θ = arctan(j/r)

3. The attempt at a solution

v1(t) = 10cos(377t + 60°)V; = (10∠60°)V; = (5 + 8.66j)V

v2(t) = 20cos(377t + 120°)V; = (20∠120°)V; = (-10 + 17.32j)V

Z = 1Ω; = (1 ∠ 0°)Ω

Vtotal = v1 - v2 = 5 + 8.66j - (-10 + 17.32j)

Vtotal = (15 - 8.66j) V

Vtotal = (17.32 ∠-30°)V

and I think these are all max values so:

I = V/Z

I = (17.32∠-30°)/(1∠0°)

I = (17.32∠-30°)A

Avg power = (1/2)(17.32)(17.32)(cos 0) = 150W

Last edited by a moderator: May 6, 2017
2. Dec 30, 2013

### Staff: Mentor

Since you're dealing with power it would be prudent to convert the voltages to RMS values. Otherwise there will be a "missing" factor of 1/2 in your power calculation.

EDIT: Never mind. I see that you're using the $\frac{1}{2}V_m I_m cos(\phi)$ formula which includes the 1/2 term. Your calculation is fine.

Last edited: Dec 30, 2013