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Homework Statement
http://imageshack.us/a/img850/788/yk72.jpg
Find the average power absorbed by the 2Ω resistor in the circuit (pictured above)
Homework Equations
Average power = (1/2)(Imax)(Vmax)(cosø)
x=rcosθ
y=rsinθ
current division, voltage division
The Attempt at a Solution
So trying to get total impedance, first I get this (2 || -2j):
http://imageshack.us/a/img30/5500/wpwn.jpg
then this (1-1j + 2j):
http://imageshack.us/a/img856/1788/u357.jpg
then (1 +1j) || 4
= (1/1+1j + 1/4)-1
= (5 + 1j)/(4 + 4j)-1
= (4 + 4j)/(5 + 1j)
multiplying by conjugate (5 - 1j)/(5 - 1j) got me
= (24 + 16j)/24
Z total = 0.923 + 0.6153j
Z total = (1.109 ∠ 33.68°)Ω
I total = (6∠0)/(1.109 ∠ 33.68°)
I total = (5.41 ∠ -33.68A)
doing current division, current in the middle branch is:
I = (5.41 ∠ -33.68)*(4 / [5 + 1j] )
I = (5.41 ∠ -33.68)*(4∠0 / 5.099 ∠ 11.3)
I = (4.243 ∠ -45°)A in mid branch
Then doing current division again where 2 || -2j, the current through the 2Ω would be
I = (4.243 ∠ -45°)A * (-2j/[2 - 2j] )
I = (4.243 ∠ -45°)A * (2 ∠ -90 / 2.82 ∠ -45)
I = (3.009 ∠ -90°)A through the 2Ω
For the voltage across the 2Ω it's
V = (6 ∠ 0)*( [1-1j]/[1 + 1j] )
V = (6 ∠ 0)*(1.41∠-45 / 1.41∠45)
V = (6∠-90°)V across the 2Ω resistor
So now I have
Pavg = (1/2)*(6)(3.009)*cosø
So my only question is, how to find ø? Is it the angle of total impedance of the whole circuit or just of the 2Ω resistor?
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