Using the superposition principle, determine the current i(t.
2. Homework Equations
Zc = 1/jwC
Zl = jwL
V = I*R
I = V/Z
The Attempt at a Solution
First, I converted inductor/capacitor to impedance:
L = 1.5H -> jwL = j * 10 * 1.5 = 15j
C = 10mf = 1/(jwC) = 1/(j*10*10*10^-3) = -0.1j
Then, I transformed the 3A DC current source and 10 ohm resistor into a DC voltage source:
I then shorted the 30V DC source to find i(t) with only Vs(t) = 10cos(10t)V
After shorting the 30V DC source, C2 and R3 are in parallel. Their impedance is:
(0.1j * 10)/(10-0.1j) = -j/(10-0.1j)
After rationalizing: (0.1 - j)/100.01
Now that, L2 and R4 are in series, total impedance:
15j + 5 + (0.1 - j)/100.01 = (500.15-1499.15j)/100.01 = 15.8 < -71.55°.
I = V/Z = (10<0°)/(15.88<-71.55° ) = 10/15.88<71.55°A
Now I short the AC source:
The DC source had a frequency of 0 rad/s, so the impedance of C3 is infinity (1/0) and L3 is 0 (j*0*L).
The inductor is shorted and the capacitor is opened. There is a voltage source of 30V and a combined resistance of 15ohms. V=IR -> I = 30/15 = 2A<0°
[(500.15/100.01 + 2) + (1499.15j/100.01)]
Is 16.54<64.96° A the correct answer?