# phasors and differential equation circuit analysis?

I am having trouble, the steady stae voltage i find by differential equations is no the same as the one i find by phasors, what is worng?

$$\V_{ss}(t) = \cos(4t)\$$ volts ........ by diff eq, just looking at thevinin voltage

phasor;$$V = 1\times\frac{-4j}{4\sqrt3 - 4j} = 0.5e^{-j\pi/3} = 1/2\cos(4t - \pi/3)$$

they are not the same, what is the problem?

Thevnin voltage gives the open circuit voltage across two terminals (capacitor removed in this case). And$V_{th} = \cos(4t)$ in here. When a load (capacitor in this case) is connected across these terminals, the voltage across the terminals A-B will change and you can use voltage division rule to obtain the output voltage. $$V_{AB} = V_{th}\times\dfrac{X_C}{R_{th}+X_C}$$ $X_C$-capacitive reactance.
The phasor method used you have used actually does that. So the answer is $0.5\cos(4t-\pi/3)$.
• Thevinin voltage measured with capacitor removed is $\cos 4t$. When capacitor is connected, the voltage across these terminals will change. – nidhin Jul 6 '14 at 1:31