# Calculating the current of an AC circuit

My goal is to calculate the current i(t) up in the right corner. Using the formula $$Z_{total} = \left( U\over I\right), I = \left( U\over Z_{total}\right)$$

Given values are

$$R_{1} = 1k \Omega,\ R_{2} = 0.5k \Omega,\ R_{3} = 0.4k \Omega,\ C_{1} = 1 \mu F,\\ C_{2} = 1 \mu F,\ L_{1} = 2H,\ e(t) = 10sin(1000t),\ \omega = 1000$$ The internal impedance of the source is $$Z_{i} = 10e^{j\pi/4}k \Omega$$

I start with the parallell connection between $$Z_{R2}//Z_{C1} = \left( Z_{R2}*Z_{C1}\over Z_{R2}+Z_{C1}\right) = \ \left( 500*\left( 1\over 0.001j\right) \over500+\left( 1\over 0.001j\right)\right) = -400 + 200j$$

Ztotal: $$Z_{total} = Z_{R1} +Z_{R2}//Z_{C1}+Z_{C2} + Z_{R3} +Z_{L1} + Z_{i}$$ Im not sure if the internal impedace of the source Z(i) is supposed to be added to the total impedance as if it is connected in series with the rest of the impedances? If it is I get the following.

$$Z_{total} = 1000 +(-400 + 200j) -1000j + 400 + 2000j + 10\times1000(\cos(\pi/4)+j\sin(\pi/4)$$ $$= (1000+5000\sqrt2)+j(1200+5000\sqrt2)$$

Now I use the formula above, since the current goes from + to - I get a negative voltage. $$I = \left( -U\over Z_{total}\right) = \left( -10\over (1000+5000\sqrt2)+j(1200+5000\sqrt2)\right)$$ $$=0.000865e^{-j0.007976}$$ $$= 0.000865\sin(1000t-0.007976)$$ Does this make any sense?

The current flows from +ve terminal to -ve terminal and the current $i(t)$ marked in the schematic is in the same direction. So the voltage you have to consider is +U. That will change the phase angle by $\pi$. ie., $$I = \left( U\over Z_{total}\right) = \left( 10\over (1000+5000\sqrt2)+j(1200+5000\sqrt2)\right)$$ $$= 0.000865e^{-j(0.007976+\pi)}$$ $$= 0.000865\sin(1000t-0.007976-\pi)$$
The ac current through the circuit is going to be a sinusoid with same frequency as the voltage source, phase shifted (lag) by $\mathbf{0.007976+\pi}$ and peak value of $0.865mA$.