- #1

zenterix

- 555

- 76

- Homework Statement
- I'd like to go through the calculations involved in finding the current in the RC circuit seen in the picture below.

- Relevant Equations
- The complex numbers in the equations are confusing me a little. Specifically, the concept of impedance.

As far as I understand, impedance is defined as the proportionality constant between complex voltage and complex current.

Let ##V_c(t)=V_0e^{i\omega t}## be complex voltage.

Again, as far as I understand, the actual (real-world) voltage we observe is a real number. Thus, this real-world voltage can be either the real or imaginary part of ##V_c(t)##.

Now let's apply KVL to the circuit.

$$\oint\vec{E}\cdot d\vec{l}=-V_c(t)+I_c(t)R+\frac{q(t)}{C}=0\tag{1}$$

Note that ##q(t)## is the charge on the top capacitor plate in the diagram above and so given the direction of positive current we have that ##I=\dot{q}##.

Now, it is not clear to me that we can write ##I_c(t)=\dot{q}(t)##, but I will use this in the next part.

Let's differentiate (1)

$$I_c(t)+RC\dot{I}_c(t)=C\dot{V}_c(t)=Ci\omega V_0e^{i\omega t}=\omega C i V_c(t)\tag{2}$$

$$\dot{I}_c(t)+\frac{1}{RC}I_c(t)=\frac{\omega i}{R}V_c(t)=\frac{\omega V_0}{R}e^{i\left (\omega t+\frac{\pi}{2}\right )}\tag{3}$$

The characteristic polynomial is

$$p(s)=s+\frac{1}{RC}\tag{4}$$

and

$$p(i\omega)=i\omega+\frac{1}{RC}=e^{i\phi}\sqrt{\omega^2+\left ( \frac{1}{RC} \right )^2}\tag{5}$$

where ##\phi=\tan^{-1}(\omega RC)##.

The solution to the differential equation in (3), using the exponential response formula is

$$I_c(t)=\frac{\frac{V_0\omega}{R}}{p(i\omega)}e^{i(\omega t+\pi/2)}\tag{6}$$

$$=(...)\tag{7}$$

$$=\frac{V_0}{\sqrt{R^2+\left (\frac{1}{\omega C}\right )^2}}e^{i(\omega t+\pi/2-\phi)}\tag{8}$$

$$=\frac{ie^{-i\phi}}{\sqrt{R^2+\left (\frac{1}{\omega C}\right )^2}}V_c(t)\tag{9}$$

Thus

$$V_c(t)=e^{i(\phi-\pi/2)}\sqrt{R^2+\left (\frac{1}{\omega C}\right )^2} I_c(t)\tag{10}$$

The impedance seems to be ##e^{i(\phi-\pi/2)}\sqrt{R^2+\left (\frac{1}{\omega C}\right )^2}##.

Note that for an AC voltage of ##V_0\sin{\omega t}## the resulting current is the imaginary part of the equation for ##I_c(t)## in (9).

$$I(t)=\frac{V_0}{\sqrt{R^2+\left (\frac{1}{\omega C}\right )^2}}\sin{(\omega t+\pi/2-\phi)}$$