# Why can we add impedances in series?

• zenterix
zenterix
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).

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

Suppose the resistance is zero. Then ##\phi=\tan^{-1}(0)=0## and for a voltage ##V(t)=V_0\sin{\omega t}## we get current

$$I(t)=V_0\omega C\sin{(\omega t+\pi/2)}$$

which is indeed what I get when I did this entire calculation but for a circuit with AC source and capacitor only.

Here is what I think I am wondering about and why I wrote the post above.

1) Given an RLC circuit, I can always go through the entire process of solving the differential equation and finding current.

However, it seems is a shortcut.

Consider three simple circuits. In each we have an AC voltage and a single circuit element: a resistor, an inductor, and a capacitor, respectively.

The complex current we find in each circuit is the reciprocal of impedance for that circuit times the complex voltage.

The impedances are ##R##, ##\omega Li##, and ##\frac{1}{\omega Ci}## for our three simple circuits.

Now consider the three possible circuits with two different circuit elements besides the AC voltage.

We have RC, RL, and LC circuits.

Apparently, the complex current is again the reciprocal of an impedance times the complex voltage.

My main question is how to compute this impedance without solving the differential equation directly.

In particular, consider the RC circuit in the OP.

The impedance for a purely resistive circuit is ##R## and for a purely capacitive circuit is ##\frac{1}{\omega Ci}##.

If there were only a resistor we'd have ##I_c(t)=\frac{1}{R}V_c(t)## and if there were only a capacitor then we'd have ##I_c(t)=\omega C i V_c(t)##.

Note that in both cases the input signal is the same, namely, ##V_c(t)=V_0e^{i\omega t}##.

If we sum the impedances then we have

$$R+\frac{1}{\omega Ci}=R-\frac{i}{\omega C}$$

$$=\sqrt{R^2+\left ( \frac{1}{\omega C}\right )^2} e^{i\gamma}$$

where $$\gamma=\tan^{-1}{\left (-\frac{1}{R\omega C}\right )}$$.

If we look at the OP we found that the impedance was

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

We can show that in fact ##\gamma=\phi-\pi/2##.

This seems to indicate that perhaps for any of the possibilities of circuits (R, L, C, RC, RL, LC, RLC) all we need to do to find the solution is sum the impedances for the circuits with just one circuit element. The solution is then ##I_c(t)=Z_{eq}V_c(t)## where ##Z_{eq}## is the equivalent impedance.

One question I have is: what is it that allows us to just sum impedances?

It doesn't seem to be superposition. After all, e.g. in the case of an RC circuit, we have two fundamentally different circuits (one with one resistor, and the other with one capacitor) with the same input voltage. Why can we essentially add the individual solutions to obtain the solution to the RC circuit?

zenterix said:
One question I have is: what is it that allows us to just sum impedances?
The same current must flow through every one of the impedances (the law of conservation of charge) and the voltage drop across all of the impedances must be the sum of the voltage drops across every one of the impedances (the law of conservation of energy).

Ohm’s law must be included too.

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