# Impedance (Resistance) in AC Circuits

concept

In order to solve problems in AC circuits we can use Ohm's Law just like we did in DC however we need a new idea of "resistance" that takes into account how capacitors and inductors affect current when an AC voltage is applied to them.
We call this new concept of resistance "impedance".
In this topic we'll see how to calculate the impedance of resistors, capacitors and inductors.
In the next topic we'll use these impedances to solve for circuit currents just like we did in DC, but this time we can find solutions to circuits with capacitors and inductors, opening a whole new world of circuit design and analysis.

fact

The common concept of resistance in DC circuits is replaced with impedance in AC circuits.

When working on electrical circuits we use the symbol \(j\) to represent \(\sqrt{-1}\) instead of \(i\) to avoid any confusion with currents.
Whenever you see a \(j\) remember that it's the imaginary unit.

fact

The impedance of a resistor with resistance \(R\) is given by:
\(Z = R\)

As we'll see in a moment the impedance of capacitors and inductors change based on the frequency of the applied voltage.
This allows us to create circuits that react differently to different frequencies, like a circuit that removes high frequency sounds from a microphone or a filter for a RADAR that only lets in signals of the correct frequency and blocks outside noise.

fact

The impedance of a capacitor with capacitance \(C\) is given by:
\(Z = \frac{1}{j\omega C} = \frac{1}{\omega C}\angle -\frac{\pi}{2}\)
When operated at radian frequency \(\omega\).
Since \(\frac{1}{j} = -j\) we can also say that \(Z = \frac{-j}{\omega C}\)

This shows that the impedance of a capacitor decreases as the frequency increases.
So at DC a capacitor acts like an open circuit but at incredibly high frequencies the capacitor acts like a short circuit.

example

Find the impedance of a capacitor with capacitance \(C = 16\mu\)F at frequency \(\omega = 120\pi\)

We can just plug these values into our formula and find: \(Z = \frac{1}{\omega C}\angle -\frac{\pi}{2}\) \(Z = 166\angle -\frac{\pi}{2}\)fact

The impedance of an inductor with inductance \(L\) is given by:
\(Z = j\omega L = \omega L\angle \frac{\pi}{2}\)
when operated at radian frequency \(\omega\).

So at low frequencies the inductor acts like a short circuit but at larger and larger frequencies the inductor acts more like an open circuit.

So we can see that a resistor has purely real impedance but the inductor and capacitor have purely imaginary impedance.

example

Find the impedance of an inductor with inductance \(L = 10\)mH at frequency \(f = 50\)Hz

Now in this case we need to convert our frequency into radian frequency first. \(\omega = 2\pi f = 100\pi\) And now we can use our formula: \(Z = \omega L \angle \frac{\pi}{2}\) \(Z = 3.14 \angle \frac{\pi}{2}\)fact

We call the imaginary part of an impedance the reactance and give it the symbol \(X\).
So we can write the impedance as:
\(Z = R + iX\)
Where \(R\) is the resistance and \(X\) is the reactance.

fact

The reactance of an inductor is given by:
\(X_L = \omega L\)

fact

The reactance of a capacitor is given by:
\(X_c = -\frac{1}{\omega C}\)

fact

Impedance, resistance and reactance are all measured in Ohms (\(\Omega\))

practice problems