Ohm's Law in AC Circuits

Ohm's Law for AC circuits is just like Ohm's Law for DC circuits, using impedance (\(Z\)) in place of resistance and phasors to represent all the values: \(V = I\cdot Z\) Which can again be rearranged into the other two forms: \(I = \frac{V}{Z}\) \(Z = \frac{V}{I}\)

The circuit below has a voltage source with \(V = 10\cos(5t + 30^\circ)\) and an element with impedance \(Z = 5\angle 10^\circ\). Find the current in the circuit.

We first need the voltage in phasor form so: \(V = 10\cos(\omega t + 30^\circ) = 10\angle 30^\circ\) Now we can use Ohm's Law to find: \(I = \frac{V}{Z} = \frac{10\angle 30^\circ}{5\angle 10^\circ}\) \(I = 2\angle 20^\circ\) Now we can convert the phasor back into a sine wave to complete our answer. \(2\angle 20^\circ = 2\cos(5t + 20^\circ)\) The sine wave has \(\omega = 5\) since that was the frequency of the source in the problem. Remember all of these phasor techniques only work if every source has the same frequency which means that EVERY sine wave in the circuit has the same frequency.

The circuit below has a voltage source with \(V = 25\cos(2\pi t)\) and a current of \(I = 10\cos(2\pi t + \frac{\pi}{4})\). Find the impedance of the circuit.

Our first step is to turn the voltage and currents into phasors. \(V = 25\angle 0\) \(I = 10\angle \frac{\pi}{4}\) Remember that we're using radians here for the phase angles since there's no degree symbol. Now Ohm's Law tells us that \(Z = \frac{V}{I}\) \(Z = \frac{25 \angle 0}{10 \angle \frac{\pi}{4}}\) \(Z = \frac{25}{10}\angle (0 - \frac{\pi}{4}) \) \(Z = 2.5\angle -\frac{\pi}{4}\)

Find the current in the circuit below with source voltage \(V = 5\cos(100\pi t\) and capacitance \(C = 1m\)F

We first put the voltage into phasor form: \(V = 5\angle 0^\circ\) Next we get the impedance of the capacitor: \(Z = \frac{1}{\omega C}\angle -90^\circ\) \(Z = 3.18\angle -90^\circ\) Now we use Ohm's Law to find the current: \(I = \frac{V}{Z} = \frac{5}{3.18} \angle 90^\circ\) \(I = 1.6\angle 90^\circ\) \(I = 1.6\cos(100\pi t + 90^\circ)\) The image below shows the voltage and current waveforms on top of each other. As we can see a capacitor causes the current to "lead" the voltage by \(90^\circ\)