Wize University Physics Textbook (Master) > AC Circuits

Impedance, RC Circuits, and RL Circuits

Impedance

In AC circuits, the voltage is usually not in phase with the current. To see how AC circuits with multiple elements behave, we need to treat the voltages like vectors.

When drawing phasor diagrams, the axes can be interpreted as follows:
The horizontal axis represents the resistor
The positive vertical axis represents the inductor
The negative vertical axis represents the capacitor
For an AC circuit in series, the vector sum of these phasors represents the source (applied) voltage.

If we have a system with resistance $R$ and reactances $X_C$ and $X_L$ , the (complex) impedance can be written as follows, where $j$ is the imaginary number:

\boxed{Z=R+j(X_L-X_C)}

The magnitude of the complex impedance $|Z|$ can be written as follows:

\begin{aligned} \boxed{|Z|^2=R^2+(X_L-X_C)^2} \\ \end{aligned}

The source voltage and current of an AC circuit are related by Ohm's Law:

\boxed{V=I|Z|}

The phase angle $\phi$ of the impedance is given by:

\boxed{\tan\phi=\frac{X_L-X_C}{R}}

RC Circuits (AC)

A purely capacitive AC circuit has a voltage that lags behind the current by $\frac{\pi}{2}$ . Adding a resistor will modify this behaviour. We call these RC circuits.

Consider an RC circuit with a resistance $R$ and a capacitance $C$ connected in series to an alternating voltage source with frequency $\omega$ and maximum voltage $V_o$ .

The phasor diagram for this circuit is as follows:

To find the relationship between current and voltage, we can use the impedance.
Since there are no inductors, we set $X_L=0$ .

\begin{aligned} Z &= R+j(X_L-X_C)\\ Z &= R+j((0)-X_C)\\ Z&=R-j\ X_C \\ \end{aligned}\\ \rightarrow \boxed{|Z_{RC}|=\sqrt{R^2+X_C^2}}

This gives the current:

I_{\max}=\frac{V_{\max}}{\left|Z_{RC}\right|}

The phase angle is found similarly:

\begin{aligned} \tan\phi&=\frac{X_L-X_C}{R}\\ \tan\phi&=\frac{(0)-X_C}{R}\\ \tan\phi&=\frac{-X_C}{R}\\ \end{aligned}

Example: RC Circuit (AC)

Consider a resistor with resistance $R=50~\Omega$ and a capacitor with capacitance $C=200~\mu F$ connected in series to an AC source $V\left(t\right)=200\sin\left(6\pi t\right)$ .

a) What is the max current passing through each element?

b) What is the max voltage across the resistor?

c) What is the maximum charge stored on the capacitor?

d) What is the current in the circuit as a function of time?

We can read from the voltage source that $V_{max}=200V$ and $\omega=6\pi~\frac{rad}{s}$ .

Part a)

Because the circuit is in series, the current is equal through all circuit elements at all times.

We can use Ohm's Law to relate the maximum voltage to the maximum current, but we'll need to find the impedance first:

\begin{aligned} |Z|&=\sqrt{R^2+X_C^2} \\ |Z|&=\sqrt{R^2+\frac1{\omega^2C^2}} \\ |Z|&=\sqrt{(50~\Omega)^2+\frac1{(6\pi)^2(200\times10^{-6}F)^2}} \\ |Z|&=269.9~\Omega \end{aligned}

Using Ohm's Law:

\begin{aligned} I_{max}&=\frac{V_{max}}{|Z|}\\ I_{max}&=\frac{200~V}{269.9~\Omega}\\ I_{max}&=0.74~A \end{aligned}

This is the maximum current through each of the circuit elements.

Part b)

The maximum voltage across the resistor is found by multiplying the maximum current by the resistance (Ohm's Law):

\begin{aligned} V_{R,max}&=I_{R,max}R\\ V_{R,max}&=(0.74~A)(50~\Omega)\\ V_{R,max}&=37~V\\ \end{aligned}

Part c)

The formula for charge stored in a capacitor is $Q=CV$ , so we'll need to find the maximum voltage (with Ohm's Law) before we can find the maximum charge stored.

The maximum voltage across the capacitor is found by multiplying the maximum current by the capacitive reactance (Ohm's Law):

\begin{aligned} V_{C,max}&=I_{C,max}X_C\\ V_{C,max}&=I_{C,max}\frac1{\omega C}\\ \end{aligned}

Finding the maximum charge stored:

\begin{aligned} Q_{max}&=CV_{C,max}\\ Q_{max}&=C(I_{C,max}\frac1{\omega C})\\ Q_{max}&=\frac{I_{C,max}}{\omega }\\ Q_{max}&=\frac{(0.74~A)}{(6\pi~\frac{rad}{s}) }\\ Q_{max}&=0.039~C \end{aligned}

Part d)

The current will be written in the form $I(t)=I_{max}\sin(\omega t+\phi)$ . The maximum current was already found in part (a) and the frequency will match the voltage source, so all we need to find is the phase angle:

\begin{aligned} \phi&=\tan^{-1}(\frac{X_L-X_C}{R})\\ \phi&=\tan^{-1}(\frac{(0)-X_C}{R})\\ \phi&=\tan^{-1}(\frac{-X_C}{R})\\ \phi&=\tan^{-1}(\frac{-1}{\omega C R})\\ \phi&=\tan^{-1}(\frac{-1}{(6\pi~\frac{rad}{s}) (200\times10^{-6}C) (50\Omega)})\\ \phi&=-1.38~rad\\ \end{aligned}

We know that for RC circuits, the current leads the voltage. We'll flip the sign of this phase angle and substitute it into the formula with the other knowns:

\begin{aligned} I(t)&=I_{max}\sin(\omega t+\phi)\\ I(t)&=0.74\sin(6\pi t+1.38)~A\\ \end{aligned}

RL Circuits (AC)

A purely inductive AC circuit has a voltage that leads the current by $\frac{\pi}{2}$ . Adding a resistor will modify this behaviour. We call these RL circuits.

Consider an RL circuit with a resistance $R$ and an inductance $L$ connected in series to an alternating voltage source with frequency $\omega$ and maximum voltage $V_o$ .

The phasor diagram for this circuit is as follows:

To find the relationship between current and voltage, we can use the impedance.
Since there are no capacitors, we set $X_C=0$ .

\begin{aligned} Z &= R+j(X_L-X_C)\\ Z &= R+j(X_L-(0))\\ Z&=R+j\ X_L \\ \end{aligned}\\ \rightarrow \boxed{|Z_{RL}|=\sqrt{R^2+X_L^2}}

This gives the current:

I_{\max}=\frac{V_{\max}}{\left|Z_{RL}\right|}

The phase angle is found similarly:

\begin{aligned} \tan\phi&=\frac{X_L-X_C}{R}\\ \tan\phi&=\frac{X_L-(0)}{R}\\ \tan\phi&=\frac{X_L}{R}\\ \end{aligned}

Practice: RL Circuit (AC)

Consider a resistor with resistance $R=100~\Omega$ and two inductors with inductances $L_1=10~H$ and $L_2=20~H$ . These are all connected in series to an AC voltage source. The source has an RMS voltage of $V_{rms}=20~V$ and frequency $f=10~Hz$ .

a) What is the maximum current of the circuit?

b) What is the rms voltage across the resistor?

Part a)

0.015 A

0.090 A

0.123 A

I don't know

Impedance, RC Circuits, and RL Circuits

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Impedance

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RC Circuits (AC)

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Example: RC Circuit (AC)

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RL Circuits (AC)

Practice: RL Circuit (AC)

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