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Wize University Physics Textbook (Master) > AC Circuits

Impedance, RC Circuits, and RL Circuits

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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.

Wize Concept
We can treat phasors as complex numbers. At any given snapshot in time, the measurements of voltage can be read from the horizontal and vertical components of the phasor.
  • If we have a system with resistance RR and reactances XCX_C and XLX_L, the (complex) impedance can be written as follows, where jj is the imaginary number:
Z=R+j(XL−XC)\boxed{Z=R+j(X_L-X_C)}
  • The magnitude of the complex impedance ∣Z∣|Z| can be written as follows:
∣Z∣2=R2+(XL−XC)2\begin{aligned} \boxed{|Z|^2=R^2+(X_L-X_C)^2} \\ \end{aligned}

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  • The source voltage and current of an AC circuit are related by Ohm's Law:
V=I∣Z∣\boxed{V=I|Z|}
  • The phase angle ϕ\phi of the impedance is given by:
tan⁡ϕ=XL−XCR\boxed{\tan\phi=\frac{X_L-X_C}{R}}

Watch Out!
The phase angle ϕ\phi is the phase difference between the current and the applied (source) voltage.
If ϕ>0\phi > 0, then the voltage is ahead of the current (or the current lags the voltage).
If ϕ<0\phi < 0, then the voltage is behind the current (or the current leads the voltage).

Exam Tip
If an exam problem asks for the phase of the current, you have to flip the sign of the value found with the above phase angle formula. This formula gives the phase of voltage relative to current; flipping the sign gives current relative to voltage.

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RC Circuits (AC)


A purely capacitive AC circuit has a voltage that lags behind the current by π2\frac{\pi}{2}. Adding a resistor will modify this behaviour. We call these RC circuits.
  • Consider an RC circuit with a resistance RR and a capacitance CC connected in series to an alternating voltage source with frequency ω\omega and maximum voltage VoV_o.
  • The phasor diagram for this circuit is as follows:

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  • To find the relationship between current and voltage, we can use the impedance.
  • Since there are no inductors, we set XL=0X_L=0 .
Z=R+j(XL−XC)Z=R+j((0)−XC)Z=R−j XC→∣ZRC∣=R2+XC2\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:
Imax⁡=Vmax⁡∣ZRC∣I_{\max}=\frac{V_{\max}}{\left|Z_{RC}\right|}
  • The phase angle is found similarly:
tan⁡ϕ=XL−XCRtan⁡ϕ=(0)−XCRtan⁡ϕ=−XCR\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}

Wize Concept
For alternating RC circuits, the voltage always lags behind the current, but the lag is less than π2\frac\pi2.

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Example: RC Circuit (AC)


Consider a resistor with resistance R=50 ΩR=50~\Omega and a capacitor with capacitance C=200 μFC=200~\mu F connected in series to an AC source V(t)=200sin⁡(6πt)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 Vmax=200VV_{max}=200V and ω=6π rads\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:
∣Z∣=R2+XC2∣Z∣=R2+1ω2C2∣Z∣=(50 Ω)2+1(6π)2(200×10−6F)2∣Z∣=269.9 Ω\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:
Imax=Vmax∣Z∣Imax=200 V269.9 ΩImax=0.74 A\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):
VR,max=IR,maxRVR,max=(0.74 A)(50 Ω)VR,max=37 V\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=CVQ=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):
VC,max=IC,maxXCVC,max=IC,max1ωC\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:
Qmax=CVC,maxQmax=C(IC,max1ωC)Qmax=IC,maxωQmax=(0.74 A)(6π rads)Qmax=0.039 C\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)=Imaxsin⁡(ωt+ϕ)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:
ϕ=tan⁡−1(XL−XCR)ϕ=tan⁡−1((0)−XCR)ϕ=tan⁡−1(−XCR)ϕ=tan⁡−1(−1ωCR)ϕ=tan⁡−1(−1(6π rads)(200×10−6C)(50Ω))ϕ=−1.38 rad\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:
I(t)=Imaxsin⁡(ωt+ϕ)I(t)=0.74sin⁡(6πt+1.38) A\begin{aligned} I(t)&=I_{max}\sin(\omega t+\phi)\\ I(t)&=0.74\sin(6\pi t+1.38)~A\\ \end{aligned}

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RL Circuits (AC)


A purely inductive AC circuit has a voltage that leads the current by π2\frac{\pi}{2}. Adding a resistor will modify this behaviour. We call these RL circuits.
  • Consider an RL circuit with a resistance RR and an inductance LL connected in series to an alternating voltage source with frequency ω\omega and maximum voltage VoV_o.
  • The phasor diagram for this circuit is as follows:

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  • To find the relationship between current and voltage, we can use the impedance.
  • Since there are no capacitors, we set XC=0X_C=0 .
Z=R+j(XL−XC)Z=R+j(XL−(0))Z=R+j XL→∣ZRL∣=R2+XL2\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:
Imax⁡=Vmax⁡∣ZRL∣I_{\max}=\frac{V_{\max}}{\left|Z_{RL}\right|}
  • The phase angle is found similarly:
tan⁡ϕ=XL−XCRtan⁡ϕ=XL−(0)Rtan⁡ϕ=XLR\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}

Wize Concept
For alternating RL circuits, the voltage always leads the current, but the lead is less than π2\frac\pi2.

Practice: RL Circuit (AC)


Consider a resistor with resistance R=100 ΩR=100~\Omega and two inductors with inductances L1=10 HL_1=10~H and L2=20 HL_2=20~H . These are all connected in series to an AC voltage source. The source has an RMS voltage of Vrms=20 VV_{rms}=20~V and frequency f=10 Hzf=10~Hz.
a) What is the maximum current of the circuit?
b) What is the rms voltage across the resistor?
Part a)