High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Physics Textbook (Master) > AC Circuits

Reactance and Phasors

Basics of AC CircuitsPrevious Section
Impedance, RC Circuits, and RL CircuitsNext Section
Popular Courses
Find My Course
0:00 / 0:00

Reactance


In an AC circuit, capacitors and inductors impede current flow, similarly to resistors. To fully analyze AC circuits, we need to be able to compare the effects of resistors, capacitors, and inductors on current.
  • Reactance (X) is the effective resistance of an AC circuit component.
  • Reactance is always measured in Ohms (Ω\Omega).
  • For resistors, reactance is equal to the resistance:
XR=R\boxed{X_R=R}

Capacitors

  • For capacitors, we define the capacitive reactance as follows:
XC=1ωC\boxed{X_C=\frac{1}{\omega C}}
  • ω\omega is the angular frequency of the AC voltage source.
Inductors
  • For inductors, we define the inductive reactance as follows:
XL=Lω\boxed{X_L=L\omega}
  • ω\omega is the angular frequency of the AC voltage source.
0:00 / 0:00

AC Circuits with Capacitors


In this section, we'll look at the simplest possible capacitive AC circuit.
  • Consider an AC voltage source with a maximum voltage of VmaxV_{max} and an angular frequency of ω\omega connected to a capacitance CC.
  • Applying Kirchhoff's loop rule (remember, the voltage across a capacitor is V=Q/CV=Q/C):
ΔV=0Vsource−Vcapacitor=0Vmaxsin⁡(ωt)−QC=0Q(t)=CVmaxsin⁡(ωt)\begin{aligned} \Delta V&=0 \\ V_{source}-V_{capacitor}&=0 \\ V_{max}\sin(\omega t)-\frac{Q}{C}&=0 \\ Q(t)&=CV_{max}\sin(\omega t) \end{aligned}
  • This gives the charge on the capacitor as a function of time.
  • To find the equation for current, we will take a derivative on both sides (remember, I=dQdtI=\frac{dQ}{dt}):
dQdt=ddt[VmaxCsin⁡(ωt)]I(t)=VmaxCωcos⁡(ωt)\begin{aligned} \frac{dQ}{dt}&=\frac{d}{dt}[V_{max}C\sin(\omega t)] \\ I(t)&=V_{max}C\omega\cos(\omega t) \\ \end{aligned}
  • It will be easier to work with this equation if we substitute XC=1ωCX_C=\frac{1}{\omega C} and cos⁡(ωt)=sin⁡(ωt+π2)\cos(\omega t)=\sin(\omega t + \frac\pi2):
I(t)=VmaxXCsin⁡(ωt+π/2)\boxed{I(t)=\frac{V_{max}}{X_C}\sin(\omega t+\pi/2)}

Wize Concept
There are two take-aways here:
  1. The capacitive reactance takes the role of resistance (instead of I=VRI=\frac{V}{R}, we see I=VXCI=\frac{V}{X_C})
  2. The current I(t)I(t) is always out of phase from the voltage source V(t)V(t).
  3. That is, the voltage lags behind the current for purely capacitive AC circuits.

0:00 / 0:00

AC Circuits with Inductors


In this section, we'll look at the simplest possible inductive AC circuit.
  • Consider an AC source with a maximum voltage of VmaxV_{max} and an angular frequency of ω\omega connected to an inductance LL.
  • Applying Kirchhoff's Loop Rule (remember, for an inductor, V=LdIdtV=L\frac{dI}{dt}):
ΔV=0Vsource−Vinductor=0Vmaxsin⁡(ωt)−LdIdt=0dIdt=VmaxLsin⁡(ωt)\begin{aligned} \Delta V&=0 \\ V_{source}-V_{inductor}&=0 \\ V_{max}\sin(\omega t)-L\frac{dI}{dt}&=0 \\ \frac{dI}{dt}&=\frac{V_{max}}{L}\sin(\omega t) \\ \end{aligned}
  • We can find current by integrating both sides:
∫[dIdt]dt=∫[VmaxLsin⁡(ωt)]dtI(t)=VmaxL(−1ωcos⁡(ωt))I(t)=−VmaxωLcos⁡(ωt)\begin{aligned} \int\bigg[\frac{dI}{dt}\bigg]dt&=\int\bigg[\frac{V_{max}}{L}\sin(\omega t)\bigg]dt \\ I(t)&=\frac{V_{max}}{ L}(-\frac{1}{\omega}\cos(\omega t)) \\ I(t)&=-\frac{V_{max}}{\omega L}\cos(\omega t) \end{aligned}
  • It will be easier to work with this equation if we substitute XL=ωLX_L=\omega L and −cos⁡(ωt)=sin⁡(ωt−π2)-\cos(\omega t)=\sin(\omega t - \frac\pi2):
I(t)=VmaxXLsin⁡(ωt−π2)\boxed{\begin{aligned} I(t)&=\frac{V_{max}}{X_L}\sin(\omega t-\frac\pi2) \end{aligned}}

Wize Concept
There are two take-aways here:
  1. The inductive reactance takes the role of resistance (instead of I=VRI=\frac{V}{R}, we see I=VXLI=\frac{V}{X_L})
  2. The current I(t)I(t) is always out of phase from the voltage source V(t)V(t).
  3. That is, the voltage leads the current for purely inductive AC circuits.

0:00 / 0:00

Example: AC Circuit with an Inductor


Consider a voltage source with a maximum voltage Vo=50 VV_o=50~V and a frequency of 10 Hz10~Hz connected to an inductor of inductance L=0.1 HL=0.1~H.
a) Write down the formulas for the voltage generated from the source, the voltage across the inductor, and the current.
b) Once the voltage has reached its maximum value across the inductor, how long will it take for the current to reach its maximum value?
c) When the current through the inductor has reached its maximum value, how long will it take for the voltage to reach zero volts?

Part a)

The formulas can be written down from the general formulas. Remember that because the elements are in series, the current is equal at all points in time for both elements (and this is out of phase from the source voltage).

Source voltage:
V(t)=Vosin⁡(ωt)V(t)=(50 V)sin⁡(20πt)\begin{aligned} V(t)&=V_o\sin(\omega t)\\ V(t)&=(50~V)\sin(20\pi t)\\ \end{aligned}
Inductor voltage:
V(t)=Vosin⁡(ωt+π2)V(t)=(50 V)sin⁡(20πt+π2)\begin{aligned} V(t)&=V_o\sin(\omega t+\frac\pi2)\\ V(t)&=(50~V)\sin(20\pi t+\frac\pi2) \end{aligned}
Inductor current:
I(t)=VoXLsin⁡(ωt+π2)I(t)=VoωLsin⁡(ωt+π2)I(t)=(50 V)(2π)(0.1)sin⁡(20πt+π2)I(t)=(79.6 V)sin⁡(20πt+π2)\begin{aligned} I(t)&=\frac{V_o}{X_L}\sin(\omega t+\frac\pi2)\\ I(t)&=\frac{V_o}{\omega L}\sin(\omega t+\frac\pi2)\\ I(t)&=\frac{(50~V)}{(2\pi )(0.1)}\sin(20\pi t+\frac\pi2)\\ I(t)&=(79.6~V)\sin(20\pi t+\frac\pi2)\\ \end{aligned}
Part b)

We know that the voltage leads the current by π/2\pi/2 for a purely inductive AC circuit. So the current will each its maximum after one-quarter of a cycle passes. You could re-arrange the above formulas, but it's easier to just take one-quarter of the period:
T=1f=0.1s→T4=0.025sT=\frac{1}{f}=0.1s \\ \rightarrow \frac{T}{4}=0.025s
Part c)

Again, the current lags behind the voltage for this circuit. When the current is at its maximum value, the voltage across the inductor is already at zero (it is one-quarter of a cycle ahead!) The answer to this part is zero seconds.

Practice: AC Circuit with a Capacitor


Consider a voltage source with a maximum voltage of Vmax=500 VV_{max}=500~V and a source frequency off=200 Hzf=200~Hz attached to a capacitor of capacitance C=650 μFC=650~\mu F.
a) Write the formula for the current across the capacitor as a function of time.
b) What is the charge stored on the capacitor when the current is equal to the RMS current? (If there is more than one possible value, list all of the possible values.)
Part a)
0:00 / 0:00

Phasors


To analyze complex AC circuits, we need to introduce a new tool to deal with the different phases of current and voltage for resistors, capacitors, and inductors.
  • Phasors are vector representations of AC voltages or currents.
  • The magnitude of a phasor is the maximum voltage or maximum current.
  • The direction of a phasor represents the phase.
Wize Concept
Phasors are always rotating counter-clockwise at the angular frequency ω\omega of the source. Phasor diagrams should be interpreted as a "snapshot" in time.

Resistive AC Circuits
  • For AC voltage sources connected only to a resistor, the current and voltage are in phase.

PAGE BREAK

Capacitive AC Circuits

  • For AC voltage sources connected only to a capacitor, the voltage lags behind the current by π/2\pi/2.

Inductive AC Circuits

  • For AC voltage sources connected only to an inductor, the voltage leads the current by π/2\pi/2.