Important points

• In AC circuits, voltage and currentvary sinusoidally with time.

• Phasors/phase diagrams are usefultools for visualizing AC voltage orcurrent levels.

Important equations

• Variation with time

v = V cos ωt

i = I cos ωt

• RMS

Vrms =V√

2

Irms =I√2

.

Figure 1: AC voltage source symbol incircuit diagrams.

Time [sec.]

0 5 10 15 20 25 30 35 40 45 50

v [v

olt

s]

100−

50−

0

50

100

Figure 2: Example AC voltage plot withV = 120 V and f = 60 Hz.

PHYS 202 Notes, Week 7Greg Christian

March 1, 2016Last updated: 03/01/2016 at 11:49:07

This week we learn about AC circuits.

Alternating Current

So far, when learning about circuits we’ve treated the emf source asone of direct current, meaning that it always delivers a constant poten-tial difference. In real world applications, many circuits are alternat-ing current, or AC, with the potential difference (and thus the currentvarying as a function of time). The reason for using AC is primarilybecause it allows voltage levels to be stepped up/down using a trans-former (recall that a transformer requires a time dependent current towork—this is what we get when dealing with AC).

Generically, we can refer to the device supplying potential in anAC circuit as an AC source. In circuit diagrams, these are drawn bythe symbol in Figure 1. These supply a sinusoidally varying potentialgiven by

v = V cos ωt, (1)

where

• v is the instantaneous potential difference;• V is the maximum potential difference; and• ω is the angular frequency.

Note that the angular frequency ω (radians/second) is related to the“regular” frequency f (Hertz, or Hz) by

ω = 2π f . (2)

An example plot of AC voltage is shown in Figure 2.Here in North America, commercial power systems use a frequency

f = 60 Hz, (angular frequency ω = 377 rad/s). In Europe and muchof the rest of the world, it’s f = 50 Hz (ω = 314 rad/sec).

Similar to the voltage, the current in AC circuits varies sinusoidally,

i = I cos ωt, (3)

where i and I are the instantaneous and maximum current, respec-tively.

phys 202 notes, week 7 2

I cos ωt

I

ωt

ω

Figure 3: A phase diagram.

Phasors

To represent alternating current (or voltage) graphically, we can use ro-tating vectors called phasors, which are drawn in phase diagrams. Theseare useful tools for visualizing current or voltage vs. time. In particu-lar, it’s useful when we need to add or subtract alternating currents orvoltages.

An example phase diagram is shown in Figure 3. Basically, what wedo it to have a fixed-length vector whose total length is the maximumcurrent I (or maximum voltage V). This vector is constantly rotatingaround the x-y axes, with the angle between the vector and the x axisbeing ωt. Then the x-axis projection of the vector is always equal to theinstantaneous current (or voltage).

So for example, at t = 0, the vector lies completely along the x-axis;the projection is simply equal to its total length I. At t = ω/2π, allof the vector lies along the y-axis, so the x-axis projection is zero andi = 0. Everywhere in between, the projection is some value betweenthese two extremes.

RMS

It’s common to refer to AC voltages or currents in terms of their root-mean-square, or RMS, values. As demonstrated in Figure 4, these aregiven by

Vrms =V√

2(4)

Irms =I√2

. (5)

Figure 4: Meaning of RMS currents &potentials.

phys 202 notes, week 7 3

Important points

• The phase describes the time relation-ship between voltage and current.

• Reactance represents the voltage tocurrent ratio through an inductor ora capacitor.

Important equations

• Resistor

i = I cos ωt

v = IR cos ωt

• Inductor

i = I cos ωt

v = IωL cos (ωt + 90)

XL = ωL

• Capacitor

i = I cos ωt

v = (I/ωC) cos (ωt− 90)

XL = 1/ωC

Figure 5: Resistor in an AC circuit.

AC circuits

Here we discuss what happens when you construct simple circuitswith AC. First let’s discuss what happens when a resistor, an inductor,and a capacitor are connected to an AC source.

Resistor

Let’s imagine a resistor is connected to an AC source delivering current

i = I cos ωt, (6)

as in Figure 5(a). The potential drop across the resistor is then

v = IR cos ωt, (7)

i.e. a sinusoidal dependence with the same frequency as the currentand maximum voltage VR = IR.

As shown in the Figure 5(b) the voltage and current have the sametime dependence, although their amplitudes are different. To describethis, we say that they are “in phase”. It follows that their phasorsare parallel to each other, just having different magnitudes, as in Fig-ure 5(c).

Inductor

Now let’s imagine an inductor is connected to an AC source (Figure 6).The changing current will generate an emf according to E = −L∆i/∆t.This means that the potential difference across the inductor is given by

vL = L∆i∆t

. (8)

What this means is that the potential drop across the inductor is equalto L times the rate of change of the current with respect to time.

The equation describing vL can be solved using calculus to give

vL = −IωL sin ωt, (9)

which is mathematically equivalent to

vL = IωL cos (ωt + 90) . (10)

As shown in Figure 6(b) and (c), the peaks in current and voltageare separated by π/2, or 90. This is referred to as the phase. The phaseis the angle describing the separation in peaks between voltage andcurrent. Typically we refer to the phase of the voltage relative to thecurrent. Here the voltage “leads”, or is ahead of the current by a phaseof π/2, or 90.

phys 202 notes, week 7 4

Figure 6: Inductor in an AC circuit.

Figure 7: Capacitor in an AC circuit.

More generally, we can talk about the phase angle φ and write thevoltage as

v = V cos (ωt + φ) . (11)

In the case of a single inductor, the phase angle is φ = 90, and theamplitude is

VL = IωL. (12)

With inductors, we can define a term called the reactance, XL whichdescribes the ratio of the voltage amplitude to current amplitude. This isgiven by

XL = ωL. (13)

Keep in mind that this is the ratio of amplitudes, or maximum values,not the instantaneous v/i ratio. As a voltage/current ratio, the units ofreactance are the same as resistance: the ohm (Ω).

Capacitor

Finally, let’s consider a capacitor connected to an AC source (Figure 7).Since we have varying potential, we also have varying charge on theplates given by

∆vC∆t

=1C

∆q∆t

=iC

. (14)

Thus the rate of change of the potential is

∆vC∆t

=1C

I cos ωt. (15)

This can again be solved using calculus to give

vC =I

ωCsin ωt, (16)

which is equivalent to

vC =I

ωCcos (ωt− 90) . (17)

Similar to the inductor case, the voltage and current are out of phaseby 90, except now the voltage is behind the current rather than ahead.

phys 202 notes, week 7 5

Important points

• Algebraic sum of instantaneous volt-age drops across R, L, C equalsthe total instantaneous voltage of thesource.

• Vector sum of R, L, C phasors givesthe total voltage phasor.

Important equations

• Reactance

X = XL − XC

• Impedance

Z =√

R2 + X2

=√

R2 + (XL − XC)2

• Voltage & current

V = IZ

= I√

R2 + (XL − XC)2

• Phase angle

φ = arctan(

ωL− 1/ωC

R

)• Average power

P = (1/2)VI cos φ

= Vrms Irms cos φ

• Resonance

ω0 = 1/√

LC

f0 = 1/[2π√

LC]

Hence we say that the voltage lags the current by π/2, or 90. This isdemonstrated in Figure 7(b) and (c). As suggested by Eq. (17), thevoltage amplitude is given by

VC =I

ωC. (18)

As with the inductor, we can define a reactance for the capacitorgiven by

XC =1

ωC. (19)

RLC Circuits

Now let’s consider an RLC circuit containing a resistor, inductor, andcapacitor all in series, as Figure 8. In this case Kirchhoff’s law stillapplies, and the algebraic sum of instantaneous potential drops equalsthe instantaneous source potential, i.e.

v = vR + vL + vC. (20)

More generally, the vector sum of the phasors VR, VL, and VC equalsthe phasor that describes the total voltage, V. This is shown in Fig-ure 8(b) and (c). To form this vector sum, first subtract VC from VL

since they point in opposite directions to each other. Then we’re justleft with the resulting vector (VL − VC) and VR. These are alwaysat a 90 angle, so we can get the magnitude of the sum using thePythogorean theorem:

V =√

V2R + (VL −VC)

2 (21)

=√(IR)2 + (IXL − IXC)2 (22)

= I√

R2 + (XL − XC)2. (23)

The quantity XL − XC is called the reactance of the circuit,

X = XL − XC. (24)

We can also define a quantity called the impedence Z,

Z =√

R2 + (XL − XC)2 (25)

=√

R2 + [ωL− (1/ωC)]2. (26)

This has the property that

V = IZ. (27)

phys 202 notes, week 7 6

Figure 8: An RLC circuit.

Finally, each RLC circuit has a phase angle describing the voltage andcurrent time-relationship. This is given by

tan φ =VL −VC

VR(28)

=I(XL − XC)

IR(29)

=XL − XC

R(30)

=XR

(31)

⇒ φ = arctan(

ωL− 1/ωC

R

). (32)

Note that if XL > XC, the voltage leads the current by an angle φ be-tween 0 and 90. Conversely, if XL < XC, the voltage lags the current.

Note that all of the relationships between voltage and current am-plitudes (e.g. Eq. (27) and the like) are equally valid for RMS values,e.g.

Vrms = IrmsZ. (33)

Power in RLC circuits

To understand the power delivered to RLC circuits, let’s first considerthe simplest case of just a resistor R and a voltage source. In this case,the average power is equal to one-half of its maximum value, or

P =12

VI. (34)

phys 202 notes, week 7 7

Equivalently, we can write this in terms of RMS values,

P =V√

2I√2

(35)

= Vrms Irms. (36)

And since Vrms = RIrms, we can also write

P = I2rmsR (37)

= V2rms

/R (38)

= Vrms Irms. (39)

Remember: these equations are only valid when there’s just a resistorin the circuit. They cannot be used if an inductor or a capacitor ispresent.

So what if an inductor and/or capacitor is present? Let’s considerthe case with both an inductor and capacitor (and a resistor) hookedup to the voltage source. As with any AC circuit, the instantaneouspower is given by

p = vi (40)

= [V cos (ωt + φ)] [I cos ωt] . (41)

From this, we can derive the expression for the average power,

P =12

VI cos φ (42)

= Vrms Irms cos φ. (43)

In other words, the average power depends on the phase angle φ.We often call the quantity cos φ the power factor of the circuit. For apure resistance circuit, φ = 0, cos φ = 1, and P = Vrms Irms. For a pureresistanceless (L or C only) circuit, φ = 90, cos φ = 0, and P = 0. Fora series RLC circuit, the power factor is equal to R/Z.

From Eq. (43) and Vrms = IrmsZ, it’s possible to derive equationssimilar to those we’ve seen before:

P = Vrms Irms cos φ (44)

= I2rmsZ cos φ (45)

=V2

rmsZ

cos φ. (46)

Series Resonance

As demonstrated by Eq. (26), the impedance of an RLC circuit dependson its frequency:

Z =√

R2 + [ωL− 1/(ωC)2]. (47)

phys 202 notes, week 7 8

Hence when the frequency is such that

ωL = 1/(ωC), (48)

the impedance is equal to its smallest possible value, which is just theresistance R. This is called the resonance angular frequency, or ω0. Itcorresponds to the maximum possible current amplitude for the circuit.To find the resonance frequency, just solve Eq. (48),

ω20 =

1LC

(49)

⇒ ω0 =1√LC

. (50)

The resonance frequency, f0 is then ω0/(2π).When a circuit is “on resonance”, a few things are true:

• The currents are the same in L and C.• The voltage in the inductor leads the current by 90.• The voltage in the capacitor lags the current by 90.• Hence the voltages across L and C always vary by 180, or a half

cycle.

– If the amplitudes of the two voltages are equal, then they alwayssum to zero. The total voltage across the LC combination is zero.

Resonance in circuits is very similar to resonance in other systemsin nature. As an example that you saw last semester, consider theforced oscillation of a harmonic oscillator. Here the amplitude of themechanical oscillation peaks when the driving frequency is close to thenatural frequency of the system. This is basically the same thing that’sgoing on in circuits, except the natural frequency is now defined bythe values of L and C.

phys 202 notes, week 7 9

Example Problems