Chapter 35 – AC Circuits

 

This chapter covers simple AC circuits, an important topic, but rather disconnected from the previous chapters.  The main results of the chapter follow directly from applications of Kirchoff’s two laws: conservation of current and the sum of potentials = 0 around a loop.  Those laws will be applied to circuits with an AC (alternating current) power source which supplies a sinusoidal emf, E(t) = E0 cos ωt, ω (omega) is the angular frequency, radians/second.  Note that ω = 2πf, where f is the frequency in cycles/second.  The electricity delivered to our homes has an f = 60 cycles/second = 60 Hertz.

 

The material in this chapter requires that you be able to picture the relationship between cosine and sine graphs and/or be able to use the relationship cos (A + B) = cosAcosB -sinAsinB.

 

1.         We can graphically represent the sinusoidally changing emf by considering E(t) to be the projection on the x-axis of a rotating vector with length E0.  The tail of the vector is at the origin.  The vector rotates counter clockwise with angular velocity = ω and at t = 0 it is parallel to the x-axis.  Convince yourself that the projection of this rotating vector on the x-axis is just E0 cos ωt.  The rotating vector is called a phasor.

 

2.         The first case considered will be a resistor in series with the emf.  Since the sum of the potential drops around this circuit are zero, E(t) - I(t)R = 0.  Therefore,

 

            IR(t) = E(t)/R = (E0/R)cos ωt = current through the resistor

            VR(t) =  E0 cos ωt = potential drop across the resistor.

 

For a resistor in an AC circuit the current through the resistor and the potential drop across the resistor are in phase with one another.  That is they each oscillate in time as cos ωt.

 

3.         Next consider a capacitor in series with E(t).  Applying Kirchoff’s loop law gives, E(t) - q(t)/C = 0, or

 

q(t) = C E0 cos ωt.

 

The current in the circuit, remember the current has to be the same everyplace in the circuit, is just dq/dt,

 

            I(t) = - Cω E0 sin ωt = Cω E0 cos (ωt + π/2) = Ic cos (ωt + π/2), where Ic = Cω E0

 

Using a resistor as our model where the maximum current is E0/R, the maximum current in this circuit is E0/Xc where Xc is the capacitive reactance and has units of ohms.  Its value is Xc = 1/Cω.

 

Note that in this circuit the current is not in phase with the emf.  Instead the current phasor is π/2 ahead of the emf phasor.  Since the voltage across the capacitor is proportional to q(t), and q(t) is in phase with the emf, the voltage across the capacitor lags the current by π/2.

 

4.         Next consider a circuit with an inductor and E (t).  Kirchoff’s law gives,

E(t) - LdI/dt = 0 or dI/dt = (E0/L) cos ωt.  To find the current in this equation we need to integrate the equation for dI/dt.  Multiply both sides by dt.  The integral of dI just gives I(t) while the integral of cos ωt is just (sin ωt)/ω.  So I(t) is given by,

 

            I(t) = (E0/ωL) sin ωt = (E0/ωL) cos (ωt - π/2),

 

 and the inductive reactance, XL is just ωL which also has units of ohms.  Note that in this circuit the current lags the emf by π/2.  Since the voltage drop across an inductor is proportional to dI/dt, the voltage across an inductor leads the current by π/2.

 

5.         Now we want to use the above information to analyze composite series circuits.  First we will consider a circuit with R and C and then one with R, C, and L.  In both cases we use the fact that the current is the same everywhere in a series circuit.  To draw a phasor diagram, pick a direction for i(t).  For example imagine i(t) is a phasor in the first quadrant.  Then vR(t) is parallel to i(t) since the voltage drop across a resistor is in phase with the current.  But the voltage drop across the capacitor, vC(t) lags the current by π/2.  So draw the vC(t) phasor π/2 behind i(t), that puts it in the fourth quadrant.  From Kirchoff’s law, the sum of E(t) - vR(t) - vC(t) = 0 or the phasor representing E(t) is the sum of  the phasors representing vR(t) and vC(t).  Note that the time dependent current and potentials are now represented by lower case letters.  The upper case  letters are reserved for the magnitude of the current and voltage phasors.

 

The length of the current phasor is I.  In terms of I, the length of the VR and the VC phasors are IR and I Xc = I/Cω.  Therefore from Pythagoras’ theorem,

 

E02 = (I R)2 + (I/Cω)2 or I = E0/[R2 + (1/Cω)2 ]˝.

 

By looking at the phasor diagram for this circuit, notice that the emf phasor is an angle φ below the current phasor where tan φ = Xc/R = 1/Rcω.  The emf phasor is defined as, E(t) = E0 cos ωt.  With this definition, the phasors representing i(t), vR(t), and vC(t) are just,

 

i(t) = E0 cos (ωt + φ)/[R2 + (1/Cω)2 ]˝

            vR(t) = E0 R cos (ωt + φ)/[R2 + (1/Cω)2 ]˝

            vC(t) = (E0/Cω) cos (ωt + φ - π/2)/[R2 + (1/Cω)2 ]˝

 

6.         As described in the book and can be seen by examining the above equations, for small frequencies,  1/Cω >> R, vR(t) -> 0 and vC(t) -> E0.  Therefore if the output of this circuit is taken across the capacitor, the circuit filters out high frequencies while letting low frequencies pass.

 

For high frequencies, it is just the opposite.  vC(t) -> 0 and vR(t) -> E0.  Therefore if the output of this circuit is taken across the resistor, the circuit filters out low frequencies while letting high frequencies pass.

 

7.         Using the same basic ideas to analyze an RCL circuit, the following results are obtained.

 

E(t) = E0 cos ωt

            i(t) = E0 cos (ωt + φ)/[R2 + (1/Cω - L ω)2 ]˝, tan φ = (Xc - XL)/R = (1/Cω - L ω)/R

vR(t) = E0 R cos (ωt + φ)/[R2 + (1/Cω - L ω)2 ]˝

vC(t) = (E0/Cω) cos (ωt + φ - π/2)/[R2 + (1/Cω - L ω)2 ]˝

vL(t) = (E0Lω) cos (ωt + φ + π/2)/[R2 + (1/Cω - L ω)2 ]˝

Z = [R2 + (1/Cω - L ω)2 ]˝

 

Notice that i(t), the current in the circuit, is a maximum when 1/Cω - L ω =0.  The frequency at which that is true, ω = [1/LC}1/2 , is called the resonance frequency.  An RCL circuit can be “tuned” to resonate at any given frequency.  This is the basic idea behind radio and television tuners.

 

The quantity [R2 + (1/Cω - L ω)2 ]˝ acts like an effective resistance and is called the impedance of the circuit, Z.

 

8.         The instantaneous power across a component of the circuit is p(t)= i(t)v(t), the current times the potential difference across that component.

 

            pemf(t) = (E0 2/Z)cos ωt cos (ωt + φ)

pR(t) = (E0 2R/Z2)cos2(ωt + φ)

pC(t) = (E0 2/CωZ2)cos(ωt + φ )cos (ωt + φ - π/2)

pC(t) = (E0 2/CωZ2)cos(ωt + φ )cos (ωt + φ + π/2).

 

Typically, the quantity of interest is the time-averaged power defined as the integral,

 

P = ∫p(t)dt/T, where the integral is over one period of oscillation, 0 to T and T = 2π/ω.  To do the integrals, use cos(A + B) = cosAcosB - sinAsinB first to get,

 

cos (ωt + φ) = cos ωt cos (φ) - sin ωt sin (φ) where A = ωt and B= φ

cos (ωt + φ - π/2)= cos (ωt + φ) cos(π/2)+ sin(ωt + φ) sin(π/2) = + sin(ωt + φ)

cos (ωt + φ + π/2)= cos (ωt + φ) cos(π/2) - sin(ωt + φ) sin(π/2) = - sin(ωt + φ)

 

When these quantities are put back into the p(t) expressions, the integrals are of the form of cos2(ωt + φ) or cos(ωt + φ) sin(ωt + φ).  Over one cycle it is easy to show that the first integral is T/2 and the second is zero.  To do the first, use the identity cos2x = (1 + cos2x)/2.  The second integral is even easier.

 

The time averaged power is then just,

 

            Pemf = ˝ (E0 2/Z) cos φ

PR = ˝(E0 2R/Z2)

PC and PL both equal zero since they do not dissipate any power, they store and release energy periodically.

 

From those equations it appears as if the power supplied by the emf is not equal to the power dissipated by the resistor but from tan φ = (1/Cω - L ω)/R it is easy to show that cos φ = R/Z.  This substitution makes the power supplied by the emf each cycle equal to the power dissipated by the resistor.