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Year 2 Electromagnetism > AC Circuits > Flashcards

Flashcards in AC Circuits Deck (26):
1

What is the relationship between E and B in an electromagnetic wave?

E = Bc

2

What is the energy density for a magnetic field in an electromagnetic wave?

um = 1/2 1/μo B²

3

What is the energy density for an electric field in an electromagnetic wave?

ue = 1/2 εo E²

4

What is the speed of an electromagnetic wave?

c = √[1/μoεo]

5

What is the relationship between energy density in the electric and magnetic fields that make up an electromagnetic wave?

ue = 1/2 εo E² = 1/2 εo (Bc)² = 1/2 εo c²B² =
1/2 εo (1/√[μoεo])² B² = 1/2 1/μo B² = um
-so the energy densities are equal

6

Wave Intensity for an Electromagnetic Waves
Energy

-consider a power source emitting waves
-lets consider a part of a wave front from this source, this part has surface area A and is travelling at speed v
-the wave front is initially at position 1 and then a small time Δt later it is at position 2
-in the time interval Δt, the wave front has moved a distance v Δt
-the small amount of volume traversed by the wavefront in going from position 1 to 2 is Av Δt
-the wave has brought energy to this volume, if the wave is electromagnetic then in this volume there will now be oscillating electric and magnetic fields that weren't there before
-the amount of energy in this small volume is:
ΔE = uav Av Δt
-where uav is the average energy density of the wave within that volume

7

Wave Intensity for an Electromagnetic Waves
Power

-we have that the energy in a small volume traversed by part of a wavefront of area A is given by:
ΔE = uav Av Δt
-the power entering this volume is given by:
P = ΔE/ Δt = uav Av Δt / Δt = uav A v

8

Wave Intensity for an Electromagnetic Waves
Intensity - General Form

-we have that the power entering a volume traversed by an area A of a wavefront in a small time Δt is given by:
P = uav A v
-the intensity of a wave is by definition the amount of power per area:
I = P/A = uav A v / A = uav v

9

Wave Intensity for Electromagnetic Waves
Intensity - Electromagnetic Waves

I = uav * v
-this is a general result for a wave, for an electromagnetic wave specifically, v=c and total energy density u is given by u = ue + um
-but we know that ue = um, so u = 2um = 2(1/2 1/μo B²)
u = 1/μo B²
-|B fluctuates with time, and we need the average energy density uav
uav = \B² /μo , where \B² indicates the mean value of B²
-we can then write this in terms of the root mean squared value
uav = (Brms * Brms) / μo
-rewrite in terms of B and E
uav = (Erms/c * Brms) / μo
-sub in for intensity
I = uav v = (Erms/c * Brms) / μo * c = Erms*Brms/ μo

10

Wave Intensity for Electromagnetic Waves
Poynting Vector

-for the intensity of an electromagnetic wave, we have:
I = Erms*Brms/ μo
-recall the Poynting vector:
|S = |Ex|B /μo , where |S is in the direction of propagation of the wave
-find the magnitude of |S
S = | |S | = | |Ex|B | / μo = |E||B|sin90/μo = EB/μo
-we can write this only in terms of B:
S = (Bc)B / μo = B²/μo * c
-since S depends on B and B fluctuates with time, S also fluctuates with time, calculate the average value of S, Sav:
Sav = \S = \B²/μo * c = uav * c = I
-therefore the average magnitude of |S is equal to the intensity of the wave

11

Oscillating Voltage
Notation

-in AC circuit theory complex numbers are used to express the voltages and currents across and through components in a circuit:
-an oscillating voltage such as
V = Vo cos(ωt)
-this can be written as the real part of a complex number;
V = Re{ Vo exp(j ωt) }
-where j²=-1 to prevent confusion of i with I (current)
-we can then write voltages and currents as complex numbers:
~V = Vo exp (jωt) and ~I = Io exp[j (ωt-𝛿)]
-where Vo and Io are real numbers, ω is angular frequency and 𝛿 is a phase difference
-the measured voltages and currents are then determined from V = Re{~V} and I = Re{~I}

12

Dealing With Components in AC Circuits

-in an AC circuit the direction of current is constantly alternating
-imagine taking a snapshot of the circuit so that the direction of current at that exact moment is constant
-apply Kirchhoff's loop rule
-substitute in complex forms of ε and I
-cancel ωt terms
-remove exponentials using Euler's formula:
exp(jθ) = cosθ + jsinθ
-separate real and imaginary parts
-solve for 𝛿 using the imaginary part
-sub this value for 𝛿 into the real part

13

Inductor in an AC Circuit

-applying Kirchhoff's loop rule, where εL is the emf across the inductor:
~ε - ~εL = 0
-recall that the emf across an inductor is given by εL=L*d~I/dt and sub in:
~ε = L * d~I/dt
-sub in complex forms of ε and I
εo exp(j ωt) = L * d(Io exp(j(ωt - 𝛿))/dt
εo exp(j ωt) = L Io jω * exp(j(ωt) - 𝛿))
-cancel the ωt terms:
εo = L Io jω * exp(-j𝛿)
-remove exponential using Euler's formula:
εo = L Io jω [cos(-𝛿) + jsin(-𝛿)]
εo = L Io jω [cos(𝛿) - jsin(𝛿)]
-move j into bracket:
εo = L Io ω [jcos(𝛿) + sin(𝛿)]
-equate real and imaginary parts to get two equations:
εo = L Io ω sin(𝛿) and 0 = - L Io ω cos(𝛿)
-solve the imaginary part to find 𝛿
𝛿 = π/2 or 3π/2
-but if 𝛿 = 3π/2, then sin(𝛿) = -1 which doesn't make sense since εo, L, Io, and ω are all positive
-therefore 𝛿 = π/2, sub in to the real equation:
εo = L Io ω sin(π/2)
εo = L Io ω

14

Complex Impedance
Inductor

~V = ~I ~Z :
-for an inductor we have εo = L Io ω and 𝛿=π/2
εo exp(j ωt) = εo/Lω * exp(j(ωt - π/2)) * Zo * exp(j𝛿)
-divide by εo exp(j ωt) :
1 = 1/Lω * exp(-j * π/2) * Zo * exp(j𝛿)
-rearrange for ~Z :
~Z = Zo exp(j𝛿) = ωL exp(j * π/2)
-thus, Zo = ωL and 𝛿 = π/2
-sub in Euler's Formula:
~Z = Zo exp(j𝛿) = ωL [cos(π/2) + jsin(π/2)]
~Z = ωLj

15

Capacitor in an AC Circuit

-applying Kirchhoff's loop rule, where Vc is the voltage across the capacitor:
~ε - ~Vc = 0
-recall that the voltage across a capacitor is given by V=Q/C and sub in:
~ε = ~Q/C
-since ~I is complex, ~Q is too
~Q = C ~ε
-sub in for ~I :
~I = d~Q/dt = d(C~ε)/dt = C * d~ε/dt
-sub in complex form of ε
~I =C * d(εo exp(j ωt))/dt
-differentiate
~I = C * εo j ω * exp(j ωt)
-replace j = exp(j * π/2) from Euler's Formula:
~I = C εo * exp(j * π/2) * ω * exp(j ωt)
~I = C εo ω * exp(j(ωt + π/2)
-equating this to the general form of ~I
~I = Io exp(j(ωt - 𝛿)) = C εo ω * exp(j(ωt + π/2)
-thus 𝛿 = -π/2 , and:
Io = C εo ω

16

Complex Impedance
Capacitor

-the definition of complex impedance:
~V = ~I ~Z
-for a capacitor, Io exp(j(ωt - 𝛿)) = Cεoω*exp(j(ωt + π/2) , so
εo exp(j ωt) = Cεoω*exp(j(ωt + π/2) * Zo * exp(j𝛿)
-divide by εo exp(j ωt) :
1 = Cω exp(j * π/2) * Zo * exp(j𝛿)
-rearrange for ~Z :
~Z = Zo * exp(j𝛿) = 1/ωC * exp(-j * π/2)
-thus Zo = 1/ωC and 𝛿 = - π/2
-sub in Euler's Formula:
~Zc = 1/ωC [cos(-π/2) + jsin(-π/2)]
~Zc = - j/ωC

17

Capacitive Reactance

-the complex impedance is given by:
~Zc = - j/ωC
-the capacitive reactance is given by Xc = 1/ωC

18

Inductive Reactance

-for an inductor we have a complex impedance:
~Z = ωLj
-the inductive reactance is XL = ωL

19

Resistor in an AC Circuit

-applying Kirchhoff's loop rule, where Vr is the voltage across the resisitor:
~ε - ~Vr = 0
-recall that the voltage across a resistor is given by Ohm's Law: V = IR and sub in:
~ε = ~I R
-sub in complex forms of ε and I
εo exp(j ωt) = RIo exp(j(ωt - 𝛿)
-cancel the ωt terms:
εo = R Io * exp(-j𝛿)
-remove exponential using Euler's formula:
εo = R Io [cos(-𝛿) + jsin(-𝛿)]
εo = R Io [cos(𝛿) - jsin(𝛿)]
-equate real and imaginary parts to get two equations:
εo = R Io cos(𝛿) and 0 = - R Io sin(𝛿)
-solve the imaginary part to find 𝛿
𝛿 = π or 0
-but if 𝛿 = π, then cos(𝛿) = -1 which doesn't make sense since εo, R, and Io are all positive
-therefore 𝛿 = 0, sub in to the real equation:
εo = R Io cos(0)
εo = R Io

20

Complex Impedance
Definition

~V = ~I ~Z
-where ~Z = Zo exp(j𝛿)

21

Complex Impedance
Resistor

-for a resistor, V=IR so the complex impedance given by:
~Vr = ~I ~Zr
-is just ~Zr = R

22

Complex Voltage Notation

~V = Vo exp(jωt)

23

Complex Current Notation

~I = Io exp(j(ωt - 𝛿))

24

Complex Impedance in Series

~Zt = ~Z1 + ~Z2 + ~Z3 + ..

25

Complex Impedance in Parallel

1/~Zt = 1/~Z1 + 1/~Z2 + 1/~Z3 + ...

26

Electric Resonance Circuit

-AC generator connected in series to a resistor, inductor and capacitor
-has a natural or resonance frequency:
ωo = 1/√(LC)
-power dissipated in the resistor is maximum when the AC generator oscillates at the natural frequency
-resonance circuits are used in radio receivers where the resonance frequency is varied by changing the capacitance in order to tune the radio