Lecture 5: AC circuits
1. Oscillations and resonance
In an LC circuit we have VL = LI˙ and VC = Q
C , and the sum of voltages round the circuit (any circuit)
Q
must be zero, giving LI˙ + C
= 0. Differentiating (as having integrals and differentials in the same equation
1
.
is messy) gives LI¨ + C1 I = 0. This is the SHM equation with angular frequency ω = √LC
Now add a resistor to the circuit. There is an extra voltage IR. (It’s still the same I in all the three
components.) So we get
1
LI¨ + RI˙ + I = 0
C
This is the damped SHM equation. Dividing by L it becomes
R
1
I =0
I¨ + I˙ +
L
LC
and we can identify ω0 =
damping and the lower the Q
√1
LC
and γ =
R
L
and hence Q =
1
R
q
L
C.
The larger the R the larger the
2. Adding a force term
Add a voltage source V (t) = V ejωt to the circuit. (j is the square root of -1 for electrical engineers.) The
current will oscillate at the same frequency - but we need to find the amplitude and the phase. I(t) = Iejωt
−ω 2 LI + jωRI + I/C = jωV
So we have
I=
jωV
jωV /L
= 2
1/C − ω 2 L + jωR
ω0 − ω 2 + jωR/L
This is the same form (resonance peak etc) as before.
3. AC circuits
Component
Quantity
Symbol
Law
Unit
Abbrev. common unit(s) Impedance
Resistor
Resistance
R
V = IR
Ohms
Ω
Ω,kΩ, MΩ
Z=R
Inductor
Inductance
L
V = L dI
Henrys
H
mH
Z = jωL
dt
Capacitor
Capacitance
C
V = Q/C Farads
F
µF, nF, pF
Z = 1/jωC
The resistor is familiar. Whan a current goes through it, there is a voltage drop, and they are related
by the resistance through Ohm’s law. The Inductor produces a voltage proportional to the rate of change
of current, V = L dI
dt , and the Capacitor a voltage proportional to the integrated current, so the current is
I = C dV
dt
In some situations you have to keep all these differentials and integrals around. But now suppose that
components have been joined together in some electrical circuit, in which there is an AC voltage with some
angular frequency ω. All the currents and voltages are going to oscillate like ejωt – with various amplitudes
and phases. Differentiating anything (wrt time) is then just a matter of multiplying by jω. Integrating
means dividing by jω.
That means we can write the differential relations involving L and C as V = jωLI and I = jωCV .
1
Generalising to the equivalent of Ohm’s law as V = IZ, we have Z = R, Z = jωL, or Z = jωC
as
appropriate.
Z is complex but don’t worry. It’s telling you not just about the numerical relation between voltage
and current in terms of volts and amps, it’s telling you about the phase relation. In a resistance, voltage
and current flow together: they are in phase. For an inductor, the voltage is greatest when the current is
changing most rapidly - which is as the current itself is zero. If the current crosses the axis, like sin(ωt), at
t = 0 then the voltage is at its maximum, like cos(ωt). The voltage leads the current (or the current lags
behind the voltage) by π2 - or 90 degrees. This chimes in with V = jωLI, remembering j = ejπ/2 . For a
capacitor, the voltage is also greatest when the current is zero because the capacitor is as charged as it’s ever
Lecture 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .AC circuits
going to be: it is fully charged which means the current is changing from positive to negative, like − sin ωt.
So the current leads the voltage (or the voltage lags the current) by π2 or 90 degrees.
1
The expressions ωL and ωC
are sometimes referred to as the reactance of the component. It does express
the equivalent resistance in Ohms of the inductor or capacitor at a particular frequency: a 10 mH inductor
has a reactance of 6.3 Ω at 100 Hz, 683 Ω at 10 KHz and 68KΩ at 1 MHz; a 1 µF capacitor has a reactance
1.6 KΩ at 100 Hz, 160 Ω at 10 KHz and 0.16 Ω at 1 MHz . But it doesn’t contain the phase information.
One has to add in the leading or lagging. Some electronics experts are happy with this leading and lagging
of phases. Non-experts can find it confusing. But the complex numbers will never let you down.
4. Adding Impedances
Adding two (or more) impedances in series is just the same as adding two resistances
Z = Z1 + Z2
as the total voltage V is V1 +V2 = I1 Z2 +I2 Z2 and the current I is the same through both, so V = I(Z1 +Z2 )
Adding two (or more) impedances in parallel is just the same as adding two resistances
1
1
1
=
+
Z
Z1
Z2
as the total current is I = I1 + I2 = V1 /Z1 + V2 /Z2 but V1 = V2 = V so I = V (1/Z1 + 1/Z2 )
It is almost always simpler to calculate this (for two, not more) as
Z=
Z1 Z2
Z1 + Z2
5. Handling complex numbers
Tip 1: Use 1/j = j/j 2 = −j
a+jb
Tip 2: Given c+jd
don’t multiply top and bottom by c − jd! Instead, express each as an amplitude and
q
2 +b2 j(Atan(b/a)−Atan(d/c))
a phase. Then divide the amplitudes and subtract the phases. It is ac2 +d
2e
6. Some Examples
1) Suppose a 1000 Hz voltage of amplitude 0.3 Volts is applied to a 10 nF capacitor. What is the
current?
1
1
3
= 2π×1000×10
First evaluate |Z| = ωC
−8 = 15.9 × 10 Ω
−5
Then use I = V /Z to get I = 1.9 × 10 = 19µA
2) A 30 mH inductor has a 1mA current flowing through it at 60 Hz. What is the voltage?
Again, first calculate |Z| = 2π × 60 × 0.03 = 11.3Ω. Then use V = IZ to get V = 0.0113V
3) A 10µF capacitor, a 10mH inductor and a 100Ω resistor are connected in series. What current flows
if a 50V, 1500 Hz voltage is applied?
Same idea as the previous ones: First find Z = 100 + 2π × 1500 × 0.010j + 1./(2π × 10−5 × 1500j) =
100 + 94.25j − 10.61j = 100
√+ 83.64j
The magnitude |Z| is 1002 + 83.642 = 130.35Ω
So the current is 50/130.35 = 0.38A
7. Series and Parallel resonance
With an L, a C and an R in series, the impedance is high if the frequency is very low (no current flows
through the capacitor) or very high (no current flows through the inductance). The frequency at which there
is the greatest current (for some fixed voltage) occurs when
√ωL = 1/ωC as the imaginary impedances cancel
and the total is just R, at this resonant frequency ω = 1/ LC
With an L and a C and maybe an R in parallel, the impedance is low at high frequencies and low
frequencies. It is maximum when ωL = 1/ωC as then 1/Z is a minimum.
√
So parallel and series LC circuits have the same ‘resonant frequency’ ω = 1/ LC, but it means the
opposite thing in the two cases.
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