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Theory of AC Circuits
1 Sinusoidal Waveforms
Let’s return to the simple series LCR circuit
E = IR + L
dI dt
C
∫ Idt
What now if the voltage applied to such a circuit (and similar circuits) is sinusoidal rather than a square wave.
E = E 0 cos ωt or E = E 0 sin ωt
However instead let us use instead a “complex” form
E = E 0 exp iωt = E 0 cos ωt + iE 0 sin ωt
Furthermore let us guess the solution to the differential equation as
I = I 0 exp i(ωt + φ)
The question is — what are the values of I 0 and φ which give us the solution. First we need to know
dI dt
= I 0
d dt exp i(ωt + φ) = iωI 0 exp i(ωt + φ) ∫ Idt = I 0
∫ exp i(ωt + φ)dt = I 0 exp i(ωt + φ) iω
Then if we substitute all of this into the differential equation we have
E 0 exp i(ωt) = I 0 R exp i(ωt + φ) + iωI 0 L exp i(ωt + φ) +
I 0
iωC exp i(ωt + φ)
⇒ E 0 = I 0 exp iφ
[ R + i
( ωL −
ωC
)]
= I 0 exp iφ
√ R^2 +
( ωL −
ωC
) 2 exp i arctan
( ωL − (^) ωC^1 R
)
= I 0
√ R^2 +
( ωL −
ωC
) 2 exp i
[ φ − arctan
( (^1) ωC −^ ωL R
)]
which means that the solutions we are looking for are
φ = arctan
( (^1) ωC −^ ωL R
)
I 0 =
√ E^0
R^2 +
( ωL − (^) ωC^1
) 2
If we use these values for I 0 and φ then the equation I(t) = I 0 exp i(ωt + φ) solves the differntial equation.
2 Complex Impedances
We do not have to use differential equations to “solve” circuits involving AC sinusoidal waveforms, instead we can use ordinary algebra but with complex numbers. In ordinary resistor (only) circuits we could use (the simple) Ohms law V = IR to solve circuits.
In series E = IRT otal = V 1 + V 2 = I(R 1 + R 2 ) ⇒ RT otal = R 1 + R 2
and in parallel
E = I 1 R 1 = I 2 R 2 = (I 1 + I 2 )RT otal ⇒
RT otal
R 1
R 2
For each of our elements in the LCR circuits we can have an equivalent of Ohm’s law for that element if we assume that the current in the circuit is given by I(t) = I 0 exp iωt and that the applied volatge is E(t) = E 0 exp iωt. Note I 0 is a complex number. Then for each of our elements we can generate an Ohm’s law equivalent V = IZ where Z is known as the impedance. For a resistor VR = IR ⇒ ZR = R
For an inductor VL = L dI dt = iωLI ⇒ ZL = iωL
For a capacitor
VC =
C
∫ Idt =
iωC
I ⇒ ZC =
iωC Using these complex impedances we can just use our results for resistances earlier to solve circuits rather than solve differential equations.
4 Parallel Circuits
ZT
Z 1
Z 2
Z 3
RC Circuit
E 0 = I 0 ZT
ZT
R
1 /iωC
R
R
1 + iωRC I 0 =
E 0
ZT
E 0
R
(1 + iωRC)
=
E 0
R
√ 1 + (ωRC)^2 exp i arctan(ωRC)
Note that the current through the resistor and capacitor will be different, it’s the voltage E 0 that will be the same. Hence
E 0 = IRR ⇒ IR =
E 0
R
and
E 0 = IC ZC = IC
iωC
⇒ IC =
E 0
ZC
= E 0 ıωC
Now Kirchoff’s first law still applies and as such we must have that
I 0 = IR + IC =
E 0
R
R
)
E 0
R
(1 + iωRC)
as before.
