A phasor is a way of representing the voltage across a component taking into account the phase difference between the voltage and the current.

As we saw in our last lecture a resistor has a voltage in phase with the current flowing through it. So if we now represent the current as vector moving in a plane we can also represent the voltage across the resistor as a vector of magnitude $V_{R}=I{R}$ which points in the same direction as the current flowing through it.

We saw in our last lecture that in a capacitor the current leads the voltage by 90^{o}.

Also we saw that the reactance $X_{C}=\frac{1}{\omega C}=\frac{1}{2\pi f C}$ depends on frequency and so the size of the voltage phasor $V_{0}=I_{0}X_{C}$ also should.

We saw in our last lecture that in an inductor the current lags the voltage by 90^{o}.

Also we saw that the reactance $X_{L}=\omega L=2\pi f L$ depends on frequency and so the size of the voltage phasor $V_{0}=I_{0}X_{L}$ also should.

To find the total voltage in the circuit at time $t$ we add the phasors for the different components together as we would vectors. We can see that for a current $I=I_{0}\cos\omega t$ flowing through through the circuit the voltage will be offset by a phase $\phi$

$V=V_{0}\cos(\omega t + \phi)$

The peak voltage $V_{0}$ is linked to the peak current $I_{0}$ through the impedance $Z$

$V_{0}=I_{0}Z$ and we can also say $V_{rms}=I_{rms}Z$

The value of $Z$ is found by considering the vector sum of the voltages

$V_{0}=\sqrt{V_{R0}^{2}+(V_{L0}-V_{C0})^{2}}=I_{0}\sqrt{R^{2}+(X_{L}-X_{C})^2}$

so

$Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}$

The phase difference between the current and voltage $\phi$ is obtained from

$\tan \phi=\frac{V_{L0}-V_{C0}}{V_{R0}}=\frac{I_{0}(X_{L}-X_{C})}{I_{0}R}=\frac{X_{L}-X_{C}}{R}$

Power in an LRC circuit is only dissipated in the resistor, and so the average power dissipated is given by

$\bar{P}=I_{RMS}^{2}R$

but we may want to express this in terms of the impedance of the circuit or the $V_{RMS}$ which is applied.

To do this we write

$\cos \phi =\frac{V_{R0}}{V_{0}}=\frac{I_{0}R}{I_{0}Z}=\frac{R}{Z}$

which means that $R=Z\cos\phi$ and

$\bar{P}=I_{RMS}^{2}Z\cos\phi$

or, as $V_{RMS}=I_{RMS}Z$

$\bar{P}=I_{RMS}V_{RMS}\cos\phi$

The RMS current in the circuit we are considering is given by

$I_{RMS}=\frac{V_{RMS}}{Z}=\frac{V_{RMS}}{\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}}$

We can see that the current should be frequency dependent and have a maximum when

$(\omega L - \frac{1}{\omega C})=0$

which gives the resonant frequency

$\omega_{0}=\sqrt{\frac{1}{LC}}$

One can get a feel for the circuit response here and here.

Here is a demo that shows how the resonant frequency is varied by the inductor and capacitor values.

The phasor approach gave us a nice visual picture of how a simple combination of circuit elements should respond to an AC field.

But a graphical addition approach would rapidly become unmanageable when more elements both in series and in parallel are added.

We deal with this by representing our current, voltage and the the impedance of circuit elements in terms of complex numbers.

Note, that we will use the math and physics symbol for $i=\sqrt{-1}$ electrical engineers frequently use $j$ instead, for example on this page from hyperphysics.

The starting point is Euler's formula.

$e^{i\omega t}=\cos\omega t+i\sin\omega t$

Now suppose we want to represent a voltage

$V=V_{0}\cos\omega t$

and a current which is offset by a phase $\phi$

$I=I_{0}\cos(\omega t - \phi)$

This can be done by instead writing

$V=V_{0}e^{i\omega t}$ and $I=I_{0}e^{i(\omega t - \phi)}$

The advantage of doing this is that we can now express the impedance of a circuit element as $Z=\frac{V_0}{I_{0}}e^{i\phi}=|Z|e^{i\phi}$.

$$Z=\frac{V_0}{I_{0}}e^{i\phi}=|Z|e^{i\phi}$$

and

$$V=IZ$$

In this scheme the impedance of a resistor, capacitor and and inductor become

$Z_{R}=R$

$Z_{C}=-\frac{i}{\omega C}$

$Z_{L}=i\omega L$

When connected in series complex impedances simply add together, the complex math is taking care of the vector addition we had do to previously

$Z_{S}=Z_{1}+Z_{2}+Z_{3}+..$

When connected in parallel, impedances, like resistors, add as reciprocals

$\frac{1}{Z_{P}}=\frac{1}{Z_{1}} +\frac{1}{Z_{2}}+\frac{1}{Z_{3}}+..$

$Z_{T}=R+i\omega L-\frac{i}{\omega C}$

$Z_{T}=|Z_{T}|e^{i\phi}$

$|Z_{T}|=\sqrt{Z\bar{Z}}$

$\phi=\tan^{-1}\frac{(\omega L-\frac{1}{\omega C})}{R}$

$V_{0}=I_{0}|Z_{T}|=\sqrt{(R+i(\omega L-\frac{1}{\omega C}))(R-i(\omega L-\frac{1}{\omega C}))}=I_{0}\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}$

$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$

Gauss's Law for electric fields and charge tells us that a charge within a surface acts as either a source or sink of electric flux through the surface.

As there are no magnetic monopoles the equivalent of Gauss's Law for magnetism is

$\oint\vec{B}\cdot d\vec{A}=0$

which means that for any closed surface the total magnetic flux is zero, an equal number of field lines must enter and leave the volume.

So far we have considered Ampere's Law as he originally formulated it

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I$

might seem somehow asymmetric when we consider Faraday's Law.

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$ (Faraday's Law)

which says that a changing magnetic field can produce an electric field.

In the original form of Ampere's Law a changing electric field does not produce a magnetic field.

Maxwell corrected this asymmetry by adding a term to Ampere's law which reflects the fact that a time varying electric field actually can produce a magnetic field.

The need for a term in Ampere's Law reflecting changing electric fields can be demonstrated by considering the environment of a charging capacitor.

If we consider the top surface $S_{1}$ then $\oint \vec{B}\cdot d\vec{l}=\mu_{0}I$ and $B=\frac{\mu_{0}I}{2\pi r}$

If we instead consider the bottom surface there is no current flowing through it so $B=0$. Measurement would show us that the first result is true.

We can reconcile these results by including a term involving the changing electric flux through the bottom surface.

This flux can be linked to the current flowing through the wire.

The charge on the capacitor

$Q=CV=(\varepsilon_{0}\frac{A}{d})Ed=\varepsilon_{0}AE$

$\frac{dQ}{dt}=\varepsilon_{0}A\frac{dE}{dt}=\varepsilon_{0}\frac{d\Phi_{E}}{dt}=I$

We can thus put the current and the changing electric flux on an equal footing by writing

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I+\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$

Ampere's Law with Maxwell's correction

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I+\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$

can be written as

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}(I+I_{D})$

Where the displacement current $I_{D}$

$I_{D}=\varepsilon_{0}\frac{d\Phi_{E}}{dt}$

Maxwell's equations, named after James Clerk Maxwell who first expressed them together are a set of four equations from which all electromagnetic theory can be derived.

The integral form of Maxwell's equations in free space (ie., in the absence of dielectric or magnetic materials) are

$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$ (Gauss's Law)

$\oint \vec{B}\cdot d\vec{A}=0$ (Magnetic equivalent of Gauss's Law)

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$ (Faraday's Law)

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I+\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$ (Modified form of Ampere's Law).