# Lecture 25 - AC Circuits II

## LRC Circuit

A real LC circuit always has some resistance and is therefore better represented by an LRC circuit.

We first consider the analogy of the damped spring.

$$F=-kx-bv$$

$$m\frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=0$$

Underdamped: $b^2\ll 4km$

Critical damping: $b^2=4km$

Overdamped: $b^2\gg 4km$

## LRC Circuit Analysis

Now Kirchoff's loop rule when the circuit is closed is

$-L\frac{dI}{dt}-IR+\frac{Q}{C}=0$

Which can be written in terms of the charge as before

$L\frac{d^{2}Q}{dt^{2}}+R\frac{dQ}{dt}+\frac{Q}{C}=0$

and we can compare this to the differential equation for a damped harmonic oscillator.

When $R^{2}<\frac{4L}{C}$ the system is underdamped and the solution to this equation is

$Q=Q_{0}e^{-\frac{R}{2L}t}\cos(\omega' t+\phi)$

where $\omega'=\sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}}$

When $R^{2}>\frac{4L}{C}$ the system is overdamped and the charge will decay slowly.

## Resistor and AC Source

If we apply a sinusoidal current to a resistor

$I=I_{0}\cos (2\pi ft)=I_{0}\cos (\omega t)$

From Ohm's Law $V=IR$ we get

$V=I_{0}R\cos (\omega t)=V_{0}\cos (\omega t)$

and we can see that the current and voltage in a resistor are in phase with each other.

## Inductor and AC Source

For an inductor directly connected to an AC source $I=I_{0}\cos (2\pi ft)=I_{0}\cos (\omega t)$

$$V-L\frac{dI}{dt}=0$$

$$V=L\frac{dI}{dt}=-\omega LI_{0}\sin (\omega t)=\omega LI_{0}\cos (\omega t+90^{0})=V_{0}\cos (\omega t+90^{o})$$

and we can see that the current lags (is behind) the voltage by $90^{o}$.

As $V_{0}=\omega LI_{0}$ we define the inductive reactance as

$$X_{L}=\omega L=2\pi f L$$

so that we can write

$$V_{0}=I_{0}X_{L}$$

## Capacitor and AC Source

For a capacitor

$$V=\frac{Q}{C}$$

and when $I=I_{0}\cos (2\pi ft)=I_{0}\cos (\omega t)$

$$I=\frac{dQ}{dt}=I_{0}\cos(\omega t)$$

$$Q=\int_{0}^{t}dQ=\int_{0}^{t}I_{0}\cos\omega t\,dt=\frac{I_{0}}{\omega}\sin\omega t$$

So the voltage is

$$V=\frac{Q}{C}=I_{0}\frac{1}{\omega C}\sin\omega t=I_{0}\frac{1}{\omega C}\cos(\omega t-90^{o})=V_{0}\cos(\omega t-90^{o})$$

As we did with inductance

As $V_{0}=I_{0}(\frac{1}{\omega C})$ we define the capacitive reactance as

$$X_{C}=\frac{1}{\omega C}=\frac{1}{2\pi f C}$$

so that we can write

$$V_{0}=I_{0}X_{C}$$

## A Common Application

Capacitors “impede” low frequencies and provide little resistance to high frequencies.

Inductors “impede” high frequencies and provide little resistance to low frequencies.