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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.



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


Which can be written in terms of the charge as before


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.

RLC Circuit Summary

Pictorial RLC Circuit Summary

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


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

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



Capacitor and AC Source

For a capacitor


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

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

leads us to

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

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



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.

figure_30_18.jpg $

LRC series circuit

We can begin our analysis of this circuit by applying Kirchoff's loop rule to the find potential at any given time.


However an important consequence of the voltages not being at the same phase in the different components is that the peak voltage is not experienced at the same time in each component, and so the peak source voltage $V_{0}$

$V_{0}\neq V_{R0}+V_{L0}+V_{C0}$

The condition of continuity of current demands that the current throughout the circuit should always be in phase, so at any point in this circuit the current will

$I=I_{0}\cos\omega t$


To understand the Phasor approach to AC circuits we can follow a very nice set of animations from Physclips.

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

Phasor for a resistor

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.

Phasor for a capacitor

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

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.

Phasor for an inductor

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

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.


Phasor approach to LRC series circuit

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



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

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

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


phy142kk/lectures/26.txt · Last modified: 2015/04/01 09:26 by kkumar
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