LC

R

Today...• RLC circuits with a sinusoidal drive

• Phase difference between current & voltage for Resistors, Capacitors, and Inductors.

• Reactance

• Phasors

• Application to frequency filters (high-, low-pass)

Driven LRC circuits• Last time we discovered that an LC

circuit was a natural oscillator:LC

+ +

- -

R

• However, the resistance of any real inductor will cause oscillations to damp out, unless we can supply energy at the rate the resistor dissipates it! How? A sinusoidally varying emf (AC generator) will sustain sinusoidal current oscillations!

1resonance in absence of resistive loss

LC

• Very useful → TV, radio, computer clocks, …

osintGeneric problem: we are given an “ac voltage source” “driving” a circuit

We need to find the current that flows: I(t)=Iosin(t-) It is also sinusoidal at the same frequency, possibly with some “phase” angle relative to the voltage source

Preview• Our goal is to understand how an AC LRC circuit works.

• Physical picture of each object:

– Source: produces an oscillating voltage (supplies whatever

current the rest of the circuit “requires”)

– Resistor: causes a voltage drop when a current flows

through. As soon as the voltage changes, so does the

current  always in phase.

– Capacitor: resists change in charge Q resists change in

voltage . voltage across capacitor lags behind (90˚) the

current (charge leaving & entering the plates).

– Inductor: resists change in magnetic flux resists change

in current. current always lags voltage (90˚).

QV

C

~

AC CircuitsSeries LRC• Statement of problem:

Given = msint, find I(t).

Everything else will follow.

• We could solve this equation with tons of algebra involving sin(t) and cos(t); or with simple complex algebra.

• We will do neither, but start by considering simple circuits with one element (R, C, or L) in addition to the driving emf.

LC

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• Procedure: start with loop equation

2

2sinm

d Q dQ QL R tdt dt C

R Circuit

• R only: Loop eqn gives:

IR

R

sinR R mV RI t

Note: this is always, always, true in LRC circuit, always…

0

0t

IRRm

Rm

0

0

VR

t

m

m

Voltage across R in phase with current through R

s inmRI t

R

But voltage across R is not always in phase with source!

C Circuit• Now consider C only: Loop eqn gives:

C IC

Voltage across C lags current through C by one-quarter cycle (90).

tC

QV mC sin tCQ m sin

cosC m

dQI C t

dt

Is this always true?

YES!

0

0 t

m

m

VC

t

0

0

mC

mC

IC

L Circuit• Now consider L: Loop eqn gives:

Voltage across L leads current through L by one-quarter cycle (90).

IL

LsinLL m

dIV L t

dt sinm

LdI tdtL

cosmL LI dI t

L

Yes, yes, but how to remember?

2/sin

tLm

t

0

0

m

m

VL

t

0

0

Lm

Lm

IL

VL leads IL VC lags IC

Hi kids, I’m Eli and I’ll help you learn

physics !

…we’ll see about that

Introducing...

Summary thus far…• LRC Circuit

– Given:– Assume solution for current: I(t) = Im sin( t LC

R

sin( )R mV RI t 1

cos( )C mV I tC

cos( )L mV LI t

sinm t

– Note that in all cases , though there may be a phase shift:V I

R mV I R 1C mV I

C L mV I L

XCXL

reactance

What is reactance?You can think of it as a frequency-dependent resistance.

For high ω, χC~0

- Capacitor looks like a wire (“short”)

For low ω, χC∞

- Capacitor looks like a break

1CX C

For low ω, χL~0

- Inductor looks like a wire (“short”)

For high ω, χL∞

- Inductor looks like a break(inductors resist change in current)

LX L

( " " )RX R

Filter Example #1

~

Consider the AC circuit shown.For very high frequencies, is Vout big or small?

Recall: capacitor resists change in voltage.High frequency more change smaller reactance smaller VC

1CX C

, Vout pulled to ground

0CX

Low frequency capacitor has time to charge up larger VC

, no current flows no voltage drop across R Vout ~ ε

0 CX

outV

0

What is ω0?Use dimensional analysis.

RC 0

2 2

time RC

So, this is a circuit that only passes low frequencies: “low-pass” filter Bass knob on radio

εVout

1

Lecture 19, Act 1• A driven RC circuit is connected as

shown.

– For what frequencies of the voltage source is the current through the resistor largest?

(a) small (b) large (c)LC

1

C

R

More Filters

~Vout

a.

b.

c.

ω=0 No currentVout ≈ 0

ω=∞ Capacitor ~ wireVout ≈ ε

~Vout

ω = ∞ No currentVout ≈ 0

ω = 0 Inductor ~ wireVout ≈ ε

ω = 0 No current because of capacitor

ω = ∞ No current because of inductor

outV

0

outV

0

outV

0

(Conceptual sketch only)

High-pass filter

Low-pass filter

Band-pass filter

~ Vout

AC Circuits, Quantitative• Our goal is to calculate the voltages across the various elements, and also the current flowing through the circuit.

• If these were just three resistors inseries, we could calculate the net resistance simply by adding the individual resistances.

• Because current and voltage are 90out of phase for the capacitor and the inductor, a straight sum does not work (unless you use complex numbers, which we don’t in P212).

• Instead we use “phasors”, a geometric way to visualize an oscillating function (avoiding nasty trig. or complex numbers).

• A phasor is a rotating “vector” whose magnitude is the maximum value of a quantity (e.g., V or I).

–The instaneous value is the projection on the y-axis.–The phasor rotates at the drive frequency . (Appendix)

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“Beam me up Scotty –

It ate my phasor!”

Phasors

Problem: Given Vdrive = εm sin(ωt),find VR, VL, VC, iR, iL, iC

Strategy:

We will use Kirchhoff’s voltage law that the (phasor)sum of the voltages VR, VC, and VL must equal Vdrive.

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Phasors, cont.

1. Draw VR phasor along x-axis (this direction is chosen for convenience). Note that since VR = iRR, this is also the direction of the

current phasor iR. Because of Kirchhoff’s current law, iL = iC = iR ≡ i (i.e., the same current flows through each element).

VR, iRR

LC

RProblem: Given Vdrive = εm sin(ωt),

find VR, VL, VC, iR, iL, iC

Phasors, cont.

LC

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Problem: Given Vdrive = εm sin(ωt),find VR, VL, VC, iR, iL, iC

2. Next draw the phasor for VL. Since the inductor current iL always lags VL draw VL 90˚ further counterclockwise. The

length of the VL phasor is iL XL = i L

VR = i R

VL = i XL

Phasors, cont.

3. The capacitor voltage VC always lags iC draw VC 90˚ further clockwise.

The length of the VC phasor is iC XC = i /C

VR = i R

VL = i XL

VC = i XC

The lengths of the phasors depend on R, L, C, and ω. The relative orientation of the VR, VL, and VC phasors is always the way we have drawn it. Memorize it!

Problem: Given Vdrive = εm sin(ωt),find VR, VL, VC, iR, iL, iC

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Phasors, cont.

• The phasors for VR, VL, and VC are added like vectors to give the drive voltage VR + VL + VC = εm :

2

• From this diagram we can now easily calculate quantities of interest, like the net current i , the maximum voltage across any of the elements, and the phase between the current the drive voltage (and thus the power).

VR

VL

VC

εm

Problem: Given Vdrive = εm sin(ωt),find VR, VL, VC, iR, iL, iC

LC

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A series RC circuit is driven by emf =m sint. Which of the following could be an appropriate phasor diagram?

Lecture 19, Act 2

(a) (c)(b)

VR

VL

VC

εm

VR

VC

εm

~

2A

For this circuit which of the following is true?

(a) The drive voltage is in phase with the current.

(b) The drive voltage lags the current.

(c) The drive voltage leads the current.

2B

VR

εmVC

RC Circuit, quantitative:

2 22( )C miR iX

22 2 2( )C mi R X

2 2

m

C

iR X

VR = iR

εm

VC = iXC

22R m

C

RV iR

R X

02 22 1

1

1

R

mC

V R

R

~

m sint

C

R

RC Circuit, cont.

02 22 1

1

1

out

C

V R

R

0

1

RC

Ex.: C = 1 μF, R = 1Ω

High-pass filter

High-pass filter

0

0.2

0.4

0.6

0.8

1

0.E+00 1.E+06 2.E+06 3.E+06 4.E+06 5.E+06 6.E+06

(Angular) frequency, omega

"tra

nsm

issi

on"

Note: this is ω,2

f

Vout

~

C

R

LRC Circuits, quantitative

LC

R

The phasor diagram gives us graphical solutions for and Im:

LX L

where . . .

CXC

1

R

XX CL tan

2222CLmm XXRI

ZXXRI m

CL

mm

22

22CL XXRZ

ImR

Im XL

Im XC

εm

εm

ImR

Im XL Im XC

)sin( tII m

Summary

• Reactances ~ frequency-dependent resistance– Capacitors

» look like wires for high frequencies

» look like breaks for low frequencies

» Voltage lags current by 90˚

– Inductors

» look like breaks for high frequencies

» look like wires for low frequencies

» Current lags voltage by 90˚

– Filters (low-pass, high-pass, band-pass)

• LRC Circuit– Apply KVL using phasors

CXC

1

LX L

ImR

Im XL

Im XC

εm

R

XX CL tan ZXXR

I m

CL

mm

22

)sin( tII m

Ex. Source = y-component of the V-phasor

sin( )mV t

Appendix: Phasors• A phasor is a “vector” whose magnitude is the maximum value

of a quantity (e.g., V) and which rotates counterclockwise in a 2-d plane with angular velocity . Recall uniform circular motion:

The projections of r (on the vertical y axis) execute sinusoidal oscillation.

trx costry sin

x

y y

1

2

4

ωt=0

V=0

ωt=90˚V=εm

ωt=270˚

V=-εm

ωt=45˚

2mV

3

When the current through the circuit is maximum, what is the potential difference across the inductor?

a) VL = 0

b) VL = VLmax/2

c) VL = VLmax

When the capacitor is fully charged, what is the magnitude of the voltage across the inductor?

a) VL = 0 b) VL = VLmax/2 c) VL = VLmax

The phasor picture corresponds to a snapshot at some time t. The projections of the phasors along the vertical axis are the actual values of the voltages at the given time. One can draw the phasors at different times, simply by rotating the entire diagram. With this understanding, other questions can be easily answered…

Explanation:

Since the current and VL are 90 degrees out of phase, when the current is at a maximum, VL will be at 0. When the capacitor is fully charged, the current through the circuit will be 0, and the magnitude of L will be at a maximum.

VL will actually be at a minimum because VC and VL are 180 degrees out of phase.

Current = max

VR

VL

VC

VR

VL

VC

Charge on C = max