Physics 1202: Lecture 16Today’s Agenda

• Announcements:– Lectures posted on:

www.phys.uconn.edu/~rcote/

– HW assignments, etc.

• Homework #5:Homework #5:– Due next FridayDue next Friday

• Midterm 1:– Answers today

– New average = 63%

LC

R

R Circuit• We begin by considering simple circuits with one element

(R,C, or L) in addition to the driving emf.

• Begin with R: Loop eqn gives:

Voltage across R in phase with current through R

iR

R

Note: this is always, always, true… always.

0 tx

m

m

0

0 t

m / R

m / R

0

RMS Values• Average values for I,V are not that helpful (they are zero).

• Thus we introduce the idea of the Root of the Mean Squared.

• In general,

So Average Power is,

C Circuit (… calculus !)• Now consider C: Loop eqn gives:

C

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

Is this always true?

YES

0 tx

m

m

0

t

0

0

Cm

Cm

Lecture 16, ACT 1• A circuit consisting of capacitor C and voltage

source is constructed as shown. The graph shows the voltage presented to the capacitor as a function of time. – Which of the following graphs best represents

the time dependence of the current i in the circuit?

(a) (b) (c)i

t

i

t t

i

t

L Circuit (… calculus !)• Now consider L: Loop eqn gives:

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

L

Yes, yes, but how to remember?

0 tx

m

m

0

tx

m L

m L0

0

Phasors

• A phasor is a vector whose magnitude is the maximum value of a quantity (eg V or I) 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.

• R: V in phase with i

• C: V lags i by 90

• L: V leads i by 90

x

y y

0

i

0

i

Phasors for L,C,Ri

t

i

t

i

t

Suppose:

0

i

• A series LCR circuit driven by emf = 0sint produces a current i=imsin(t-). The phasor diagram for the current at t=0 is shown to the right.– At which of the following times is VC, the

magnitude of the voltage across the capacitor, a maximum?

Lecture 16, ACT 2

i

t=0

(a) (b) (c)i

t=0

i

t=tb

i

t=tc

Series LCR

AC Circuit• Consider the circuit shown here: the loop equation gives:

• Here all unknowns, (im,) , must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be: m sint.

• To solve this problem graphically, first write down expressions for the voltages across R,C, and L and then plot the appropriate phasor diagram.

LC

R

• Assume a solution of the form:

C= -Q/CL= -L I / t

R= -RI

Phasors: LCR

• Assume:

• From these equations, we can draw the phasor diagram to the right.

LC

R

• This picture corresponds to a snapshot at t=0. The projections of these phasors along the vertical axis are the actual values of the voltages at the given time.

• Given:

im

Phasors: LCR

• The phasor diagram has been relabeled in terms of the reactances defined from:

LC

R

The unknowns (im,) can now be solved for graphically since the vector sum of the voltages VL + VC + VR must sum to the driving emf.

Phasors:LCR

Phasors:Tips• This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis.

y

x

imR

imXL

imXC

m

“Full Phasor Diagram”

From this diagram, we can also create a triangle which allows us to calculate the impedance Z:

• Sometimes, in working problems, it is easier to draw the diagram at a time when the current is along the x-axis (when i=0).

“ Impedance Triangle”

Z

|

R

| XL-XC |

Resonance• For fixed R,C,L the current im will be a maximum at the

resonant frequency 0 which makes the impedance Z purely resistive.

the frequency at which this condition is obtained is given from:

• Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself!

• At this frequency, the current and the driving voltage are in phase!

ie:

reaches a maximum when: XL=XC

ResonanceThe current in an LCR circuit depends on the values

of the elements and on the driving frequency through the relation

Suppose you plot the current versus , the source voltage frequency, you would get:

“ Impedance Triangle”

Z

|

R

| XL-XC |

1 2x

im

00

o

R=Ro

m / R0

R=2Ro

Power in LCR Circuit• The power supplied by the emf in a series LCR circuit

depends on the frequency . It will turn out that the maximum power is supplied at the resonant frequency 0.

• The instantaneous power (for some frequency, ) delivered at time t is given by:

• The most useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle.

• To evaluate the average on the right, we first expand the sin(t-) term.

Remember what this stands for

Power in LCR Circuit• Expanding,

• Taking the averages,

• Generally:

sin2t

t0

0

+1

-1

• Putting it all back together again,

01/2

(Integral of Product of even and odd function = 0)sintcost

t0

0

+1

-1

Power in LCR Circuit• The power can be expressed in term of i max:

• Power delivered depends on the phase, the“power factor”

• phase depends on the values of L, C, R, and

• This result is often rewritten in terms of rms values:

Fields from Circuits?• We have been focusing on what happens within the circuits we have been

studying (eg currents, voltages, etc.)

• What’s happening outside the circuits??– We know that:

» charges create electric fields and » moving charges (currents) create magnetic fields.

– Can we detect these fields?– Demos:

» We saw a bulb connected to a loop glow when the loop came near a solenoidal magnet.

» Light spreads out and makes interference patterns.Do we understand this?