The Ups and Downs of Circuits

The End is Near! Quiz – Nov 18 th – Material since last quiz. (Induction) Exam #3 – Nov 23 rd – WEDNESDAY LAST CLASS – December 2 nd FINAL EXAM – 12/5 10:00-12:50 Room MAP 359 Grades by end of week. Hopefully Maybe.

A circular region in the xy plane is penetrated by a uniform magnetic field in the positive direction of the z axis. The field's magnitude B (in teslas) increases with time t (in seconds) according to B = at, where a is a constant. The magnitude E of the electric field set up by that increase in the magnetic field is given in the Figure as a function of the distance r from the center of the region. Find a. [0.030] T/s VG r

For the next problem, recall that i

RLRL

For the circuit of Figure 30-19, assume that = 11.0 V, R = 6.00 , and L = 5.50 H. The battery is connected at time t = 0. (a) How much energy is delivered by the battery during the first 2.00 s? [23.9] J (b) How much of this energy is stored in the magnetic field of the inductor? [7.27] J (c) How much of this energy is dissipated in the resistor? [16.7] J 66 5.5H

Let’s put an inductor and a capacitor in the SAME circuit.

At t=0, the charged capacitor is connected to the inductor. What would you expect to happen??

Current would begin to flow…. High Low High Energy Density in Capacitor Energy Flows from Capacitor to the Inductor’s Magnetic Field

Energy Flow Energy

LC Circuit High Low High

When t=0, i=0 so B=0 When t=0, voltage across the inductor = Q 0 /C

The Math Solution:

Energy

Inductor

The Capacitor

Add ‘em Up …

Add Resistance

Actual RLC:

New Feature of Circuits with L and C These circuits can produce oscillations in the currents and voltages Without a resistance, the oscillations would continue in an un-driven circuit. With resistance, the current will eventually die out.  The frequency of the oscillator is shifted slightly from its “natural frequency”  The total energy sloshing around the circuit decreases exponentially There is ALWAYS resistance in a real circuit!

Types of Current Direct Current  Create New forms of life Alternating Current  Let there be light

Alternating emf Sinusoidal DC

Sinusoidal Stuff “Angle” Phase Angle

Same Frequency with PHASE SHIFT 

Different Frequencies

Note – Power is delivered to our homes as an oscillating source (AC)

Producing AC Generator x x x x x x x x x x x x x x x x x x x x x x x

The Real World

A

The Flux:

OUTPUT WHAT IS AVERAGE VALUE OF THE EMF ??

Average value of anything: Area under the curve = area under in the average box T h

Average Value For AC:

So … Average value of current will be zero. Power is proportional to i 2 R and is ONLY dissipated in the resistor, The average value of i 2 is NOT zero because it is always POSITIVE

Average Value

RMS

Usually Written as:

Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit: E R ~

Power

More Power - Details

Resistive Circuit We apply an AC voltage to the circuit. Ohm’s Law Applies

Consider this circuit CURRENT AND VOLTAGE IN PHASE

Alternating Current Circuits  is the angular frequency (angular speed) [radians per second]. Sometimes instead of  we use the frequency f [cycles per second] Frequency  f [cycles per second, or Hertz (Hz)]  f V = V P sin (  t -  v ) I = I P sin (  t -  I ) An “AC” circuit is one in which the driving voltage and hence the current are sinusoidal in time.  vv  V(t) tt VpVp -V p

 vv  V(t) tt VpVp -V p V = VP sin (wt -  v ) Phase Term

V p and I p are the peak current and voltage. We also use the “root-mean-square” values: V rms = V p / and I rms =I p /  v and  I are called phase differences (these determine when V and I are zero). Usually we’re free to set  v =0 (but not  I ). Alternating Current Circuits V = V P sin (  t -  v ) I = I P sin (  t -  I )  vv  V(t) tt VpVp -V p V rms I/I/ I(t) t IpIp -Ip-Ip I rms

Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so V p =V rms = 170 V.

Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so V p =V rms = 170 V. This 60 Hz is the frequency f: so  =2  f=377 s -1.

Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so V p =V rms = 170 V. This 60 Hz is the frequency f: so  =2  f=377 s -1. So V(t) = 170 sin(377t +  v ). Choose  v =0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).

Resistors in AC Circuits E R ~ EMF (and also voltage across resistor): V = V P sin (  t) Hence by Ohm’s law, I=V/R: I = (V P /R) sin(  t) = I P sin(  t) (with I P =V P /R) V and I “In-phase” V tt  I 

This looks like I P =V P /R for a resistor (except for the phase change). So we call X c = 1/(  C) the Capacitive Reactance Capacitors in AC Circuits E ~ C Start from: q = C V [V=V p sin(  t)] Take derivative: dq/dt = C dV/dt So I = C dV/dt = C V P  cos (  t) I = C  V P sin (  t +  /2) The reactance is sort of like resistance in that I P =V P /X c. Also, the current leads the voltage by 90 o (phase difference). V tt   I V and I “out of phase” by 90º. I leads V by 90º.

I Leads V??? What the does that mean?? I V Current reaches it’s maximum at an earlier time than the voltage! 1 2 I = C  V P sin (  t +  /2) 

Capacitor Example E ~ C A 100 nF capacitor is connected to an AC supply of peak voltage 170V and frequency 60 Hz. What is the peak current? What is the phase of the current? Also, the current leads the voltage by 90o (phase difference).

Again this looks like I P =V P /R for a resistor (except for the phase change). So we call X L =  L the Inductive Reactance Inductors in AC Circuits L V = V P sin (  t) Loop law: V +V L = 0 where V L = -L dI/dt Hence: dI/dt = (V P /L) sin(  t). Integrate: I = - (V P / L  cos (  t) or I = [V P /(  L)] sin (  t -  /2) ~ Here the current lags the voltage by 90 o. V tt   I V and I “out of phase” by 90º. I lags V by 90º.

Phasor Diagrams VpVp IpIp  t Resistor A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current.

Phasor Diagrams VpVp IpIp  t VpVp IpIp ResistorCapacitor A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current.

Phasor Diagrams VpVp IpIp  t VpVp IpIp VpVp IpIp ResistorCapacitor Inductor A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current.

+ + i i i i i i LC Circuit time

Analyzing the L-C Circuit Total energy in the circuit: Differentiate : N o change in energy

Analyzing the L-C Circuit Total energy in the circuit: Differentiate : N o change in energy

Analyzing the L-C Circuit Total energy in the circuit: Differentiate : N o change in energy

Analyzing the L-C Circuit Total energy in the circuit: Differentiate : N o change in energy The charge sloshes back and forth with frequency  = (LC) -1/2 The charge sloshes back and forth with frequency  = (LC) -1/2