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# alternating currents & electromagnetic waves - PowerPoint PPT Presentation

alternating currents & electromagnetic waves. PHY232 – Spring 2007 Jon Pumplin http://www.pa.msu.edu/~pumplin/PHY232 (Ppt courtesy of Remco Zegers). R. L. V. Question. At t=0, the switch is closed. After that: a) the current slowly increases from I = 0 to I = V/R

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### alternating currents & electromagnetic waves

PHY232 – Spring 2007

Jon Pumplin

http://www.pa.msu.edu/~pumplin/PHY232

(Ppt courtesy of Remco Zegers)

L

V

Question

At t=0, the switch is closed. After that:

a) the current slowly increases from I = 0 to I = V/R

b) the current slowly decreases from I = V/R to I = 0

c) the current is a constant I = V/R

I

PHY232 - Pumplin - alternating currents and electromagnetic waves

L

V

The coil opposes the flow of current due to self-inductance, so the

current cannot immediately become the maximum I=V/R. It will

slowly rise to this value (characteristic time Tau = L/R).

• At t=0, the switch is closed. After that:

a) the current slowly increases from I=0 to I=V/R

b) the current slowly decreases from I=V/R to I=0

c) the current is a constant I=V/R

I

PHY232 - Pumplin - alternating currents and electromagnetic waves

R

R

• Previously, we look at DC circuits: the voltage delivered by the source is constant, as on the left.

• Now, we look at AC circuits, in which case the source is sinusoidal. A is used in circuits to denote this.

I

I

V

V

PHY232 - Pumplin - alternating currents and electromagnetic waves

R

• The voltage over the resistor is the same as the voltage delivered by the source: VR(t) = V0 sint = V0 sin(2ft)

• The current through the resistor is: IR(t)= (V0/R)sint

• Since V(t) and I(t) have the same behavior as a function of time, they are said to be ‘in phase’.

• V0 is the maximum voltage

• V(t) is the instantaneous voltage

•  is the angular frequency; =2f f: frequency (Hz)

IR(A)

I

V0=10 V

R=2 Ohm

V(t)=V0sint

PHY232 - Pumplin - alternating currents and electromagnetic waves

• To understand energy consumption by the circuit, it doesn’t matter what the sign of the current/voltage is. We need the absolute average currents and voltages (root-mean-square values) :

• Vrms=Vmax/2

• Irms=Imax/2

• The following hold:

• Vrms=IrmsR

• Vmax=ImaxR

IR(A)

Vrms

|IR|(A) |VR|(V)

Irms

PHY232 - Pumplin - alternating currents and electromagnetic waves

• We already know for DC

P = V I = V2/R = I2 R

• For AC circuits with a single resistor:

P(t) = V(t) * I(t) = V0 I0 (sint)2

• Average power consumption:

Pave= Vrms* Irms = V2rms/R = I2rms R

where

Vrms = Vmax/2)

Irms = Imax/2

Vrms

|IR|(A) |VR|(V)

Irms

P(W)

PHY232 - Pumplin - alternating currents and electromagnetic waves

V0

=t

V

time (s)

-V0

The voltage or current as a function of time can be

described by the projection of a vector rotating with

constant angular velocity on one of the axes (x or y).

PHY232 - Pumplin - alternating currents and electromagnetic waves

C

I

I(A)

V(t)=V0sint

Vc = V0sint

Qc = CVc= C V0 sint

Ic = Qc/t =  C V0 cost

So, the current peaks ahead of the voltage:

There is a difference in phase of /2 (900).

Why? When there is not much charge on the capacitor it readily accepts more

and current easily flows. However, the E-field and potential between the plates

increase and consequently it becomes more difficult for current to flow and

the current decreases. If the potential over C is maximum, the current is zero.

PHY232 - Pumplin - alternating currents and electromagnetic waves

C

I

I(A)

V(t) = V0 sint

Note: Imax=  C V0

For a resistor we have I = V0/R so ‘1/C’ is similar to ‘R’

And we write: I=V/Xc with Xc= 1/C thecapacitive reactance

Units of Xc are Ohms. The capacitive reactance acts like a resistance

in this circuit.

PHY232 - Pumplin - alternating currents and electromagnetic waves

There is no power consumption in a purely capacitive circuit:

Energy (1/2 C V2) gets stored when the (absolute) voltage over the

capacitor is increasing, and released when it is decreasing.

Pave = 0 for a purely capacitive circuit

PHY232 - Pumplin - alternating currents and electromagnetic waves

L

I

I(A)

V(t) = V0 sint

VL= V0 sint = L I/t

I= -(V0/(L)) cost

(no proof here: you need calculus…)

the current peaks later in time than the voltage:

there is a difference in phase of /2 (900)

Why? As the potential over the inductor rises, the magnetic flux produces a

current that opposes the original current. The voltage across the inductor

peaks when the current is just beginning to rise.

PHY232 - Pumplin - alternating currents and electromagnetic waves

L

IL(A)

I

I(A)

V(t) = V0 sint

Note: Imax= V0/(L)

For a resistor we have I = V0/R so ‘L’ is similar to ‘R’

And we write: I = V/XL with XL = L theinductive reactance

Units of XL are Ohms. The inductive reactance acts as a resistance

in this circuit.

PHY232 - Pumplin - alternating currents and electromagnetic waves

There is no power consumption in a purely inductive circuit:

Energy (1/2 L I2) gets stored when the (absolute) current through the

inductor is increasing, and released when it is decreasing.

Pave = 0 for a purely inductive circuit

PHY232 - Pumplin - alternating currents and electromagnetic waves

The inductive reactance (and capacitive reactance) are like the resistance of a normal resistor, in that you can calculate the current, given the voltage, using I = V/XL (or I = V/XC ).

This works for the Maximum values, or for the RMS average values.

But I and V are “out of phase”, so the maxima occur at different times.

PHY232 - Pumplin - alternating currents and electromagnetic waves

L

C

R

• Things to keep in mind when analyzing this system:

• 1) The current in the system has the same value everywhere I = I0 sin(t-)

• 2) The voltage over all three components is equal to the source voltage at any point in time: V(t) = V0 sin(t)

I

V(t)=V0sint

PHY232 - Pumplin - alternating currents and electromagnetic waves

L

C

R

I

• For the resistor: VR = IR and VR and I are in phase

• For the capacitor: Vc = I Xc (“Vc lags I by 900”)

• For the inductor: VL= I XL (“VL leads I by 900”)

• at any instant: VL+Vc+VR=V0 sin(t). But the maximum values of VL+Vc+VR do NOT add up to V0 because they have their maxima at different times.

VR

VC

I

VL

V(t)=V0sint

PHY232 - Pumplin - alternating currents and electromagnetic waves

L

C

R

• Define X = XL-Xc = reactance of RLC circuit

• Define Z = [R2+(XL-Xc)2]= [R2+X2] = impedance of RLC cir

• Then Vtot = I Z looks like Ohms law!

I

V(t)=V0sint

PHY232 - Pumplin - alternating currents and electromagnetic waves

• If the maximum voltage over the capacitor equals the maximum voltage over the inductor, the difference in phase between the voltage over the whole circuit and the voltage over the resistor is:

• a) 00

• b)450

• c)900

• d)1800

In this case, XL

PHY232 - Pumplin - alternating currents and electromagnetic waves

• Even though the capacitor and inductor do not consume energy on the average, they affect the power consumption since the phase between current and voltage is modified.

• P = I2rms R

PHY232 - Pumplin - alternating currents and electromagnetic waves

R=250 Ohm

L=0.6 H

C=3.5 F

f=60 Hz

V0=150 V

Example

L

C

R

• questions:

• what is the angular frequency of the system?what are the inductive and capacitive reactances?

• what is the impedance, what is the phase angle 

• what is the maximum current and peak voltages over each element

• compare the algebraic sum of peak voltages with V0. Does this make sense?

• what are the instantaneous voltages and rms voltages over each element?

• what is power consumed by each element and total power consumption

I

V(t)=V0sint

PHY232 - Pumplin - alternating currents and electromagnetic waves

R=250 Ohm

L=0.6 H

C=3.5 F

f=60 Hz

V0=150 V

• a) angular frequency  of the system?

• b) Reactances?

• XC=1/C=1/(377 x 3.5x10-6)=758 Ohm

• XL= L=377x0.6=226 Ohm

• c) Impedance and phase angle

• Z=[R2+(XL-Xc)2]=[2502+(226-758)2]=588 Ohm

• d) Maximum current and maximum component voltages:

• Imax=Vmax/Z=150/588=0.255 A

• VR=ImaxR=0.255x250=63.8 V

• VC=ImaxXC=0.255x758=193 V

• VL=ImaxXL=0.255x266=57.6 V

• Sum: VR+VC+VL=314 V. This is larger than the maximum voltage delivered by the source (150 V). This makes sense because the relevant sum is not algebraic: each of the voltages are vectors with different phases.

PHY232 - Pumplin - alternating currents and electromagnetic waves

• Imax=Vmax/Z=0.255 A

• VR=ImaxR=63.8 V

• VC=ImaxXC=193 V

• VL=ImaxXL=57.6 V

• Vtot=150 V

• f) instantaneous voltages over each element (Vtot has 0 phase)?

• VR(t)=63.8sin(t+1.13) (note the phase relative to Vtot)

• VC(t)=193sin(t-0.44) phase angle : 1.13-/2=-0.44

• VL(t)=57.6sin(t+2.7) phase angle : 1.13+/2=2.7

• rms voltages over each element?

• VR,rms=63.8/2=45.1 V

• VC,rms=193/2=136 V

• VL,rms=57.6/2=40.7 V

PHY232 - Pumplin - alternating currents and electromagnetic waves

• g) power consumed by each element and total power consumed?

• PC=PL=0 no energy is consumed by the capacitor or inductor

• PR=Irms2R=(Imax/2)2R=0.2552R/2=0.2552*250/2)=8.13 W

• or: PR=Vrms2/R=(45.1)2/250=8.13 W (don’t use Vrms=V0/2!!)

• or: PR=VrmsIrmscos=(150/2)(0.255/2)cos(-64.80)=8.13 W

• total power consumed=power consumed by resistor!

PHY232 - Pumplin - alternating currents and electromagnetic waves

• Reactance of capacitor: Xc= 1/C

• Reactance of inductor: XL= L

• Current through circuit: same for all components

• ‘Ohms’ law for LRC circuit: Vtot=I Z

• Impedance: Z=[R2+(XL-Xc)2]

• phase angle between current and source voltage:

tan=(|VL|-|Vc|)/VR=(XL-Xc)/R

• Power consumed (by resistor only): P=I2rmsR=IrmsVR

P=VrmsIrmscos

• VR=ImaxR in phase with current I, out of phase by  with Vtot

• VC=ImaxXC behind by 900 relative to I (and VR)

• VL=ImaxXL ahead of 900 relative to I (and VR)

PHY232 - Pumplin - alternating currents and electromagnetic waves

• The sum of maximum voltages over the resistor, capacitor and inductor in an LRC circuit cannot be higher than the maximum voltage delivered by the source since it violates Kirchhoff’s 2nd rule (sum of voltage gains equals the sum of voltage drops).

• a) true

• b) false

The maximum voltages in each component are

not achieved at the same time!

PHY232 - Pumplin - alternating currents and electromagnetic waves

• If we chance the (angular) frequency the reactances will change since:

• Reactance of capacitor: Xc= 1/C

• Reactance of inductor: XL= L

• Consequently, the impedance Z=[R2+(XL-Xc)2] changes

• Since I=Vtot/Z, the current through the circuit changes

• If XL=XC (I.e. 1/C= L or2=1/LC), Z is minimal, I is maximum)

• = (1/LC) is the resonance angular frequency

• At the resonance frequency =0

PHY232 - Pumplin - alternating currents and electromagnetic waves

Using the same given parameters as the earlier problem,

what is the resonance frequency?

Given:

R=250 Ohm

L=0.6 H

C=3.5 F

f=60 Hz

V0=150 V

f= /2=110 Hz

PHY232 - Pumplin - alternating currents and electromagnetic waves

• An LRC circuit has R=50 Ohm, L=0.5 H and C=5x10-3 F. An AC source with Vmax=50V is used. If the resistance is replaced with one that has R=100 Ohm and the Vmax of the source is increased to 100V, the resonance frequency will:

• a) increase

• b)decrease

• c) remain the same

answer c) the resonance frequency only depends

on L and C

PHY232 - Pumplin - alternating currents and electromagnetic waves

transformers are used to convert

voltages to lower/higher levels

PHY232 - Pumplin - alternating currents and electromagnetic waves

primary circuit

with Np loops in

coil

secondary

circuit with Ns

loops in coil

Vp

Vs

iron core

If an AC current is applied to the primary circuit: Vp=-NpB/t

The magnetic flux is contained in the iron and the changing flux acts

in the secondary coil also: Vs=-NsB/t

Therefore: Vs=(Ns/Np)Vp if Ns<Np then Vs<Vp

A perfect transformer is a pure inductor (no resistance), so no power

loss: Pp=PS and VpIp=VsIs ; if Ns<Np then Vs<Vp and IS>Ip

PHY232 - Pumplin - alternating currents and electromagnetic waves

a transformer is used to bring down the high-voltage delivered

by a powerline (10 kV) to 120 V. If the primary coil has 10000

windings, a) how many are there in the secondary coil?

b) If the current in the powerline is 0.1 A, what is the maximum

current at 120 V?

• Vs=(Ns/Np)Vp or Ns=(Vs/Vp)Np = 120 windings

• VpIp=VsIs so Is=VpIp/Vs=8.33 A

PHY232 - Pumplin - alternating currents and electromagnetic waves

• Is it more economical to transmit power from the power station to homes at high voltage or low voltage?

• a) high voltage

• b) low voltage

If the voltage is high, the current is low

If the current is low, the voltage drop over the power

line (with resistance R) is low, and thus the power

dissipated in the line ([V]2/R=I2R) also low

PHY232 - Pumplin - alternating currents and electromagnetic waves

• James Maxwell formalized the basic equations governing electricity and magnetism ~1870:

• Coulomb’s law

• Magnetic force

• Ampere’s Law (electric currents make magnetic fields)

• Faraday’s law (magnetic fields make electric currents)

• Since changing fields electric fields produce magnetic fields and vice versa, he concluded:

• electricity and magnetism are two aspects of the same phenomenon. They are unified under one set of laws: the laws of electromagnetism

PHY232 - Pumplin - alternating currents and electromagnetic waves

Maxwell found that electric and magnetic waves travel

together through space with a velocity of 1/(00)

v=1/(00)=1/(4x10-7 x 8.85x10-12)=2.998x108 m/s

which is just the speed of light (c)

PHY232 - Pumplin - alternating currents and electromagnetic waves

• Consider the experiment performed by Herz (1888)

I

Herz made an RLC circuit with L=2.5 nH, C=1.0nF

The resonance frequency is = (1/LC)=6.32x108 rad/s

f= /2=100 MHz.

Recall that the wavelength of waves =v/f=c/f=3x108/100x106=3.0 m

wavelength: =v/f

PHY232 - Pumplin - alternating currents and electromagnetic waves

charges and currents vary sinusoidally in the primary and secondary circuits. The charges in the two branches also oscillate at the same frequency f

He then constructed an antenna

dipole antenna

I

PHY232 - Pumplin - alternating currents and electromagnetic waves

++++++ secondary circuits. The charges in the two branches also oscillate at the same frequency f

----------

----------

++++++

producing the electric field wave

antenna

PHY232 - Pumplin - alternating currents and electromagnetic waves

++++++ secondary circuits. The charges in the two branches also oscillate at the same frequency f

----------

I

I

I

----------

++++++

I

producing the magnetic field wave

E and B are in phase

and E=cB with

c: speed of light

The power/m2=0.5EmaxBmax/0

The energy in the wave is

shared between the

E-field and the B-field

antenna

PHY232 - Pumplin - alternating currents and electromagnetic waves

question secondary circuits. The charges in the two branches also oscillate at the same frequency f

Can a single wire connected to the + and – poles of a

DC battery act as a transmitter of electromagnetic waves?

• yes

• no

answer: no: there is no varying current and hence no