Physics for Scientists & Engineers 2

Wednesday, March 26, 2014 RC Circuits . Going around the circuit in a counterclockwise direction we can write

. We can rewrite this equation remembering that i = dq/dt

. The solution is The term Vc is negative since the top plate of the capacitor is connected to the positive - higher potential - terminal of the battery. Thus analyzing . where q = CV and τ = RC 0 emf counter-clockwise leads to a drop in voltage across the capacitor!

Wednesday, March 26, 2014 RL Circuits

. Thus we can write the sum of the potential drops around the circuit as

. The solution to this differential equation is

. We can see that the time constant of this circuit is τL = L/R

University Physics, Chapter 29 52

Wednesday, March 26, 2014 Summary: LC Circuit (1) . Consider a circuit consisting of an inductor L and a capacitor C . The charge on the capacitor as a function of time is given by

. The current in the inductor as a function of time is given by

. where φ is the phase and ω0 is the angular frequency

Wednesday, March 26, 2014 5

Wednesday, March 26, 2014 RLC Circuit (1)

. Now let’s consider a single loop circuit that has a capacitor C and an inductance L with an added resistance R . We observed that the energy of a circuit with a capacitor and an inductor remains constant and that the energy translated from electric to magnetic and back gain with no losses . If there is a resistance in the circuit, the current flow in the circuit will produce ohmic losses to heat . Thus the energy of the circuit will decrease because of these losses

Wednesday, March 26, 2014 RLC Circuit (2)

. The rate of energy loss is given by

. We can rewrite the change in energy of the circuit as a function time as

. Remembering that i = dq/dt and di/dt = d 2q/dt2 we can write

Wednesday, March 26, 2014 RLC Circuit (3)

. We can then write the differential equation

. The solution of this differential equation is

(damped harmonic oscillation!), where

Wednesday, March 26, 2014 RLC Circuit (4)

. If we charge the capacitor then hook it up to the circuit, we will observe a charge in the circuit that varies sinusoidally with time and while at the same time decreasing in amplitude . This behavior with time is illustrated below

Wednesday, March 26, 2014 RLC Circuit (5)

.Observations: •The charge varies sinusoidally with but the amplitude is damped out with time •After some time, no charge remains in the circuit . We can study the energy in the circuit as a function of time by calculating the energy stored in the electric field of the capacitor

. We can see that the energy stored in the capacitor decreases exponentially and oscillates in time

Wednesday, March 26, 2014 Alternating currents

Wednesday, March 26, 2014 DC and AC Motors and Generators

Wednesday, March 26, 2014 Direct and Alternating Current Generators . In a direct current generator the rotating coil is connected to an external circuit using a split commutator ring . As the coil turns, the connection is reversed such that the induced voltage always has the same sign . In alternating current generator, each end of the loop is connected to the external circuit through a slip ring • Thus this generator produces an induced voltage that varies from positive to negative and back, and is called an alternator . The voltages and currents produced by these generators are illustrated below Direct voltage/ Alternating voltage/ current current

Wednesday, March 26, 2014 Alternating Current (1)

. Now we consider a single loop circuit containing a capacitor, an inductor, a resistor, and a source of emf . This source of emf is capable of producing a time varying voltage as opposed to the sources of emf we have studied in previous chapters . We will assume that this source of emf provides a sinusoidal voltage as a function of time given by

. where ω is that angular frequency

of the emf and Vmax is the amplitude or maximum value of the emf

Wednesday, March 26, 2014 Alternating Current (2)

. The current induced in the circuit will also vary sinusoidally with time . This time-varying current is called alternating current . However, this current may not always remain in phase with the time- varying emf • The sinusoidal wave may crest earlier or later than that for emf . We can express the induced current as

where the angular frequency of the time-varying current is the same as the driving emf but the phase φ is not zero

Wednesday, March 26, 2014 Alternating Current (3)

. Note that traditionally the phase enters here with a negative sign

. Thus the voltage and the current in the circuit are not necessarily in phase

. Notation: instantaneous values are denoted by small letters (v, i ), amplitudes by capital letters (V, I)

Wednesday, March 26, 2014 Circuit with Resistor (1) . To begin our analysis of RLC circuits, let’s start with a circuit containing only a resistor and a source of time-varying emf as shown to the right . Applying Kirchhoff’s loop rule to this circuit we get

. where vR is the voltage drop across the resistor . Substituting into our expression for the emf as a function of time we get

. Remembering Ohm’s Law, V = iR, we get

Wednesday, March 26, 2014 Circuit with Resistor (2)

. Thus we can relate the current amplitude and the voltage amplitude by

. We can represent the time varying current by a phasor IR and the time-varying voltage by a phasor VR as shown below

. Phase difference is 0

Wednesday, March 26, 2014 Circuit with Capacitor (1)

. Now let’s address a circuit that contains a capacitor and a time varying emf as shown to the right . The voltage across the capacitor is given by Kirchhoff’s loop rule

. Remembering that q = CV for a capacitor we can write

. We would like to know the current as a function of time rather than the charge so we can write

Wednesday, March 26, 2014 Circuit with Capacitor (2): Capacitive Reactance . We can rewrite the last equation by defining a quantity that is similar to resistance and is called the capacitive reactance . Which allows us to write 1 Effective resistivity of a capacitor ωC . We can now express the current in the circuit as

. We can see that the current and the time varying emf are out of phase by 90°

Wednesday, March 26, 2014 Circuit with Capacitor (2): Capacitive Reactance . We can rewrite the last equation by defining a quantity that is similar to resistance and is called the capacitive reactance compare with . Which allows us to write V i = 1 R Effective resistivity of a capacitor ωC . We can now express the current in the circuit as

. We can see that the current and the time varying emf are out of phase by 90°

Wednesday, March 26, 2014 Circuit with Capacitor (3): Phasor

. We can represent the time varying current by a

phasor IC and the time-varying voltage by a phasor VC as shown below

. The current flowing this circuit with only a capacitor is similar to the expression for the current flowing in a circuit with only a resistor except that the current is out of phase with the emf by 90°

Wednesday, March 26, 2014 Circuit with Capacitor (4)

. We can also see that the amplitude of voltage across the capacitor and the amplitude of current in the capacitor are related by

. This equation resembles Ohm’s Law with the capacitive reactance replacing the resistance

. One major difference between the capacitive reactance and the resistance is that the capacitive reactance depends on the angular frequency of the time-varying emf

Wednesday, March 26, 2014 Circuit with Inductor (1)

. Now let’s consider a circuit with a source of time-varying emf and an inductor as shown to the right . We can again apply Kirchhoff’s Loop Rule to this circuit to obtain the voltage across the inductor as

. A changing current in an inductor will induce an emf given by

. So we can write

Wednesday, March 26, 2014 Circuit with Inductor (2)

. We are interested in the current rather than its time derivative so we integrate

. We define inductive reactance as

which, like the capacitive reactance, is similar to a resistance . We can then write

which again resembles Ohm’s Law except that the inductive reactance depends on the angular frequency of the time-varying emf

Wednesday, March 26, 2014 Circuit with Inductor (3)

. The current in the inductor can then be written as

. Thus the current flowing in a circuit with an inductor and a source time-varying emf will be -90° out of phase with the emf

. We can write the relationship between the amplitude of the current and the amplitude of the voltage as

Wednesday, March 26, 2014 Summary: RLC Circuit

. If we have a single loop RLC circuit, the charge in the circuit as a function of time is given by

where

. The energy stored in the capacitor as a function of time is given by

Wednesday, March 26, 2014 Summary: Resistance and Reactance

. Time-varying emf

. Time-varying emf VR with resistor

Resistance R

. Time-varying emf V with capacitor C Capacitive

Reactance XC

. Time-varying emf VL with inductor Inductive

Reactance XL

Wednesday, March 26, 2014 Summary: Phase and Phasors

Wednesday, March 26, 2014 Average powers

= IV < sin ωt sin ωt >= 2

= IV < sin ωt cos ωt >=0

Wednesday, March 26, 2014 Series RLC Circuit (1)

. Consider a single loop circuit that has a resistor, a capacitor, an inductor, and a source of time-varying emf . We can describe the time-varying currents in these circuit elements using a phasor I V . The projection of I on the vertical axis represents the current flowing in the circuit as a function of time • The angle of the phasor is given by ωt - φ . We can also describe the voltage in terms of a phasor V . The time-varying currents and voltages in the circuit can have different phases

Wednesday, March 26, 2014 Series RLC Circuit (2)

. We can describe the current flowing in the circuit and the voltage across the various components • Resistor iR

• The voltage vR and current iR are in phase with VR each other and the voltage phasor vR is in phase with the current phasor I • Capacitor

• The current iC leads the voltage vC by 90° so iC that the voltage phasor vC will have an angle 90° VC less than I and vR V • Inductor L

• The current iL lags behind the voltage vL by 90° iL so that voltage phasor vL will have an angle 90°

greater than I and vR

Wednesday, March 26, 2014 Series RLC Circuit (3)

. The voltage phasors for an RLC circuit are shown below

. The instantaneous voltages across each of the components are represented by the projections of the respective phasors on the vertical axis

Wednesday, March 26, 2014 Series RLC Circuit (4)

. Kirchhoff’s loop rules tells that the voltage drops across all the devices at any given time in the circuit must sum to zero, which gives us

. The voltage can be thought of as the projection of the vertical axis of the phasor

Vmax representing the time- varying emf in the circuit as shown below

. In this figure we have replaced the sum of the two

phasors VL and VC with the phasor VL - VC

Wednesday, March 26, 2014 Series RLC Circuit (5): Impedance

. The sum of the two phasors VL - VC and VR must equal Vmax so

. Now we can put in our expression for the voltage across the components in terms of the current and resistance or reactance

. We can then solve for the current in the circuit

. The denominator in the equation is called the impedance

. The impedance of a circuit depends on the frequency of the time- varying emf

Wednesday, March 26, 2014 Series RLC Circuit

impedance

1 2 Z = R2 + ωL − ωC active reactive resistance resistance (reactance) Only active resistance determines losses!

Reactive resistance can be 0, at resonance

Wednesday, March 26, 2014 Series RLC Circuit (6): Phase

. The current flowing in an alternating current circuit depends on the difference between the inductive reactance and the capacitive reactance . We can express the difference between the inductive reactance and the capacitive reactance in terms of the phase constant φ . This phase constant is defined as the phase difference

between voltage phasors VR and VL - VC

Wednesday, March 26, 2014 Series RLC Circuit (7)

. Thus we have three conditions for an alternating current circuit

• For XL > XC, φ is positive, and the current will lag behind the voltage in the circuit • This circuit will be similar to a circuit with only an inductor, except that the phase constant is not necessarily 90°

• For XL < XC, φ is negative, and the current will lead the voltage in the circuit • This circuit will be similar to a circuit with only a capacitor, except that the phase constant is not necessarily -90°

• For XL = XC, φ is zero, and the current in phase with the voltage in the circuit • This circuit is similar to a circuit with only a resistance • When φ = 0 we say that the circuit is in resonance

Wednesday, March 26, 2014 Series RLC Circuit (8)

XL > XC XL < XC XL = XC

. For XL = XC and φ = 0 we get the maximum current in the circuit and we can define a resonant frequency

Wednesday, March 26, 2014 Resonances in RLC circuit

. If R=0 V V I = max = max ωL 1/(ωC) ωL 1 ω2/ω2 | − | | − 0 | 1 ω2 = 0 LC

. Infinite current for ω = ω0 . Resonance!

Wednesday, March 26, 2014 Near resonance

. The smallest R will make the resonance finite

. If XL=XC, I =Vmax/R . At resonance, maximal current for given voltage

Wednesday, March 26, 2014 Resonant Behavior of RLC Circuit

. The resonant behavior of an RLC circuit resembles the response of a damped oscillator . Here we show the calculated maximum current as a function of the ratio of the angular frequency of the time varying emf divided by the resonant angular frequency, for a circuit with

Vmax = 7.5 V, L = 8.2 mH, C = 100 µF, and three resistances . One can see that as the resistance is lowered, the maximum current at the resonant angular frequency increases and there is a more pronounced resonant peak

Wednesday, March 26, 2014 iClicker

. Above the resonance the phase delay is

A : positive,φ>0 ω>ω0 B : negative,φ<0 C : φ =0

1 2 Z = R2 + ωL − ωC

Wednesday, March 26, 2014 iClicker

. Above the resonance the phase delay is A : positive,φ>0

ω>ω0 B : negative,φ<0 C : φ =0

1 2 1 Z = R2 + ωL ωL > ,XL >XC − ωC Cω φ>0

Wednesday, March 26, 2014 44

Wednesday, March 26, 2014 iClicker . At the resonance in an RLC circuit, the current is

. A: behind voltage by 90o . B: ahead of voltage by 90o . C: is in phase with voltage . D: relative phase of current and voltage undetermined

Wednesday, March 26, 2014 iClicker . At the resonance in an RLC circuit, the current is

. A: behind voltage by 90o . B: ahead of voltage by 90o . C: is in phase with voltage . D: relative phase of current and voltage undetermined

1 2 Z = R2 + ωL − ωC

Wednesday, March 26, 2014 iClicker . In RLC circuit, for given {V,R,L,C}, at the resonance the current is . A: Maximal . B: Minimal . C: Undetermined

Wednesday, March 26, 2014 iClicker . In RLC circuit, for given {V,R,L,C}, at the resonance the current is . A: Maximal . B: Minimal . C: Undetermined V I = Z 1 2 Z = R2 + ωL − ωC

Wednesday, March 26, 2014 Reminder: in RLC circuit current is out of phase with voltage

XL > XC XL < XC XL = XC

. For XL = XC and φ = 0 we get the maximum current in the circuit and we can define a resonant frequency

Wednesday, March 26, 2014 Complex impedace 1 Z = j ωL + R − ωC Imaginary = j -1 1 2 Z = R2 + ωL | | − ωC . sin φ Imaginary unit: phase φ shift by 90o Real jωt cos φ 1 v = V0e jωt+∆φ i = I0e v = iZ ejφ = cos φ + j sin φ

Wednesday, March 26, 2014 Energy and Power in RLC Circuits (1)

. When an RLC circuit is in operation, some of the energy in the circuit is stored in the electric field of the capacitor, some of the energy is stored in the magnetic field of the inductor, and some energy is dissipated in the form of heat in the resistor . The energy stored in the capacitor and inductor do not change in steady state operation . Therefore the energy transferred from the source of emf to the circuit is transferred to the resistor . The rate at which energy is dissipated in the resistor is the power P given by . The average power is given by

2 since < sin ωt >=1/2

Wednesday, March 26, 2014 Energy and Power in RLC Circuits (2)

. We define the root-mean-square (rms) current to be

. So we can write the average power as

. We can make similar definitions for other time-varying quantities • rms voltage:

• rms time-varying emf:

. The currents and voltages measured by an alternating current ammeter or voltmeter are rms values

Wednesday, March 26, 2014 Energy and Power in RLC Circuits (3)

. For example, we normally say that the voltage in the wall socket is 110 V . This rms voltage would correspond to a maximum voltage of

. We can then re-write our formula for the current as

. Which allows us to express the average power dissipated as

Wednesday, March 26, 2014 Energy and Power in RLC Circuits (3)

. For example, we normally say that the voltage in the wall socket is 110 V . This rms voltage would correspond to a maximum voltage of

. We can then re-write our formula for the current as

. Which allows us to express the average power dissipated as

Wednesday, March 26, 2014 Energy and Power in RLC Circuits (4)

. We can relate the phase constant to the ratio of the maximum value of the voltage across the resistor divided by the maximum value of the time-varying emf

. We can see that the maximum power is dissipated when φ = 0

. We call cos(φ) the power factor

Wednesday, March 26, 2014 Energy and Power in RLC Circuits (4)

. We can relate the phase constant to the ratio of the maximum value of the voltage across the resistor divided by the maximum value of the time-varying emf

. We can see that the maximum power is dissipated when φ = 0

. We call cos(φ) the power factor

If both L =0 and C =0, or at resonance, R/Z=1. Maximal power

Wednesday, March 26, 2014 Applications

Wednesday, March 26, 2014 Frequency filters

. Response of RLC circuit strongly depends on frequency. . Hi-Fi stereos have different speakers for low, middle and high frequencies

1 2 Z = R2 + ωL − ωC

Impedance Impedance dominated by Xc dominated by XL

Wednesday, March 26, 2014 Frequency filters

1 2 Z = R2 + ωL − ωC

High frequency Low frequency pass filter (at low pass filter (at high frequencies capacitance is frequencies inductance is like a broken circuit) like a broken circuit)

Wednesday, March 26, 2014 Transformers (1)

. When using or generating electrical power, high currents and low voltages are desirable for convenience and safety . When transmitting electric power, high voltages and low currents are desirable • The power loss in the transmission wires goes as P = I2R . The ability to raise and lower alternating voltages would be very useful in everyday life

Wednesday, March 26, 2014 Transformers

. Same idea as with the CD player and disconnected speakers, but with coils.

. Mutual inductance depends on “linked” magnetic flux . Use metal bar to contain the magnetic flux

Wednesday, March 26, 2014 Transformers

. Changing current in loop 1 induces varying B-field and magnetic flux through loop 1 and 2 . Varying magnetic flux through loop 2 induces EMF in the loop 2.

Wednesday, March 26, 2014 Transformers (2)

. To transform alternating currents and voltages from high to low one uses a transformer . A transformer consists of two sets of coils wrapped around an iron core. Huge inductor.

. Consider the primary windings with NP turns connected to a source of emf

. We can assume that the primary windings act as an inductor . The current is out of phase with the voltage (for R=0) and no power is delivered to the transformer

Wednesday, March 26, 2014 Transformers (3)

. A transformer that takes voltages from lower to higher is called a step-up transformer and a transformer that takes voltages from higher to lower is called a step-down transformer . Now consider the second coil with

NS turns . The time-varying emf in the primary coil induces a time- varying magnetic field in the iron core . This core passes through the secondary coil

Wednesday, March 26, 2014 Transformers (4) . Thus a time-varying voltage is induced in the secondary coil described by Faraday’s Law

. Because both the primary and secondary coils experience the same changing magnetic field we can write

. "stepped up" Ns > Np,

. "stepped down" Ns < Np

Wednesday, March 26, 2014 Transformers (5)

. If it’s just the inductor, power is 0. . If we now connect a resistor R across the secondary windings, a current will begin to flow through the secondary coil

. The power in the secondary circuit is then PS = ISVS . This current will induce a time-varying magnetic field that will induce an emf in the primary coil R . The emf source then will produce enough current IP to maintain the original emf . This induced current will not be 90° out of phase with the emf, thus power can be transmitted to the transformer . Energy conservation tells that the power produced by the emf source in the primary coil will be transferred to the secondary coil so we can write

Wednesday, March 26, 2014 . Increase in voltage equals decrease in current

Wednesday, March 26, 2014 Transformers (6) . When the secondary circuit begins to draw current, then current must be supplied to the primary circuit

. We can define the current in the secondary circuit as VS = ISR . We can then write the primary current as:

. With an effective primary resistance of

Wednesday, March 26, 2014 Transformers (7)

. Note that these equations assume no losses in the transformers and that the load is purely resistive • Real transformers have small losses

Wednesday, March 26, 2014 Impedance matching

. Power source Vs with internal impedance Zs. . Rs part of Zs will dissipate power: decease Rs. . For given Rs maximal power at

the load is when ZL*=ZS

VZs

VRs

VRl 2 2 1 VLCs VLCl Z = R + ωL − ωC

Wednesday, March 26, 2014 69

Wednesday, March 26, 2014 Let there be light!

Wednesday, March 26, 2014 EM field likes to oscillate . Changing magnetic field produces electric field (Faraday’s law of induction). . Changing electric field also produces magnetic field (e.g. LC circuit). . Analogue with a swing: potential energy changes into kinetic energy, kinetic energy changes into potential energy

Wednesday, March 26, 2014 Maxwell’s equations

E = ρ/ electric charge produces E-field ∇· 0 B =0 no magnetic charge ∇· B = µ J + µ ∂ E electric current and ∇× 0 0 0 t changing E-field E = ∂ B produce B-field ∇× − t changing B-field also produce E-field

= ∂ ,∂ ,∂ ∇ { x y z}

Wednesday, March 26, 2014 Maxwell’s equations in vacuum

E = ρ/ ∇· 0 B =0 ∇· B = µ J + µ ∂ E ∇× 0 0 0 t E = ∂ B ∇× − t

Wednesday, March 26, 2014 Maxwell’s equations in vacuum

E = ρ/ ∇· 0 B =0 ∇· B = µ J + µ ∂ E ∇× 0 0 0 t E = ∂tB ∇×E =0− ∇· B =0 ∇· B = µ ∂ E ∇× 0 0 t E = ∂tB

∇× − 73

Wednesday, March 26, 2014 E =0 ∇· B =0 ∇· B = µ ∂ E ∇× 0 0 t E = ∂ B ∇× − t

Wednesday, March 26, 2014 E =0 ∇· B =0 ∇· B = µ00∂tE ∇× ∇× E = ∂ B ∇× − t

Wednesday, March 26, 2014 E =0 ∇· B =0 ∇· B = µ00∂tE ∇× ∇× ∂ E = ∂tB t ∇× −

Wednesday, March 26, 2014 E =0 ∇· B =0 ∇· B = µ00∂tE ∇× ∇× ∂ E = ∂tB t ∇× − B = µ ∂ E ∇×∇× 0 0∇× t ∂ E = ∂2B t∇× − t

Wednesday, March 26, 2014 Wave equation B = µ ∂ E ∇×∇× 0 0∇× t ∂ E = ∂2B t∇× − t B = ( B) ∆B = ∆B ∇×∇× ∇ ∇· − − 1 2 2 2 µ = ∆B = ∂x + ∂y + ∂z B 0 0 c2 1 ∆B ∂2B =0 − c2 t B-field behaves like pendulum!

Wednesday, March 26, 2014 Wave equation 1 ∆B ∂2B =0 − c2 t 2 2 2 ∆B = ∂x + ∂y + ∂z B

By(z,t)= B0 sin(ωt kx) hump is moving − with velocity c 2π y k = wave length c λ λ ω2 x k2 =0 2 By − c z ω = kc

Wednesday, March 26, 2014 B

- E & B oscillate (in phase) - Changing E-field generates B-field - Changing B-field generates E-field - Fixed phase (humps or troughs) propagate with speed of light - All in vacuum, no carrier (no aether)

Wednesday, March 26, 2014 78

Wednesday, March 26, 2014 Impedance matching

. Power source Vs with internal resistance Zs. . Rs part of Zs will dissipate power: decease Rs. . For given Rs maximal power

at the load is when ZL*=ZS

VZs

VRs

VRl

VLCs VLCl

Wednesday, March 26, 2014 Example: Transformer (1)

A model train transformer plugs into 120 V ac and draws a current of 0.35 A while supplying 7.5 A to the train.

Wednesday, March 26, 2014 Example: Transformer (1)

A model train transformer plugs into 120 V ac and draws a current of 0.35 A while supplying 7.5 A to the train.

Wednesday, March 26, 2014 Example: Transformer (1)

A model train transformer plugs into 120 V ac and draws a current of 0.35 A while supplying 7.5 A to the train.

Questions:

Wednesday, March 26, 2014 Example: Transformer (1)

A model train transformer plugs into 120 V ac and draws a current of 0.35 A while supplying 7.5 A to the train.

Questions: .A) What voltage is present across the tracks?

Wednesday, March 26, 2014 Example: Transformer (1)

A model train transformer plugs into 120 V ac and draws a current of 0.35 A while supplying 7.5 A to the train.

Questions: .A) What voltage is present across the tracks? .B) Is the transformer step-up or step-down?