General Physics (PHY 2140) Lecture 10 6/12/2007 Electricity and Magnetism Induced voltages and induction Self-Inductance RL Circuits Energy in magnetic fields AC circuits and EM waves Resistors, capacitors and inductors in AC circuits The RLC circuit Power in AC circuits Chapter 20-21 http://www.physics.wayne.edu/~alan/2140website/main.htm 1
Reminder: Exam 2 this Wednesday 6/13 12-14 14 questions. Show your work for full credit. Closed book. You may bring a page of notes. Bring a calculator. Bring a pen or pencil. 6/12/2007 2
Lightning Review Last lecture: 1. Induced voltages and induction Generators and motors Self-induction Review Problem: Charged particles passing through a bubble chamber leave tracks consisting of small hydrogen gas bubbles. These bubbles make visible the particles trajectories. In the following figure, the magnetic field is directed into the page, and the tracks are in the plane of the page, in the directions indicated by the arrows. (a) Which of the tracks correspond to positively charged particles? (b) If all three particles have the same mass and charges of equal magnitude, which is moving the fastest? mv r = qb Φ = BAcosθ ΔI E = L Δt L = N Φ I 6/12/2007 3
S S N v Review example Determine the direction of current in the loop for bar magnet moving down. Initial flux Final flux By Lenz s law, the induced field is this change 6/12/2007 4
20.6 Self-inductance When a current flows through a loop, the magnetic field created by that current has a magnetic flux through the area of the loop. If the current changes, the magnetic field changes, and so the flux changes giving rise to an induced emf. This phenomenon is called self-induction because it is the loop's own current, and not an external one, that gives rise to the induced emf. Faraday s law states E = N ΔΦ Δ t 6/12/2007 5
The magnetic flux is proportional to the magnetic field, which is proportional to the current in the circuit Thus, the self-induced EMF must be proportional to the time rate of change of the current ΔI E = L Δt where L is called the inductance of the device Units: SI: henry (H) 1H = 1 V s A If flux is initially zero, N L N ΔΦ Φ = = Δ I I 6/12/2007 6
Example: solenoid A solenoid of radius 2.5cm has 400 turns and a length of 20 cm. Find (a) its inductance and (b) the rate at which current must change through it to produce an emf of 75mV. N N B = μ0ni = μ0 I Φ B = BA =μ0 IA l l L NΦ I 2 B N A = = μ = (4π x 10 0-7 )(160000)(2.0 x 10-3 )/(0.2) = 2 mh l ΔI ΔI E E = L = = (75 x 10-3 )/ (2.0 x 10-3 ) = 37.5 A/s Δ t Δ t L 6/12/2007 7
Inductor in a Circuit Inductance can be interpreted as a measure of opposition to the rate of change in the current Remember resistance R is a measure of opposition to the current As a circuit is completed, the current begins to increase, but the t inductor produces an emf that opposes the increasing current Therefore, the current doesn t t change from 0 to its maximum instantaneously Maximum current: I max = E R 6/12/2007 8
20.7 RL Circuits Recall Ohm s s Law to find the voltage drop on R Δ V = IR (voltage across a resistor) We have something similar with inductors E L = L Δ Δ Similar to the case of the capacitor, we get an equation for the current as a function of time (series circuit). I E = R ( Rt / L 1 e ) I t τ = (voltage across an inductor) L R 6/12/2007 9
RL Circuit (continued) I V = R ( Rt/ L 1 e ) 6/12/2007 10
20.8 Energy stored in a magnetic field The battery in any circuit that contains a coil has to do work to produce a current Similar to the capacitor, any coil (or inductor) would store potential energy PEL = 1 2 LI Summary of the properties of circuit elements. Resistor Capacitor Inductor 2 6/12/2007 units ohm, Ω = V / A farad, F = C / V henry, H = V s / A symbol R C L relation V = I R Q = C V emf = -L (ΔI / Δt) power dissipated P = I V = I² R = V² / R 0 0 energy stored 0 PE C = C V² / 2 PE L = L I² / 2 11
Example: stored energy A 24V battery is connected in series with a resistor and an inductor, where R = 8.0Ω and L = 4.0H. Find the energy stored in the inductor when the current reaches its maximum value. 6/12/2007 12
A 24V battery is connected in series with a resistor and an inductor, where R = 8.0Ω and L = 4.0H. Find the energy stored in the inductor when the current c reaches its maximum value. Given: V = 24 V R = 8.0 Ω L = 4.0 H Recall that the energy stored in the inductor is PEL = 1 2 LI 2 Find: PE L =? The only thing that is unknown in the equation above is current. The maximum value for the current is I max V 24V = = = 3.0A R 8.0Ω Inserting this into the above expression for the energy gives 6/12/2007 PE 1 ( 4.0 )( 3.0 ) 2 L = H A = 18J 2 13
Chapter 21 Alternating Current Circuits and Electromagnetic Waves
AC Circuit An AC circuit consists of a combination of circuit elements and an AC generator or source The output of an AC generator is sinusoidal and varies with time according to the following equation Δv v = ΔV max sin 2πƒt2 Δv v is the instantaneous voltage ΔV max is the maximum voltage of the generator ƒ is the frequency at which the voltage changes, in Hz 6/12/2007 15
Resistor in an AC Circuit Consider a circuit consisting of an AC source and a resistor The graph shows the current through and the voltage across the resistor The current and the voltage reach their maximum values at the same time The current and the voltage are said to be in phase 6/12/2007 16
More About Resistors in an AC Circuit The direction of the current has no effect on the behavior of the resistor The rate at which electrical energy is dissipated in the circuit is given by P = i 2 R= (I max sin 2πƒt) 2 2 R where i is the instantaneous current the heating effect produced by an AC current with a maximum value of I max is not the same as that of a DC current of the same value The maximum current occurs for a small amount of time 6/12/2007 17
rms Current and Voltage The rms current is the direct current that would dissipate the same amount of energy in a resistor as is actually dissipated by the AC current I max I rms = = 2 0.707 I max Alternating voltages can also be discussed in terms of rms values ΔV = ΔV 0.707 ΔV max rms max 6/12/2007 18 2 =
Ohm s s Law in an AC Circuit rms values will be used when discussing AC currents and voltages AC ammeters and voltmeters are designed to read rms values Many of the equations will be in the same form as in DC circuits Ohm s s Law for a resistor, R, in an AC circuit ΔV rms rms = I rms rms R Also applies to the maximum values of v and i 6/12/2007 19
Example: an AC circuit An ac voltage source has an output of ΔV V = 150 sin (377 t). Find An ac voltage source has an output of Find (a) the rms voltage output, (b) the frequency of the source, and (c) the voltage at t = (1/120)s. (d) Find the rms current in the circuit when the generator is connected to a 50.0Ω resistor. ΔV 0.707 0.707 x 150V 106 V 2 ω = 377 rad/sec, ω = 2 π f, f = ω/ 2 π = 377/ 2π = 60 Hz max Δ Vrms = = Δ Vmax = = ΔV = 150 sin (377 x 1/120) = 0 V ΔV rms = I rms R thus, I rms = ΔV rms /R = 2.12 A 6/12/2007 20
Capacitors in an AC Circuit Consider a circuit containing a capacitor and an AC source The current starts out at a large value and charges the plates of the capacitor There is initially no resistance to hinder the flow of the current while the plates are not charged As the charge on the plates increases, the voltage across the plates increases and the current flowing in the circuit decreases 6/12/2007 21
More About Capacitors in an AC Circuit The current reverses direction The voltage across the plates decreases as the plates lose the charge they had accumulated The voltage across the capacitor lags behind the current by 90 6/12/2007 22
Capacitive Reactance and Ohm s s Law The impeding effect of a capacitor on the current in an AC circuit is called the capacitive reactance and is given by X C When ƒ is in Hz and C is in F, X C will be in ohms Ohm s s Law for a capacitor in an AC circuit ΔV rms = 1 2πƒC rms = I rms rms X C 6/12/2007 23
Inductors in an AC Circuit Consider an AC circuit with a source and an inductor The current in the circuit is impeded by the back emf of the inductor The voltage across the inductor always leads the current by 90 6/12/2007 24
Inductive Reactance and Ohm s s Law The effective resistance of a coil in an AC circuit is called its inductive reactance and is given by X L = 2πƒL2 When ƒ is in Hz and L is in H, X L will be in ohms Ohm s s Law for the inductor ΔV rms rms = I rms rms X L 6/12/2007 25
Example: AC circuit with capacitors and inductors A 2.40mF capacitor is connected across an alternating voltage with an rms value of 9.00V. The rms current in the capacitor is 25.0mA. (a) What is the source frequency? (b) If the capacitor is replaced by an ideal coil with an inductance of 0.160H, what is the rms current in the coil? ΔV rms = I rms X C, first we find X C : ΔV rms / I rms = 9.00V/25.0 x 10-3 A = 360 ohms Now, solve for ƒ: ƒ = 1/ 2π X C C = 0.184 Hz X C = 1 2π ƒc For and inductor X L = 2πƒL, 2 try solving for I rms = ΔV rms / X L 6/12/2007 26
The RLC Series Circuit The resistor, inductor, and capacitor can be combined in a circuit The current in the circuit is the same at any time and varies sinusoidally with time 6/12/2007 27
Current and Voltage Relationships in an RLC Circuit The instantaneous voltage across the resistor is in phase with the current The instantaneous voltage across the inductor leads the current by 90 The instantaneous voltage across the capacitor lags the current by 90 6/12/2007 28
Phasor Diagrams To account for the different phases of the voltage drops, vector techniques are used Represent the voltage across each element as a rotating vector, called a phasor The diagram is called a phasor diagram 6/12/2007 29
Phasor Diagram for RLC Series Circuit The voltage across the resistor is on the +x axis since it is in phase with the current The voltage across the inductor is on the +y since it leads the current by 90 The voltage across the capacitor is on the y axis since it lags behind the current by 90 6/12/2007 30
Phasor Diagram, cont The phasors are added as vectors to account for the phase differences in the voltages ΔV L and ΔV C are on the same line and so the net y component is ΔV L - ΔV C 6/12/2007 31
ΔV From max the Phasor Diagram The voltages are not in phase, so they cannot simply be added to get the voltage across the combination of the elements or the voltage source 2 2 ΔV = ΔV + ( ΔV ΔV ) max R ΔVL ΔVC tanφ = ΔVR φ is the phase angle between the current and the maximum voltage L C 6/12/2007 32
Impedance of a Circuit The impedance, Z, can also be represented in a phasor diagram Z = R tanφ = 2 X + (X L L X R C X C ) 2 6/12/2007 33
Impedance and Ohm s s Law Ohm s s Law can be applied to the impedance ΔV max max = I max max Z 6/12/2007 34
Summary of Circuit Elements, Impedance and Phase Angles 6/12/2007 35
Problem Solving for AC Circuits Calculate as many unknown quantities as possible For example, find X L and X C Be careful of units -- use F, H, Ω Apply Ohm s s Law to the portion of the circuit that is of interest Determine all the unknowns asked for in the problem 6/12/2007 36
Power in an AC Circuit No power losses are associated with capacitors and pure inductors in an AC circuit In a capacitor, during one-half of a cycle energy is stored and during the other half the energy is returned to the circuit In an inductor, the source does work against the back emf of the inductor and energy is stored in the inductor, but when the current begins to decrease in the circuit, the energy is returned to the circuit 6/12/2007 37
Power in an AC Circuit, cont The average power delivered by the generator is converted to internal energy in the resistor P av = I rms ΔV R = I rms ΔV rms cos φ cos φ is called the power factor of the circuit Phase shifts can be used to maximize power outputs 6/12/2007 38
Resonance in an AC Circuit Resonance occurs at the frequency, ƒ o, where the current has its maximum value To achieve maximum current, the impedance must have a minimum value This occurs when X L = X C ƒ o = 2π 1 LC 6/12/2007 39