General Physics (PHY 2140)

Size: px

Start display at page:

Download "General Physics (PHY 2140)"

Gregory Daniel
6 years ago
Views:

General Physics (PHY 40) eminder: Exam this Wednesday 6/3 ecture 0-4 4 questions.

1 General Physics (PHY 40) eminder: Exam this Wednesday 6/3 ecture questions. Electricity and Magnetism nduced voltages and induction Self-nductance Circuits Energy in magnetic fields AC circuits and EM waves esistors, capacitors and inductors in AC circuits The C circuit Power in AC circuits Show your work for full credit. Closed book. You may bring a page of notes. Bring a calculator. Bring a pen or pencil. 6//007 Chapter //007 ast lecture:. nduced voltages and induction Generators and motors Self-induction ightning eview eview Problem: Charged particles passing through a bubble chamber leave tracks consisting of small hydrogen gas bubbles. These bubbles make visible the particles trajectories. n 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) f all three particles have the same mass and charges of equal magnitude, which is moving the fastest? mv r = qb Φ = BAcosθ Δ E = Δt N Φ = S S N v eview example Determine the direction of current in the loop for bar magnet moving down. change nitial flux Final flux By enz s law, the induced field is this 6// //007 4

0.6 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.

2 0.6 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. f 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 ΔΦ E = N Δ t 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 Δ E = Δt where is called the inductance of the device Units: S: henry (H) f flux is initially zero, H = V s A N N ΔΦ Φ = = Δ 6// //007 6 Example: solenoid nductor in a Circuit A solenoid of radius.5cm has 400 turns and a length of 0 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 = μ0n = μ0 Φ B = BA =μ0 A l l NΦB N A = = μ = (4π x )(60000)(.0 x 0-3 )/(0.) = mh l Δ Δ E E = = = (75 x 0-3 )/ (.0 x 0-3 ) = 37.5 A/s Δ t Δ t nductance can be interpreted as a measure of opposition to the rate of change in the current emember resistance 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 imum instantaneously Maximum current: = E 6// //007 8

3 0.7 Circuits Circuit (continued) ecall Ohm s s aw to find the voltage drop on Δ V = We have something similar with inductors Δ E = Δ t Similar to the case of the capacitor, we get an equation for the current as a function of time (series circuit). E = t/ ( e ) (voltage across a resistor) τ = (voltage across an inductor) V = t / ( e ) 6// // 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 PE = Summary of the properties of circuit elements. esistor Capacitor nductor Example: stored energy A 4V battery is connected in series with a resistor and an inductor, where = 8.0Ω and = 4.0H. Find the energy stored in the inductor when the current reaches its imum value. 6//007 units symbol relation power dissipated energy stored ohm, Ω = V / A V = P = V = ² = V² / 0 farad, F = C / V C Q = C V 0 PE C = C V² / henry, H = V s / A emf = - (Δ / Δt) 0 PE = ² / 6//007 3

4 A 4V battery is connected in series with a resistor and an inductor, where = 8.0Ω and = 4.0H. Find the energy stored in the inductor when the current c reaches its imum value. Given: V = 4 V = 8.0 Ω = 4.0 H ecall that the energy stored in the inductor is PE = Chapter Find: PE =? The only thing that is unknown in the equation above is current. The imum value for the current is V 4V = = = 3.0A 8.0Ω nserting this into the above expression for the energy gives Alternating Current Circuits and Electromagnetic Waves 6//007 PE ( 4.0 )( 3.0 ) = H A = 8J 3 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 sin πƒt Δv v is the instantaneous voltage ΔV is the imum voltage of the generator ƒ is the frequency at which the voltage changes, in Hz 6//007 5 esistor 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 imum values at the same time The current and the voltage are said to be in phase 6//

5 More About esistors 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 = ( sin πƒt) where i is the instantaneous current the heating effect produced by an AC current with a imum value of is not the same as that of a DC current of the same value The imum current occurs for a small amount of time 6//007 7 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 rms = = Alternating voltages can also be discussed in terms of rms values ΔV ΔV = = ΔV rms 6//007 8 Ohm s s aw in an AC Circuit Example: 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 aw for a resistor,, in an AC circuit ΔV rms = rms Also applies to the imum values of v and i 6// //007 9 An ac voltage source has an output of ΔV V = 50 sin (377 t). Find (a) the rms voltage output, (b) the frequency of the source, and (c) the voltage at t = (/0)s. (d) Find the rms current in the circuit when the generator is connected to a 50.0Ω resistor. ΔV Δ Vrms = = Δ V = x 50V = 06 V ω = 377 rad/sec, ω = π f, f = ω/ π = 377/ π = 60 Hz ΔV = 50 sin (377 x /0) = 0 V ΔV rms = rms thus, rms = ΔV rms / =. A 5

6 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//007 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//007 Capacitive eactance and Ohm s s aw The impeding effect of a capacitor on the current in an AC circuit is called the capacitive reactance and is given by X C = πƒc When ƒ is in Hz and C is in F, X C will be in ohms Ohm s s aw for a capacitor in an AC circuit ΔV rms = rms X C nductors 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// //

nductive eactance and Ohm s s aw The effective resistance of a coil in an AC circuit is called its inductive reactance and is given by X = πƒ When ƒ is in Hz and is in H, X will be in ohms Ohm s s aw

7 nductive eactance and Ohm s s aw The effective resistance of a coil in an AC circuit is called its inductive reactance and is given by X = πƒ When ƒ is in Hz and is in H, X will be in ohms Ohm s s aw for the inductor ΔV rms = rms X 6//007 5 Example: AC circuit with capacitors and inductors A.40mF capacitor is connected across an alternating voltage with an rms value of 9.00V. The rms current in the capacitor is 5.0mA. (a) What is the source frequency? (b) f the capacitor is replaced by an ideal coil with an inductance of 0.60H, what is the rms current in the coil? ΔV rms = rms X C, first we find X C : ΔV rms / rms Now, solve for ƒ: ƒ = / π X C C = 0.84 Hz = 9.00V/5.0 x 0-3 A = 360 ohms = π ƒc For and inductor X = πƒ, try solving for rms = ΔV rms / X 6//007 6 X C The C 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//007 7 Current and Voltage elationships in an C 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//

8 Phasor Diagrams To account for the different phases of the voltage drops, vector techniques are used epresent the voltage across each element as a rotating vector, called a phasor The diagram is called a phasor diagram 6//007 9 Phasor Diagram for C 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// Phasor Diagram, cont The phasors are added as vectors to account for the phase differences in the voltages ΔV and ΔV C are on the same line and so the net y component is ΔV - ΔV C 6//007 3 ΔV From 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 ΔV = ΔV + ( ΔV ΔV ) ΔV ΔVC tanφ = ΔV φ is the phase angle between the current and the imum voltage 6//007 3 C 8

9 mpedance of a Circuit The impedance, Z, can also be represented in a phasor diagram mpedance and Ohm s s aw Ohm s s aw can be applied to the impedance ΔV = Z Z = + (X X X tanφ = X C C ) 6// // Summary of Circuit Elements, mpedance and Phase Angles Problem Solving for AC Circuits Calculate as many unknown quantities as possible For example, find X and X C Be careful of units -- use F, H, Ω Apply Ohm s s aw to the portion of the circuit that is of interest Determine all the unknowns asked for in the problem 6// //

10 Power in an AC Circuit No power losses are associated with capacitors and pure inductors in an AC circuit n a capacitor, during one-half of a cycle energy is stored and during the other half the energy is returned to the circuit n 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// Power in an AC Circuit, cont The average power delivered by the generator is converted to internal energy in the resistor P av = rms ΔV = rms ΔV rms cos φ cos φ is called the power factor of the circuit Phase shifts can be used to imize power outputs 6// esonance in an AC Circuit esonance occurs at the frequency, ƒ o, where the current has its imum value To achieve imum current, the impedance must have a minimum value This occurs when X = X C ƒ o = π C 6//

General Physics (PHY 2140)

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