Electromagnetic Induction

Electromagnetic Induction
Chapter 29 opener. One of the great laws of physics is Faraday’s law of induction, which says that a changing magnetic flux produces an induced emf. This photo shows a bar magnet moving inside a coil of wire, and the galvanometer registers an induced current. This phenomenon of electromagnetic induction is the basis for many practical devices, including generators, alternators, transformers, tape recording, and computer memory.

Induced EMF Faraday’s Law of Induction; Lenz’s Law EMF Induced in a Moving Conductor Electric Generators Back EMF and Counter Torque; Eddy Currents

Transformers and Transmission of Power
A Changing Magnetic Flux Produces an Electric Field

Induced EMF Almost 200 years ago, Faraday looked for evidence that a magnetic field would induce an electric current with this apparatus: Figure Faraday’s experiment to induce an emf.

Induced EMF He found no evidence when the current was steady, but did see a current induced when the switch was turned on or off. Figure (a) A current is induced when a magnet is moved toward a coil, momentarily increasing the magnetic field through the coil. (b) The induced current is opposite when the magnet is moved away from the coil ( decreases). Note that the galvanometer zero is at the center of the scale and the needle deflects left or right, depending on the direction of the current. In (c), no current is induced if the magnet does not move relative to the coil. It is the relative motion that counts here: the magnet can be held steady and the coil moved, which also induces an emf.

Induced EMF Therefore, a changing magnetic field induces an emf.
Faraday’s experiment used a magnetic field that was changing because the current producing it was changing; the previous graphic shows a magnetic field that is changing because the magnet is moving.

Magnetic Flux The induced emf in a wire loop is proportional to the rate of change of magnetic flux through the loop. Magnetic flux: Unit of magnetic flux: weber, Wb: 1 Wb = 1 T·m2.

Magnetic Flux This drawing shows the variables in the flux equation:
Figure Determining the flux through a flat loop of wire. This loop is square, of side l and area A = l2.

Magnetic Flux The magnetic flux is analogous to the electric flux – it is proportional to the total number of magnetic field lines passing through the loop. Figure Magnetic flux ΦB is proportional to the number of lines of B that pass through the loop.

Magnetic Flux Determining flux.
A square loop of wire encloses area A1. A uniform magnetic field B perpendicular to the loop extends over the area A2. What is the magnetic flux through the loop A1? Solution: Assuming the field is zero outside A2, the flux is BA2.

Faraday’s Law of Induction
Faraday’s law of induction: the emf induced in a circuit is equal to the rate of change of magnetic flux through the circuit: or

A loop of wire in a magnetic field. A square loop of wire of side l = 5.0 cm is in a uniform magnetic field B = 0.16 T. What is the magnetic flux in the loop (a) when B is perpendicular to the face of the loop and (b) when B is at an angle of 30° to the area A of the loop? (c) What is the magnitude of the average current in the loop if it has a resistance of Ω and it is rotated from position (b) to position (a) in 0.14 s? Solution: a. The flux is BA = 4.0 x 10-4 Wb. b. The flux is BA cos θ = 3.5 x 10-4 Wb. c. The emf is ΔΦB/Δt = 3.6 x 10-4 V; then I = emf/R = 30 mA.

Lenz’s Law The minus sign gives the direction of the induced emf:
A current produced by an induced emf moves in a direction so that the magnetic field it produces tends to restore the changed field. or: An induced emf is always in a direction that opposes the original change in flux that caused it.

Magnetic flux will change if the area of the loop changes. Figure A current can be induced by changing the area of the coil, even though B doesn’t change. Here the area is reduced by pulling on its sides: the flux through the coil is reduced as we go from (a) to (b). Here the brief induced current acts in the direction shown so as to try to maintain the original flux (Φ = BA) by producing its own magnetic field into the page. That is, as the area A decreases, the current acts to increase B in the original (inward) direction.

Magnetic flux will change if the angle between the loop and the field changes. Figure A current can be induced by rotating a coil in a magnetic field. The flux through the coil changes from (a) to (b) because θ (in Eq. 29–1a, Φ = BA cos θ) went from 0° (cos θ = 1) to 90° (cos θ = 0).

Induction stove. In an induction stove, an ac current exists in a coil that is the “burner” (a burner that never gets hot). Why will it heat a metal pan but not a glass container? Solution: The magnetic field created by the current induces a current in the metal pan, which heats due to resistance. Very little current is induced in a glass pan (or in your hand, which is why it does not feel hot).

Lenz’s Law Problem Solving: Lenz’s Law
Determine whether the magnetic flux is increasing, decreasing, or unchanged. The magnetic field due to the induced current points in the opposite direction to the original field if the flux is increasing; in the same direction if it is decreasing; and is zero if the flux is not changing. Use the right-hand rule to determine the direction of the current. Remember that the external field and the field due to the induced current are different.

Lenz’s Law Practice with Lenz’s law.
In which direction is the current induced in the circular loop for each situation? Solution: a. Pulling the loop to the right out of a magnetic field which points out of the page. The flux through the loop is outward and decreasing; the induced current will be counterclockwise. b. Shrinking a loop in a magnetic field pointing into the page. The flux through the loop is inward and decreasing; the induced current will be clockwise. c. N magnetic pole moving toward the loop into the page. The flux through the loop is inward and increasing; the induced current will be counterclockwise. d. N magnetic pole moving toward loop in the plane of the page. There is no flux through the loop, and no induced current. e. Rotating the loop by pulling the left side toward us and pushing the right side in; the magnetic field points from right to left. The flux through the loop is to the left and increasing; the induced current will be counterclockwise.

Lenz’s Law Pulling a coil from a magnetic field.
A 100-loop square coil of wire, with side l = 5.00 cm and total resistance 100 Ω, is positioned perpendicular to a uniform T magnetic field. It is quickly pulled from the field at constant speed (moving perpendicular to B) to a region where B drops abruptly to zero. At t = 0, the right edge of the coil is at the edge of the field. It takes s for the whole coil to reach the field-free region. Find (a) the rate of change in flux through the coil, and (b) the emf and current induced. (c) How much energy is dissipated in the coil? (d) What was the average force required (Fext)? Solution: a. The flux goes from BA to zero in s, so  ΦB/ t = BA/t = x 10-2 Wb/s. b. The emf is –N  ΦB/ dt = 1.50 V. The current is emf/R = 15.0 mA. c. E = Pt = I2Rt = 2.25 x 10-3 J. d. F = W/d = N.

EMF Induced in a Moving Conductor
This image shows another way the magnetic flux can change: Figure 29-12a. A conducting rod is moved to the right on a U-shaped conductor in a uniform magnetic field B that points out of the page. The induced current is clockwise.

Force on the rod. To make the rod move to the right at speed v, you need to apply an external force on the rod to the right. (a) Explain and determine the magnitude of the required force. (b) What external power is needed to move the rod? Solution: a. The external force needs to be equal and opposite to the magnetic force (IlB) if the rod is to move at a constant speed. I = Blv/R, so F = B2l2v/R. b. The external power is Fv = B2l2v2/R, which is equal to the power dissipated in the resistance of the rod (I2R).

Electric Generators A generator is the opposite of a motor – it transforms mechanical energy into electrical energy. This is an ac generator: The axle is rotated by an external force such as falling water or steam. The brushes are in constant electrical contact with the slip rings. Figure An ac generator.

Electric Generators If the loop is rotating with constant angular velocity ω, the induced emf is sinusoidal: For a coil of N loops, Figure An ac generator produces an alternating current. The output emf E = E0 sin ωt, where E0 = NABω.

Electric Generators An ac generator.
The armature of a 60-Hz ac generator rotates in a 0.15-T magnetic field. If the area of the coil is 2.0 x 10-2 m2, how many loops must the coil contain if the peak output is to be E0 = 170 V? Solution: N = E0/BAω = 150 turns. Remember to convert 60 Hz to angular units.

Electric Generators A dc generator is similar, except that it has a split-ring commutator instead of slip rings. Figure (a) A dc generator with one set of commutators, and (b) a dc generator with many sets of commutators and windings.

Electric Generators Automobiles now use alternators rather than dc generators, to reduce wear. Figure (a) Simplified schematic diagram of an alternator. The input current to the rotor from the battery is connected through continuous slip rings. Sometimes the rotor electromagnet is replaced by a permanent magnet. (b) Actual shape of an alternator. The rotor is made to turn by a belt from the engine. The current in the wire coil of the rotor produces a magnetic field inside it on its axis that points horizontally from left to right, thus making north and south poles of the plates attached at either end. These end plates are made with triangular fingers that are bent over the coil—hence there are alternating N and S poles quite close to one another, with magnetic field lines between them as shown by the blue lines. As the rotor turns, these field lines pass through the fixed stator coils (shown on the right for clarity, but in operation the rotor rotates within the stator), inducing a current in them, which is the output.

Back EMF and Counter Torque
An electric motor turns because there is a torque on it due to the current. We would expect the motor to accelerate unless there is some sort of drag torque. That drag torque exists, and is due to the induced emf, called a back emf.

Back emf in a motor. The armature windings of a dc motor have a resistance of 5.0 Ω. The motor is connected to a 120-V line, and when the motor reaches full speed against its normal load, the back emf is 108 V. Calculate (a) the current into the motor when it is just starting up, and (b) the current when the motor reaches full speed. Solution: a. At startup, I = V/R = 24 A. b. The back emf means that the total emf in the circuit is 12 V, so the current is 2.4 A.

Motor overload. When using an appliance such as a blender, electric drill, or sewing machine, if the appliance is overloaded or jammed so that the motor slows appreciably or stops while the power is still connected, the device can burn out and be ruined. Explain why this happens. The motor is designed to operate at a particular speed, which means there will be a particular back emf. If the appliance slows or stops, the back emf becomes much less, the current becomes much more than designed, and the appliance may burn out.

A similar effect occurs in a generator – if it is connected to a circuit, current will flow in it, and will produce a counter torque. This means the external applied torque must increase to keep the generator turning.

Eddy Currents Induced currents can flow in bulk material as well as through wires. These are called eddy currents, and can dramatically slow a conductor moving into or out of a magnetic field. Figure Production of eddy currents in a rotating wheel. The grey lines in (b) indicate induced current.

Origins of Induced EMFs
Emf is the work done per unit charge by the source: Induced electric fields Motional emf

Motional EMF The induced current is in a direction that tends to slow the moving bar – it will take an external force to keep it moving. Figure 29-12b. Upward force on an electron in the metal rod (moving to the right) due to B pointing out of the page.

The induced emf has magnitude This equation is valid as long as B, l, and v are mutually perpendicular (if not, it is true for their perpendicular components).

Motional EMF Does a moving airplane develop a large emf?
An airplane travels 1000 km/h in a region where the Earth’s magnetic field is about 5 x 10-5 T and is nearly vertical. What is the potential difference induced between the wing tips that are 70 m apart? Solution: E = Blv = 1 V.

Induced Electric Fields
E produced by changing B. A magnetic field B between the pole faces of an electromagnet is nearly uniform at any instant over a circular area of radius r0. The current in the windings of the electromagnet is increasing in time so that B changes in time at a constant rate dB/dt at each point. Beyond the circular region (r > r0), we assume B = 0 at all times. Determine the electric field E at any point P a distance r from the center of the circular area due to the changing B. Figure (a) Side view of nearly constant B. (b) Top view, for determining the electric field E at point P. (c) Lines of E produced by increasing B (pointing outward). (d) Graph of E vs. r. Example 29–14. Solution: Because of symmetry, E will be perpendicular to B and constant at radius r. Integrate around a circle of radius r as shown. For r < r0, the enclosed flux is Bπr2, and E = r/2 dB/dt. For r > r0, the enclosed flux is Bπr02, and E = r02/2r dB/dt.

Applications of Induction: Sound Systems, Computer Memory, Seismograph
This microphone works by induction; the vibrating membrane induces an emf in the coil. Figure Diagram of a microphone that works by induction.

Differently magnetized areas on an audio tape or disk induce signals in the read/write heads. Figure (a) Read/Write (playback/recording) head for tape or disk. In writing or recording, the electric input signal to the head, which acts as an electromagnet, magnetizes the passing tape or disk. In reading or playback, the changing magnetic field of the passing tape or disk induces a changing magnetic field in the head, which in turn induces in the coil an emf that is the output signal. (b) Photo of a hard drive showing several platters and read/write heads that can quickly move from the edge of the disk to the center.

A seismograph has a fixed coil and a magnet hung on a spring (or vice versa), and records the current induced when the Earth shakes. Figure One type of seismograph, in which the coil is fixed to the case and moves with the Earth. The magnet, suspended by springs, has inertia and does not move instantaneously with the coil (and case), so there is relative motion between magnet and coil.

Summary Magnetic flux: Changing magnetic flux induces emf:
Induced emf produces current that opposes original flux change.

Summary Changing magnetic field produces an electric field.
General form of Faraday’s law: Electric generator changes mechanical energy to electrical energy; electric motor does the opposite. .

Inductance AC Circuits
Chapter 30 Opener. A spark plug in a car receives a high voltage, which produces a high enough electric field in the air across its gap to pull electrons off the atoms in the air–gasoline mixture and form a spark. The high voltage is produced, from the basic 12 V of the car battery, by an induction coil which is basically a transformer or mutual inductance. Any coil of wire has a self-inductance, and a changing current in it causes an emf to be induced. Such inductors are useful in many circuits.

Mutual Inductance Self-Inductance Energy Stored in a Magnetic Field LR Circuits LC Circuits and Electromagnetic Oscillations LC Circuits with Resistance (LRC Circuits) AC Circuits with AC Source

LRC Series AC Circuit Resonance in AC Circuits Impedance Matching Three-Phase AC

Inductance Induced emf in one circuit due to changes in the magnetic field produced by the second circuit is called mutual induction. Induced emf in one circuit associated with changes in its own magnetic field is called self-induction.

Inductance Unit of inductance: the henry, H: 1 H = 1 V·s/A = 1 Ω·s.

Mutual Inductance Mutual inductance: magnetic flux through coil2 due to current in coil 1 Induced emf due to mutual induction:

Mutual Inductance Solenoid and coil.
A long thin solenoid of length l and cross-sectional area A contains N1 closely packed turns of wire. Wrapped around it is an insulated coil of N2 turns. Assume all the flux from coil 1 (the solenoid) passes through coil 2, and calculate the mutual inductance. Solution: Assuming all the flux from the solenoid stays within the secondary coil, the flux is BA = μ0 (N1/l) I1A, and M = μ0 (N1N2/l) A.

Mutual Inductance Reversing the coils.
How would the previous example change if the coil with turns was inside the solenoid rather than outside the solenoid? Solution: The only difference is that the area A would be the area of the coil with turns rather than the area of the solenoid.

Self-Inductance Self-inductance: magnetic flux through the coil due to the current in the coil itself: A changing current in a coil will also induce an emf in itself:

Self-Inductance Solenoid inductance.
(a) Determine a formula for the self-inductance L of a tightly wrapped and long solenoid containing N turns of wire in its length l and whose cross-sectional area is A. (b) Calculate the value of L if N = 100, l = 5.0 cm, A = 0.30 cm2, and the solenoid is air filled. Solution: a. B = μ0nI and is constant; the flux is μ0NIA/l and the self-inductance is μ0N2A/l. b. Plugging in the numbers gives L = 7.5 μH.

Self-Inductance Direction of emf in inductor.
Current passes through a coil from left to right as shown. (a) If the current is increasing with time, in which direction is the induced emf? (b) If the current is decreasing in time, what then is the direction of the induced emf? Figure Example 30–4. The + and - signs refer to the induced emf due to the changing current, as if points A and B were the terminals of a battery (and the coiled loops were the inside of the battery). Solution: a. The induced emf opposes the change that caused it, so the induced emf acts to oppose the current. b. Now the induced emf acts to reinforce the current (as it is decreasing).

Self-Inductance Coaxial cable inductance.
Determine the inductance per unit length of a coaxial cable whose inner conductor has a radius r1 and the outer conductor has a radius r2. Assume the conductors are thin hollow tubes so there is no magnetic field within the inner conductor, and the magnetic field inside both thin conductors can be ignored. The conductors carry equal currents I in opposite directions. Solution: The flux through a rectangle of width dr and length l is dΦB = (μ0I/2πr) l dr. Integrating over r to find the total flux gives ΦB = (μ0Il/2π) ln (r2/r1). Therefore, L = (μ0l/2π) ln (r2/r1).

LR Circuits A circuit consisting of an inductor and a resistor will begin with most of the voltage drop across the inductor, as the current is changing rapidly. With time, the current will increase less and less, until all the voltage is across the resistor. Figure (a) LR circuit; (b) growth of current when connected to battery.

LR Circuits

LR Circuits If the circuit is then shorted across the battery, the current will gradually decay away: . Figure The switch is flipped quickly so the battery is removed but we still have a circuit. The current at this moment (call it t = 0) is I0. Figure Decay of the current in Fig. 30–7 in time after the battery is switched out of the circuit.

LR Circuits

LR Circuits An LR circuit.
At t = 0, a 12.0-V battery is connected in series with a 220-mH inductor and a total of 30-Ω resistance, as shown. (a) What is the current at t = 0? (b) What is the time constant? (c) What is the maximum current? (d) How long will it take the current to reach half its maximum possible value? (e) At this instant, at what rate is energy being delivered by the battery, and (f) at what rate is energy being stored in the inductor’s magnetic field? Solution: a. At t = 0 the current is zero. b. The time constant is L/R = 7.3 ms. c. The maximum current is V0/R = 0.40 A. d. When I = ½ Imax, e-t/τ = ½ then t = 5.0 ms. e. P = IV = 2.4 W. f. U = ½ LI2, so dU/dt = LI dI/dt = I(V0-RI) = 1.2 W. The rest of the power is dissipated in the resistor.

Solution:

Solution: Remark: PR, the power dissipated from the resistor
PL, the power stored in the inductor

Energy Density of a Magnetic Field
Just as we saw that energy can be stored in an electric field, energy can be stored in a magnetic field as well, in an inductor, for example. Analysis shows that the energy density of the field is given by

Energy Stored in an Inductor
The equation governs the LR circuit is Multiplying each term by the current i leads to

Energy Stored in an Inductor
Therefore, the third term represents the rate at which the energy is stored in the inductor The total energy stored from i=0 to i=I is

Energy Density of a Magnetic Field
The self-inductance of a solenoid is L=μ0nA2l. The magnetic field inside it is B=μ0nI. The energy stored thus is Since Al is the volume of the solenoid, the energy per volume is This is the energy density of a magnetic field in free space.

LC Circuits and Electromagnetic Oscillations
An LC circuit is a charged capacitor shorted through an inductor. Figure An LC circuit.

Electromagnetic Oscillations

LC Circuits Across the capacitor, the voltage is raised by Q/C. As the current passes through the inductor, the induced emf is –L(dI/dt). The Kirchhof’s loop rule gives The current causes the charge in the capacitor to decreases so I=-dQ/dt. Thus the differential equation becomes

The equation describing LC circuits has the same form as the SHO equation: The charge therefore oscillates with a natural angular frequency .

Electromagnetic Oscillations
The charge varies as The current is sinusoidal as well: Remark: When Q=Q0 at t=t0, we have φ=0.

The charge and current are both sinusoidal, but with different phases. Figure Charge Q and current I in an LC circuit. The period T = 1/f = 2π/ω = 2π(L/C)1/2.

The total energy in the circuit is constant; it oscillates between the capacitor and the inductor: Figure Energy UE (red line) and UB (blue line) stored in the capacitor and the inductor as a function of time. Note how the energy oscillates between electric and magnetic. The dashed line at the top is the (constant) total energy U = UE + UB.

A 1200-pF capacitor is fully charged by a 500-V dc power supply. It is disconnected from the power supply and is connected, at t = 0, to a 75-mH inductor. Determine: (a) the initial charge on the capacitor; (b) the maximum current; (c) the frequency f and period T of oscillation; and (d) the total energy oscillating in the system. Solution: a. Q0 = CV = 6.0 x 10-7 C. b. Imax = ωQ0 = 63 mA. c. f = ω/2π = 17 kHz; T = 1/f = 6.0 x 10-5 s. d. U = Q02/2C = 1.5 x 10-4 J.

Solution:

LRC Circuits Any real (nonsuperconducting) circuit will have resistance. Figure An LRC circuit.

LRC Circuits Adding a resistor in an LC circuit is equivalent to adding –IR in the equation of LC oscillation Initially Q=Q0, and the switch is closed at t=0, the current is I=-dQ/dt. The differential equation becomes

LRC Circuits The equation describing LRC circuits now has the same form as the equation for the damped oscillation: The solution to LRC circuits therefore is

LRC Circuits The damped angular frequency is where ω02=1/LC.
The system will be underdamped for R2 < 4L/C, and overdamped for R2 > 4L/C. Critical damping will occur when R2 = 4L/C.

LRC Circuits This figure shows the three cases of underdamping, overdamping, and critical damping. Figure Charge Q on the capacitor in an LRC circuit as a function of time: curve A is for underdamped oscillation (R2 < 4L/C), curve B is for critically damped (R2 = 4L/C), and curve C is for overdamped (R2 > 4L/C).

LRC Circuits Damped oscillations.
At t = 0, a 40-mH inductor is placed in series with a resistance R = 3.0 Ω and a charged capacitor C = 4.8 μF. (a) Show that this circuit will oscillate. (b) Determine the frequency. (c) What is the time required for the charge amplitude to drop to half its starting value? (d) What value of R will make the circuit nonoscillating? Solution: a. R2 is less than 4L/C, so this circuit will oscillate. b. f = ω/2π = 360 Hz. c. t = 2L/R ln 2 = 18 ms. d. For R2 ≥ 4L/C, R must be ≥ 180 Ω.

Solution:

AC Source AC generator can produce The instantaneous current
the emf the current varying sinusoidally in time The instantaneous current The instantaneous voltage

AC Source ω=2πf, the angular frequency; f, the frequency of the source. ψ, the phase difference between current and voltage. Resistors, capacitors, and inductors have different phase relationships between current and voltage when placed in an ac circuit.

A Resistor in an AC Circuit
The current through a resistor is in phase with the voltage. Figure (a) Resistor connected to an ac source. (b) Current (blue curve) is in phase with the voltage (red) across a resistor.

Instantaneous power dissipated in the resistor average value over a complete cycle

root mean square (rms) current rms voltage rms power

An Inductor in an AC Circuit
Figure (a) Inductor connected to an ac source. (b) Current (blue curve) lags voltage (red curve) by a quarter cycle or 90°. Therefore, the current through an inductor lags the voltage by 90°.

The voltage across the inductor is related to the current through it in the form of Ohm’s law: The quantity XL is called the inductive reactance, and has units of ohms:

The instantaneous power supplied to the inductor is: The average power over a complete circle is zero.

Reactance of a coil. A coil has a resistance R = 1.00 Ω and an inductance of H. Determine the current in the coil if (a) 120-V dc is applied to it, and (b) 120-V ac (rms) at 60.0 Hz is applied. Solution: a. For dc current, there is no magnetic induction, so the current is determined by the resistance: I = V/R = 120 A. b. The inductive reactance is XL = 2πfL = 113 Ω. The current is Irms = Vrms/XL = 1.06 A.

Solution: a. For dc current, there is no magnetic induction, so the current is determined by the resistance b. The inductive reactance is The current is

A Capacitor in an AC Circuit
The current in the circuit is charging the capacitor, so i=+dq/dt. Thus dq=idt: The constant depends on the initial condition and we choose it to be zero. The quantity XC is called the capacitive reactance, and (just like the inductive reactance) has units of ohms:

According to the loop rule, v-vc=0: Figure (a) Capacitor connected to an ac source. (b) Current leads voltage by a quarter cycle, or 90°. Therefore, the current leads the voltage by 90°.

The voltage across the inductor is related to the current through it in the form of Ohm’s law: The quantity XC is called the reactance of the capacitor, and has units of ohms:

An Capacitor in an AC Circuit
The instantaneous power supplied to the capacitor is: The average power over a complete circle is zero.

Capacitor reactance. What is the rms current in the circuit shown if C = 1.0 μF and Vrms = 120 V? Calculate (a) for f = 60 Hz and then (b) for f = 6.0 x 105 Hz. Solution: a. XC = 1/2πfC = 2.7 kΩ; I = V/XC = 44 mA. b. Now XC = 0.27 Ω and I = 440 A.

Solution:

AC Circuits with AC Source
This figure shows a high-pass filter (allows an ac signal to pass but blocks a dc voltage) and a low-pass filter (allows a dc voltage to be maintained but blocks higher-frequency fluctuations).

Power in AC Circuits The instantaneous power delivered by the source emf is The average power over a complete circle is

Power in AC Circuits The first term in the integrand contributes because the average of sin2(ωt) is ½ whereas the average of sin(ωt)cos(ωt) is zero. The average power is, therefore, since The rms power becomes

Instantaneous and peak values of current and voltage
Root mean square (rms) values of current and voltage

Ohm’s law in LRC circuits
The impedance and reactance The phase angle

The resonance angular frequency
The rms power delivered by the source emf

Transformers The transformer is a device that can raise or lower the amplitude of ac potential differences. It consists of two coils, either interwoven or linked by an iron core. A changing emf in one induces an emf in the other. The primary coil is connected to the source. While the secondary coil is connected to the load.

Transformers The flux through both coils are the same. The emf’s are
Take the ratio of the emf’s, we find Figure Step-up transformer (NP = 4, NS = 12).

Transformers Energy must be conserved; in the absence of losses, the power supplied by the primary coil and loaded to the secondary coil are the same. Therefore, the ratio of the currents must be the inverse of the ratio of turns:

Transformers Cell phone charger.
The charger for a cell phone contains a transformer that reduces 120-V ac to 5.0-V ac to charge the 3.7-V battery. (It also contains diodes to change the 5.0-V ac to 5.0-V dc.) Suppose the secondary coil contains 30 turns and the charger supplies 700 mA. Calculate (a) the number of turns in the primary coil, (b) the current in the primary, and (c) the power transformed. Solution: a. NP = NS VP/VS = 720 turns. b. IP = IS NS/NP = 29 mA. c. P = ISVS = 3.5 W.

Transformers Transformers work only if the current is changing; this is one reason why electricity is transmitted as ac. Figure The transmission of electric power from power plants to homes makes use of transformers at various stages.

Transformers Transmission lines.
An average of 120 kW of electric power is sent to a small town from a power plant 10 km away. The transmission lines have a total resistance of 0.40 Ω. Calculate the power loss if the power is transmitted at (a) 240 V and (b) 24,000 V. Solution: a. The total current is 120 kW/240 V = 500 A. Then the power loss is I2R = 100 kW. b. Same reasoning, different numbers: the current is 5.0 A, and the power loss is 10 W. This is why electricity is transmitted at very high voltages.

Impedance Matching When one electrical circuit is connected to another, maximum power is transmitted when the output impedance of the first equals the input impedance of the second. The power delivered to the circuit will be a minimum when dP/dt = 0; this occurs when R1 = R2. Figure Output of the circuit on the left is input to the circuit on the right.

Three-Phase AC Transmission lines usually transmit three-phase ac power, with the phases being separated by 120°. This makes the power flow much smoother than if a single phase were used. Figure The three voltages, out of phase by 120° (= 2/3 π radians) in a three-phase power line.

Three-Phase AC Three-phase circuit.
In a three-phase circuit, 266 V rms exists between line 1 and ground. What is the rms voltage between lines 2 and 3? Solution: V0 = 376 V. To find V3 – V2, use trigonometric identities. Result: (V3 – V2)rms = √2 V0 sin π/3 = 460 V.

Solution:

Summary Mutual inductance: Self-inductance:
Energy density stored in magnetic field:

Electromagnetic Induction

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