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Chapter 31B - Transient Currents and Inductance A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

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Objectives: After completing this module, you should be able to: Define and calculate inductance in terms of a changing current.Define and calculate inductance in terms of a changing current. Discuss and solve problems involving the rise and decay of current in capacitors and inductors.Discuss and solve problems involving the rise and decay of current in capacitors and inductors. Calculate the energy stored in an inductor and find the energy density.Calculate the energy stored in an inductor and find the energy density.

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Self-Inductance R Increasing I Consider a coil connected to resistance R and voltage V. When switch is closed, the rising current I increases flux, producing an internal back emf in the coil. Open switch reverses emf. R Decreasing I Lenz’s Law: The back emf (red arrow) must oppose change in flux:

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Inductance The back emf E induced in a coil is proportional to the rate of change of the current I/ t. An inductance of one henry (H) means that current changing at the rate of one ampere per second will induce a back emf of one volt. R Increasing i/ t

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Example 1: A coil having 20 turns has an induced emf of 4 mV when the current is changing at the rate of 2 A/s. What is the inductance? L = 2.00 mH Note: We are following the practice of using lower case i for transient or changing current and upper case I for steady current. R i/ t = 2 A/s 4 mV

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Calculating the Inductance Recall two ways of finding E: Setting these terms equal gives: Thus, the inductance L can be found from: Increasing i/ t R Inductance L

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Inductance of a Solenoid The B-field created by a current I for length l is: and = BA Combining the last two equations gives: R Inductance L l B Solenoid

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Example 2: A solenoid of area 0.002 m 2 and length 30 cm, has 100 turns. If the current increases from 0 to 2 A in 0.1 s, what is the inductance of the solenoid? First we find the inductance of the solenoid: R l A L = 8.38 x 10 -5 H Note: L does NOT depend on current, but on physical parameters of the coil.

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Example 2 (Cont.): If the current in the 83.8- H solenoid increased from 0 to 2 A in 0.1 s, what is the induced emf? R l A L = 8.38 x 10 -5 H

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Energy Stored in an Inductor At an instant when the current is changing at i/ t, we have: Since the power P = Work/t, Work = P t. Also the average value of Li is Li/2 during rise to the final current I. Thus, the total energy stored is: Potential energy stored in inductor: R

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Example 3: What is the potential energy stored in a 0.3 H inductor if the current rises from 0 to a final value of 2 A? U = 0.600 J This energy is equal to the work done in reaching the final current I; it is returned when the current decreases to zero. L = 0.3 H I = 2 A R

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Energy Density (Optional) R l A The energy density u is the energy U per unit volume V Substitution gives u = U/V :

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Energy Density (Continued) R l A Energy density: Recall formula for B-field:

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Example 4: The final steady current in a solenoid of 40 turns and length 20 cm is 5 A. What is the energy density? R l A B = 1.26 mT u = 0.268 J/m 3 Energy density is important for the study of electro- magnetic waves.

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The R-L Circuit R L S2S2 S1S1 V E An inductor L and resistor R are connected in series and switch 1 is closed: i V – E = iR Initially, i/ t is large, making the back emf large and the current i small. The current rises to its maximum value I when rate of change is zero.

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The Rise of Current in L At t = 0, I = 0 At t =, I = V/R At t = , I = V/R The time constant In an inductor, the current will rise to 63% of its maximum value in one time constant = L/R. Time, t I i Current Rise 0.63 I

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The R-L Decay R L S2S2 S1S1 V Now suppose we close S 2 after energy is in inductor: E = iR Initially, i/ t is large and the emf E driving the current is at its maximum value I. The current decays to zero when the emf plays out. For current decay in L: E i

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The Decay of Current in L At t = 0, i = V/R At t =, i = 0 At t = , i = 0 The time constant In an inductor, the current will decay to 37% of its maximum value in one time constant Time, t I i Current Decay 0.37 I

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Example 5: The circuit below has a 40-mH inductor connected to a 5- resistor and a 16-V battery. What is the time constant and what is the current after one time constant? 5 5 L = 0.04 H 16 V R Time constant: = 8 ms After time i = 0.63(V/R) i = 2.02 A

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The R-C Circuit R C S2S2 S1S1 V E Close S 1. Then as charge Q builds on capacitor C, a back emf E results: i V – E = iR Initially, Q/C is small, making the back emf small and the current i is a maximum I. As the charge Q builds, the current decays to zero when E b = V.

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Rise of Charge t = 0, Q = 0, I = V/R t =, i =, Q m = C V t = , i = , Q m = C V The time constant In a capacitor, the charge Q will rise to 63% of its maximum value in one time constant Of course, as charge rises, the current i will decay. Time, t Q max q Increase in Charge Capacitor 0.63 I

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The Decay of Current in C At t = 0, i = V/R At t =, i = 0 At t = , i = 0 The time constant The current will decay to 37% of its maximum value in one time constant the charge rises. Time, t I i Current Decay Capacitor 0.37 I As charge Q increases

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The R-C Discharge Now suppose we close S 2 and allow C to discharge: E = iR Initially, Q is large and the emf E driving the current is at its maximum value I. The current decays to zero when the emf plays out. For current decay in L: R S2S2 S1S1 V i C E

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Current Decay At t = 0, I = V/R At t =, I = 0 At t = , I = 0 In a discharging capacitor, both current and charge decay to 37% of their maximum values in one time constant = RC. Time, t I i Current Decay Capacitor 0.37 I As the current decays, the charge also decays:

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Example 6: The circuit below has a 4- F capacitor connected to a 3- resistor and a 12-V battery. The switch is opened. What is the current after one time constant ? Time constant: = 12 s After time i = 0.63(V/R) i = 2.52 A 3 3 C = 4 F 12 V R = RC = (3 )(4 F)

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Summary R l A Potential Energy Energy Density:

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Summary In an inductor, the current will rise to 63% of its maximum value in one time constant = L/R. Time, t I i Current Rise 0.63I Inductor The initial current is zero due to fast-changing current in coil. Eventually, induced emf becomes zero, resulting in the maximum current V/R.

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Summary (Cont.) The current will decay to 37% of its maximum value in one time constant = L/R. Time, t I i Current Decay 0.37I Inductor The initial current, I = V/R, decays to zero as emf in coil dissipates.

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Summary (Cont.) When charging a capacitor the charge rises to 63% of its maximum while the current decreases to 37% of its maximum value. Time, t Q max q Increase in Charge Capacitor 0.63 I Time, t I i Current Decay Capacitor 0.37 I

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CONCLUSION: Chapter 31B Transient Current - Inductance

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