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**Chapter 31B - Transient Currents and Inductance**

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. Calculate the energy stored in an inductor and find the energy density. Discuss and solve problems involving the rise and decay of current in capacitors and inductors.

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**The back emf (red arrow) must oppose change in flux:**

Self-Inductance 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 Increasing I 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 DI/Dt. R Increasing Di/ Dt 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.

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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? R Di/ Dt = 2 A/s 4 mV 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.

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**Calculating the Inductance**

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

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**Inductance of a Solenoid**

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

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**First we find the inductance of the solenoid:**

Example 2: A solenoid of area m2 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**

Example 2 (Cont.): If the current in the 83.8-mH 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 Di/Dt, we have: R Since the power P = Work/t, Work = P Dt. 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:

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**Example 3: What is the potential energy stored in a 0**

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? L = 0.3 H I = 2 A R U = J This energy is equal to the work done in reaching the final current I; it is returned when the current decreases to zero.

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**Energy Density (Optional)**

The energy density u is the energy U per unit volume V R l A Substitution gives u = U/V :

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**Energy Density (Continued)**

l A 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 Energy density is important for the study of electro-magnetic waves. u = J/m3

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The R-L Circuit R L S2 S1 V An inductor L and resistor R are connected in series and switch 1 is closed: E i V – E = iR Initially, Di/Dt 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 t i I At t = 0, I = 0 Current Rise**

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

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**The R-L Decay V Now suppose we close S2 after energy is in inductor:**

E = iR For current decay in L: Initially, Di/Dt 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.

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**The Decay of Current in L**

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

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

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The R-C Circuit V Close S1. Then as charge Q builds on capacitor C, a back emf E results: S1 E S2 i V – E = iR R C 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 Eb = V.

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**Rise of Charge t q Capacitor Qmax t = 0, Q = 0, I = V/R**

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

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**The Decay of Current in C**

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

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**The R-C Discharge V Now suppose we close S2 and allow C to discharge:**

E = iR For current decay in L: 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.

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**Current Decay t i Capacitor I Current Decay At t = 0, I = V/R Time, t**

As the current decays, the charge also decays: In a discharging capacitor, both current and charge decay to 37% of their maximum values in one time constant t = RC.

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

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**Potential Energy Energy Density:**

Summary R l A Potential Energy Energy Density:

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**Summary Time, t I i Current Rise t 0.63I Inductor**

In an inductor, the current will rise to 63% of its maximum value in one time constant t = L/R. 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.) Time, t I i Current Decay t 0.37I Inductor**

The initial current, I = V/R, decays to zero as emf in coil dissipates. The current will decay to 37% of its maximum value in one time constant t = L/R.

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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 Qmax q Increase in Charge Capacitor t 0.63 I Time, t I i Current Decay Capacitor t 0.37 I

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**CONCLUSION: Chapter 31B Transient Current - Inductance**

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