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Chapter 31 Inductance Circuits
- 1. Chapter 31: Inductance 1 1. Define (Self) Inductance 2. Inductance of a Solenoid 3. RL Circuits 4. Energy Stored in an Inductor With Current 5. LC Oscillation Circuit 6. The RLC Circuit ₢
- 2. Magnetism 1. Inductor 2. Solenoid Electricity 1. Capacitor 2. Parallel Plate 𝐼 Area: A Distance: d Ideal capacitor Infinite area Draw // plate Draw a solenoid Length: L Number of turns: N Ideal solenoid Length: infinite Number of turns density: n Parallel Plate and Solenoid 2
- 3. Magnetism 1. Inductor 2. Solenoid Electricity 1. Capacitor 2. Parallel Plate 𝐶 Inductor 𝐿 Capacitor ** When current in a solenoid changes, it plays a role like a battery. 𝜀𝐿 = −𝐿 Δ𝐼 Δ𝑡 CΔV = Q 3 Capacitor and Inductor Circuit Elements Circuit symbol Circuit symbol Key equation Key equation
- 4. 𝐼 𝐵 = 𝜇0𝑛𝐼 𝐼 ↑ 𝐵 ↑ ∆Φ𝐵 ≠ 0 𝜀𝐿 Induced EMF (self-induced) When current in a solenoid changes, it plays a role like a battery. 𝜀𝐿 = −𝐿 Δ𝐼 Δ𝑡 𝜀 = −𝑁 ∆Φ𝐵 ∆𝑡 𝑁Φ𝐵 = 𝐿𝐼 4 Define (Self) Inductance 𝜀 = −𝑁 ∆Φ𝐵 ∆𝑡 Key equation in inductor circuit This equation is usefulfor calculate the inductance of a given inductor.
- 5. Each Circuit Element has a Key Equation Magnetism 1. Inductor 2. Solenoid Electricity 1. Capacitor 2. Parallel Plate 𝐶 Inductor 𝐿 Capacitor 𝐼 𝐵 = 𝜇0𝑛𝐼 𝐼 ↑ 𝐵 ↑ ∆Φ𝐵 ≠ 0 𝜀𝐿 Induced emf (self-induced) ** When current in a solenoid changes, it plays a role like a battery. 𝜀𝐿 = −𝐿 d𝐼 d𝑡 𝜀 = −𝑁 𝑑Φ𝐵 𝑑𝑡 𝑁Φ𝐵 = 𝐿𝐼 CΔV = Q
- 6. 𝐼 𝐵 = 𝜇0𝑛𝐼 𝐼 𝐵 Φ𝐵 = 𝐵𝐴 𝐿 𝑁Φ𝐵 = 𝐿𝐼 6 Inductance of a Solenoid This equation is usefulfor calculate the inductance of a given inductor. 𝑁Φ𝐵 = 𝐿𝐼 𝐿 = 𝜇0𝑛2 𝑉 𝑉 = 𝐴𝑙
- 7. Self Inductance give 𝐼 𝐵 Φ𝑩 = න𝐵 ∙ 𝑑𝐴 𝐿 𝑁Φ𝐵 = 𝐿𝐼 𝑁Φ𝐵 = 𝐿𝐼 EX-U01 Coaxial cables are often used to connect electrical devices. Model a long coaxial cable as two thin, concentric, cylindrical conducting shells of radii a and b and length as in the Figure. The conducting shells carry the same current I in opposite directions. Calculate the inductance L of this cable. Calculate the inductance in Coaxial Cable (Calculus) Ch29 page 46: EX-U04 𝐵 = 𝜇0𝐼 2𝜋𝑟 𝑑𝛷𝐵 = 𝐵𝑑𝐴 = 𝐵(𝑏𝑑𝑟)
- 8. 𝐿 = 𝜇0𝑛2 𝑉 𝑉 = 𝐴𝑙 a) Calculate the inductance of a solenoid containing 300 turns if the length of the solenoid is 25.0 cm and its cross-sectional area is 4.00 × 10−4 𝑚2 . b) Calculate the self-induced EMF in the solenoid described in part (a) if the current in the solenoid decreases at the rate of 50.0 A/s. EX-11 8 Inductance of a Solenoid 𝜀𝐿 = −𝐿 d𝐼 d𝑡
- 9. 𝐿 = 𝜇0𝑛2 𝑉 𝑉 = 𝐴𝑙 9 Inductance of a Solenoid 𝜀𝐿 = −𝐿 Δ𝐼 Δ𝑡 A solenoid of radius 2.5 cm 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 75 mV. HW
- 10. Chapter 31: Inductance 10 1. Define (Self) Inductance 2. Inductance of a Solenoid 3. RL Circuits 4. Energy Stored in an Inductor With Current 5. LC Oscillation Circuit 6. The RLC Circuit
- 11. Inductor: used to “store current”. When “charging an inductor, we can think of an inductor as magnetic container used to store electric current; like a water container to store water. Water source connect to a container When the water pipeline opens at 𝑡 = 0, the water begins to fill the container. The amount of water 𝑞 will change as time 𝑡 changes. 𝑡 = 0 𝑞 = 0 𝑡 = ∞ (Infinite) 𝑞 = 𝑞𝑚𝑎𝑥 = 𝑄 𝑡 (any time) 𝑞(𝑡) The water filling rate will decay as the water level in the container increases. In the beginning, the amount of water in the container = 0 Since the source is big and its level never changes, this is the maximum storagecapacity of the container. RL Circuit: “Charging an Inductor” 11
- 12. Charging an inductor is similar to storing water in a container. Here ➢ Battery ↔ water source ➢ Inductor ↔ Container ➢ Current↔ the amount of water ➢ Currentin the inductor ↔ the amount of water in the container Inductor: used to “store current”. When “charging an inductor, we can think of an inductor as a magnetic container used to store electric current, like water containers used to store water. Because RL circuits are structurally similar to RC circuits, we choose the similar language in describing their functions (“charging an inductor” or “discharging an inductor”). But unlike capacitors, inductors cannot hold currentstatically. RL Circuit: “Charging an inductor” 12
- 13. 𝑡 = 0 𝐼 = 0 𝑡 = ∞ (Infinite) 𝐼 = 𝐼𝑚𝑎𝑥 = 𝐼𝑚 𝑡 (any time) 𝐼(𝑡) The function of current vs. time mathematicalbehavior should be as the following: 𝐼 𝑡 𝐼 𝑡 𝐼𝑚 𝐼 = 0 D𝐫𝐚𝐰 𝐭𝐡𝐞 𝐂𝐮𝐫𝐯𝐞 𝐨𝐟 𝐼(𝑡) 𝐼 𝑡 𝑡 = 0 𝐼 = 0 𝑡 = ∞ 𝐼 = 𝐼𝑚 𝐼𝑚 The charging rate decays (slope decrease). RL Circuit: “Charging an inductor” 13
- 14. Mathematicalform 𝐼(𝑡) 𝐼 𝑡 = 𝐼𝑚(1 − 𝑒− 𝑡 𝜏) Check 1. t = 0, 𝐼 = 0; 2. t = ∞, 𝐼 = 𝐼𝑚; 3. t = τ, 𝐼 = 0.63𝐼𝑚 = 63% 𝐼𝑚; 𝜏 called time constant (this constanttells us how fast it takes to “charge the inductor) When time reaches 𝜏 (time constant) Current in the inductor fills to 63% of its maximum value 𝐼𝑚 Charging Curve 𝐼(𝑡) The function of current vs. time mathematicalbehavior should be as the following: 𝐼 𝑡 𝑡 = 0 𝐼 = 0 𝑡 = ∞ 𝐼 = 𝐼𝑚 𝐼𝑚 The charging rate decays (slope decrease). RL Circuit: “Charging an inductor” 14
- 15. RC Charging Circuit 𝜀 𝑅 𝐶 𝐴 𝐵 𝐷 𝐶 RC Charging Circuit Equation Apply to Loop rule to the circuit 𝜀 − 𝐼𝑅 − 𝑞 𝐶 = 0 (C1) RL Charging Circuit RC Charging Circuit Equation Apply to Loop rule to the circuit 𝜀 − 𝐼𝑅 − 𝐿 d𝐼 d𝑡 = 0 (𝐿1) 𝜀 𝑅 𝐿 𝐴 𝐵 𝐷 𝐶 𝜀 − 𝑅 d𝑞 d𝑡 − 𝑞 𝐶 = 0 (C1) 15 RC Charging Vs. RL Charging Circuit 𝜀 − 𝐿 d𝐼 d𝑡 − 𝐼𝑅 = 0 (𝐿1) Same Mathematical form Solutions are the same form
- 16. RC Charging Circuit RL time constant 𝜏 = 𝐿 𝑅 When time reaches 𝜏 (time constant) Current in the inductor fills 63% of its maximum value 𝐼𝑚 𝜀 𝑅 𝐶 𝐴 𝐵 𝐷 𝐶 𝐼𝑚 = 𝜀 𝑅 𝐼 𝑡 = 𝐼𝑚(1 − 𝑒− 𝑡 𝜏) 16 RC Charging Vs. RL Charging Circuit RL Charging Circuit 𝜀 𝑅 𝐿 𝐴 𝐵 𝐷 𝐶 RC time constant 𝜏 = 𝑅𝐶 𝑄 = 𝜀𝑅 𝑞 𝑡 = 𝑄(1 − 𝑒− 𝑡 𝜏) When time reaches 𝜏 (time constant) Charge in the capacitor fills 63% of its maximum value 𝑄
- 17. RC discharging Circuit 𝑅 𝐶 𝐴 𝐵 𝐷 𝐶 RC Charging Circuit Equation Apply Loop rule to the circuit −𝐼𝑅 − 𝑞 𝐶 = 0 (C1) RL “discharging” Circuit RC Charging Circuit Equation Apply Loop rule to the circuit −𝐼𝑅 − 𝐿 d𝐼 d𝑡 = 0 (𝐿1) 𝑅 𝐿 𝐴 𝐵 𝐷 𝐶 −𝑅 d𝑞 d𝑡 − 𝑞 𝐶 = 0 (C1) 17 RC Discharging Vs. RL Discharging Circuit −𝐿 d𝐼 d𝑡 − 𝐼𝑅 = 0 (𝐿1) Same Mathematical form Solutions are the same form
- 18. RC discharging Circuit RL time constant 𝜏 = 𝐿 𝑅 𝑅 𝐶 𝐴 𝐵 𝐷 𝐶 Current: 𝐼 𝑡 = 𝐼𝑚 𝑒− 𝑡 𝜏 RL “discharging” Circuit 𝑅 𝐿 𝐴 𝐵 𝐷 𝐶 When time reaches to 𝜏 (time constant) 63% of current is gone (37% left) 18 RC Discharging Vs. RL Discharging Circuit RC time constant 𝜏 = 𝑅𝐶 Current: 𝑞 𝑡 = 𝑄 𝑒− 𝑡 𝜏 When time reaches to 𝜏 (time constant) 63% of charge is gone (37% left)
- 19. EX-12 A 12.0 V battery is in a circuit with a 30.0-mH inductor and a 0.150-Ω resistor, as in shown in the Figure. The switch is closed at 𝑡 = 0. a) Find the time constant of the circuit. b) Find the current after one time constant has elapsed. c) Find the voltage drops across the resistor when 𝑡 = 0 and 𝑡 = one time constant. d) What’s the rate of change of the current after one time constant? a 𝜏 = 𝐿 𝑅 = 0.20𝑠 b 𝐼𝑚 = 𝜀 𝑅 = 80.0𝐴 c 𝐼 𝜏 = 0.63𝐼𝑚 = 50𝐴 ∆𝑉𝑅 = 𝐼 𝜏 𝑅 = 50 × 0.15 = 7.5𝑉 d Δ𝐼 Δ𝑡 =? 𝜀 − 𝐼 𝜏 𝑅 − 𝐿 Δ𝐼 Δ𝑡 = 0 12.0 − 7.5 − 0.030 Δ𝐼 Δ𝑡 = 0 19 𝜀 𝑅 𝐿 𝐴 𝐵 𝐷 𝐶 RL Charging Circuit
- 20. a 𝜏 = 𝐿 𝑅 = 2.00𝑚𝑠 b 𝐼 250𝑚𝑠 =? 𝐼 250𝑚𝑠 = 𝐼𝑚 c 𝐼𝑚 = 𝜀 𝑅 d 𝐼 𝑡 = 𝐼𝑚 1 − 𝑒− 𝑡 𝜏 𝐼 𝑡 = 0.80 𝐼𝑚 Consider the circuit shown in the Figure. Take 𝜀 = 6.00 𝑉, 𝐿 = 8.00 𝑚𝐻, and 𝑅 = 4.00 Ω. a) What is the inductive time constant of the circuit? b) Calculate the current in the circuit 250 ms after the switch is closed. c) What is the value of the final steady-state current? d) How long does it take the current to reach 80.0% of its maximum value? HW 20 𝜀 𝑅 𝐿 𝐴 𝐵 𝐷 𝐶
- 21. a 𝜏 = 𝐿 𝑅 b 𝐼𝑚 = 𝜀 𝑅 b 𝐼 𝑡 = 𝐼𝑚 1 − 𝑒− 𝑡 𝜏 c 𝐼 𝑡 𝑅 = 𝜀𝐿 = −𝐿 d𝐼 d𝑡 As shown in the Figure. Suppose the circuit elements have the following values: 𝜀 = 12.0, 𝑅 = 6.00 Ω, and 𝐿 = 30.0 𝑚𝐻. (A) Find the time constant of the circuit. (B) Switch S2 is at position a, and switch S1 is thrown closed at t = 0. Calculate the current in the circuit at t = 2.00 ms. (C) Compare the potential difference across the resistor with that across the inductor. EX-U The RL circuit as shown in the Figure, with switch 𝑆2 at position a and the current having reached its steady-state value. When 𝑆2 is thrown to position b, the current in the right-hand loop decays exponentially with time according to the expression 𝐼 = 𝐼𝑖𝑒 − 𝑡 𝜏, where 𝐼𝑖 = 𝜀/𝑅 is the initial current in the circuit and 𝜏 = 𝐿/𝑅 is the time constant. Show that all the energy initially stored in the magnetic field of the inductor appears as internal energy in the resistor as the current decays to zero. EX-U02
- 22. HW-U A 12.0-V battery is connected into a series circuit containing a 10.0-Ω resistor and a 2.00-H inductor. In what time interval will the current reach (a) 50.0% and (b) 90.0% of its final value? Consider the circuit in the Figure, taking 𝜀 = 6.00 𝑉, 𝐿 = 8.00𝑚𝐻, and 𝑅 = 4.00 Ω. (a) What is the inductive time constant of the circuit? (b) Calculate the current in the circuit 250 ms after the switch is closed. (c) What is the value of the final steady-state current? (d) After what time interval does the current reach 80.0% of its maximum value? A 12.0-V battery is connected in series with a resistor and an inductor. The circuit has a time constant of 500 μs, and the maximum current is 200 mA. What is the value of the inductance of the inductor?
- 23. Chapter 31: Inductance 23 1. Define (Self) Inductance 2. Inductance of a Solenoid 3. RL Circuits 4. Energy Stored in an Inductor With Current 5. LC Oscillation Circuit 6. The RLC Circuit
- 24. Magnetism Electricity 𝐿 = 𝜇0𝑛2 𝑉 𝑉 = 𝐴𝑙 𝐵 = 𝜇0𝑛𝐼 Self-inductance 𝜀 = −𝐿 𝑑𝐼 𝑑𝑡 𝑁Φ𝐵 = 𝐿𝐼 Solenoid inductor 𝐿 = 𝜇0𝑛2 𝑉 𝑈𝐵 = 1 2 𝐿𝐼2 V: Volume Magnetic Energy 𝑈𝐵 = 𝑢𝐵𝑉 𝑢𝐵 = 1 2𝜇0 𝐵2 Capacitance 𝐶∆𝑉 = 𝑄 𝐶 = 𝜀0 𝐴 𝑑 Energy 𝑈𝑒 = 1 2 𝑄2 𝐶 𝑈𝑒 = 𝑢𝑒𝑉 𝑉: 𝑣𝑜𝑙𝑢𝑚𝑒 𝑢𝑒 = 1 2 𝜀0𝐸2 𝑈𝐵 = 1 2 𝐿𝐼2 24 Magnetic Energy Stored in an Inductor Capacitor with charge 𝑄 Store electric energy 𝑈𝑒 = 1 2 𝑄2 𝐶 Inductor with current 𝐼 store magnetic energy
- 25. A 300-turn solenoid has a radius of 5.00 cm and a length of 20.0 cm. Find (a) the inductance of the solenoid and (b) the energy stored in the solenoid when the current in its windings is 0.500 A. HW 25
- 26. LC Oscillation Circuit 𝑚 𝑘 𝐸𝑡𝑜𝑡𝑎𝑙 = 1 2 𝑚𝑣2 + 1 2 𝑘𝑥2 𝑈𝑡𝑜𝑡𝑎𝑙 = 1 2 𝐿𝐼2 + 1 2𝐶 𝑄2 = 1 2 𝐿𝐼𝑚 2 = 1 2𝐶 𝑄𝑚 2 𝑑𝐸𝑡𝑜𝑡𝑎𝑙 𝑑𝑡 = 0 = 1 2 𝑚𝑣𝑚 2 = 1 2 𝑘𝐴2 𝑑2 𝑥 𝑑𝑡 + 𝜔2 𝑥 = 0 𝑥 = 𝐴 cos(𝜔𝑡 + 𝜑) 𝑑2𝑄 𝑑𝑡 + 𝜔2𝑄 = 0 𝑄 = 𝑄𝑚 cos(𝜔𝑡 + 𝜑) 𝜔 = 1 𝐿𝐶 natural frequency of LC circuit 𝜔 = 𝑘 𝑚 angular frequency 𝑑𝑈𝑡𝑜𝑡𝑎𝑙 𝑑𝑡 = 0
- 27. LC Oscillation Circuit 𝑈𝑡𝑜𝑡𝑎𝑙 = 1 2 𝐿𝐼2 + 1 2𝐶 𝑄2 = 1 2 𝐿𝐼𝑚 2 = 1 2𝐶 𝑄𝑚 2 𝑄 = 𝑄𝑚 cos(𝜔𝑡 + 𝜑) 𝜔 = 1 𝐿𝐶 natural frequency of LC circuit EX-U03 As shown in the Figure, the battery has an emf of 12.0 V, the inductance is 2.81 mH, and the capacitance is 9.00 pF. The switch has been set to position a for a long time so that the capacitor is charged. The switch is then thrown to position b, removing the battery from the circuit and connecting the capacitor directly across the inductor. (A) Find the frequency of oscillation of the circuit. (B) What are the maximum values of charge on the capacitor and current in the circuit?
- 28. LC Oscillation Circuit An LC circuit consists of a 20.0-mH inductor and a 0.500-μF capacitor. If the maximum instantaneous current is 0.100 A, what is the greatest potential difference across the capacitor? An LC circuit like the one in the Figure contains an 82.0- mH inductor and a 17.0-mF capacitor that initially carries a 180-𝜇𝐶 charge. The switch is open for t < 0 and then thrown closed at t = 0. (a) Find the frequency (Hz) of the resulting oscillations. (b) At t = 1.00 ms, find the charge on the capacitor and (c) the current in the circuit. HW-U 28 ₢