AP Physics C Inductance Free Response Problems 1. Two toroidal solenoids are wounded around the same frame. Solenoid 1 has 800 turns and solenoid 2 has 500 turns. When the current 7.23 A flows through solenoid 1, the average flux through each turn in solenoid 2 is 0.043 Wb. a. What is the mutual inductance of the system of two solenoids? b. When the current in solenoid 2 is 3.45 A, what is the average flux though each turn in solenoid 1? c. In the new trial with the same solenoids, the current in solenoid 1 changes at the rate of - 0.25 A/s, what is the induced emf in solenoid 2? 2. A long solenoid 1 is wounded with N 1 turns of wire. A coil with N2 turns surrounds the solenoid 1. When the current in solenoid 1 decreases at a rate of -0.34 A/s, the induced emf in the second coil is 0.0051 V. a. What is the mutual inductance of two coils? b. If the second coils has 50 turns and the current in the first coil is 1.70 A, what is the flux through each turn in the second coil? c. If the current in the second coil decreases at a rate of 0.450 A/s, what is the induced emf in the first coil?
3. An inductor has inductance 0.43 H and carries an electric current in the direction shown on the diagram. The current varies with time at a constant rate. The potential difference between points a and b is 0.78 V. a. Which point a or b has higher potential? Explain. b. Is the current increasing or decreasing? Explain. c. If the current at t=0 is 1.8 A, what is the current at t=1.5 s? 4. In the circuit above Ɛ=12 V, R = 24 Ω and L = 0.15 H. Initially, there is no current in the circuit and the switch is open. a. Just after the switch is closed, what is the voltage across the resistor and the inductor? b. Long time after the switch closed, what is the voltage across the resistor and the inductor? c. What is the current in the circuit when the equilibrium is reached? d. How much energy is stored the inductor when the equilibrium is reached?
5. In the circuit above, Ɛ=30 V, R = 120 Ω and L = 0.24 H. Initially, switch 2 is open and switch 1 is closed until a constant current is reached. Then at time t = t 1 switch 1 is open and switch 2 is closed. a. What is the current in the resistor at time t 1? b. What is the voltage across the resistor at time t 1? c. What is the current in the resistor at time t = 1.0 10-3 s? d. What is the voltage across the inductor at time t 1? e. How long will it take the current to decrease to half its initial value? 6. In the circuit above, the switch is closed for a long time. a. What is the current in the circuit? After being closed for a long time, the switch is reopen at time t = 0. b. Determine the current in the circuit after the switch has been open for a long time. c. On the axes below, graph the current as a function of time from t = 0.
d. By applying Kirchhoff s loop rule, write the equation to relate potential difference around the circuit. e. Find the equation for the current as a function of time from t = 0. 7. In the circuit above the switch is closed for a long time. a. Determine the current flowing through point A. A 5 µf capacitor is connected to the circuit between point A and B. The switch is closed for a long time and equilibrium is reached. b. Determine the current flowing through point A. c. Determine the charge on the capacitor. The capacitor is replaced with a 4 mh inductor with zero resistance. The switch is closed for a long time and equilibrium is reached. d. Determine the current flowing through the inductor. e. Determine the energy stored in the inductor.
8. In the circuit above, the switch is initially open and all currents are zero. For the time immediately after the switch is closed answer the following: a. The voltage across the 25 Ω resistor. b. The rate of change of current in the inductor. The switch is closed for a long time. Determine the following: c. The current in the inductor d. The energy stored in the inductor. After a long time the switch is reopened. e. Find the instantaneous voltage across 25 Ω resistor. f. On the axes below, graph the voltage as a function of time across 25 Ω resistor from the time when the switch is reopened.
9. In the circuit above the switch is initially open and all currents are zero. At time t = 0, the switch is closed to position A. a. Determine the current immediately after the switch is closed. b. Determine the current when the switch is closed for a longtime. c. On the axes below, graph the current as a function of time after the switch is closed. d. Determine the energy stored in the inductor when the equilibrium is reached. e. Determine the current immediately after the switch position is changed. f. Determine the voltage across the inductor immediately after the switch position is changed.
10. In the circuit above, the switch is initially open all currents are zero and the capacitor is uncharged. a. At time t = 0, the switch is closed to position A. Determine the voltage across 15 Ω resistor. b. After a long time when the switch is closed to position A, determine the voltage across 15 Ω resistor. c. After a long time when the switch is closed to position A, determine the charge on the capacitor. d. At new time t = t 1, the switch is closed to position B, determine the voltage across 15 Ω resistor at time t 1. e. Determine the current in 15 Ω resistor after long time from t 1. f. Determine the energy stored in the inductor long time after t 1.
11. In the circuit above, the switch is initially open and all currents are zero, the capacitor is uncharged. a. Determine the current and voltage distribution in each part of the circuit immediately after the switch is closed. b. Determine the current and voltage distribution in each part of the circuit after the switch is closed for a long time. c. Determine the charge on the capacitor and energy stored in the inductor after the switch is closed for a long time. d. On the axes below, graph voltage as a function of time across the capacitor and the current as a function of time in the inductor after the switch is closed.
12. In the circuit above, the switches are initially open and all currents are zero. At time t = 0 switch 1 is closed and switch 2 is left open. a. Determine the current in 6 Ω resistor at time t = 0. b. Determine the voltage across 6 Ω resistor at time t = 0. Then switch 1 has been closed for a long time. c. Determine the current in 6 Ω resistor. d. Determine the voltage across 6 Ω resistor. e. Derive an expression for the current through 6 Ω resistor from time t = 0. At new time t = t 1, switch 1 is open and switch 2 is closed. f. Determine the current and voltage across 6 Ω resistor at time t 1.
Answer Key 1. 2. 3. 4. 5. 6. a. M = N 2Φ 2 b. Φ= M i 2 N 1 = = (500)(0.043Wb) i 1 7.23 A (2.97 H)(3.45A) 800 =2.97 H =0.013 Wb c. Ɛ 2 =-M di =-(2.97 H)(-0.25A/s)=0.74 V dt a. Ɛ 2 =- M di 1, M= - Ɛ 2 = - 0.0051V =.015 H dt di 1 /dt ( 0.34A ) b. N 2 Φ 2 =Mi 1, Φ 2 = Mi 1 N 2 = s (0.015 H)(1.7A) 50 =5.1 10-4 Wb c. Ɛ 1 =-M di 2 dt =-(0.015 H)(-0.45 A/s)=6.75 10-3 V a. Point b has higher potential since potential difference is positive and the current is flowing to the left. b. The current is increasing because the potential drops from point b to point a and the potential difference is positive. c. V=L di, dt Δi=VΔt L =(0.78 V)(1.5 s) =2.21 A, i 0.43 H 2 =i 1 + Δi=2.21 A+1.8 A=4.01 A a. V R =0, V L =12V b. V R =12 V, V L =0 c. I = Ɛ V =12 =0.5 A R 24Ω d. U = 1/2LI 2 =0.019 J a. I = Ɛ R = 0.25 A b. V = 30 V c. I = I 0 e t τ = 0.25 A(e 0.001s 0.002s )=0.15 A d. V = I R = 18 V e. I = I 0 e t τ, solve for t when I = ½ I 0, t = 1.38 10-3 s a. I = Ɛ/R1 b. I = Ɛ/(R1=R2)
7. 8. c. d. Ɛ-i(R1+R2)-L di dt =0 e. I= Ɛ [(1 (1 R 1+R 2 R 1 +R 2 R 1 )e (R1+R2)t L ] a. R net= 2.8 + 12/2=8.8 Ω, I net=30v/8.8 Ω=3.4 A, I A = I net/2= 1.7 A b. When t>>0 I AB = 0 and capacitor can be taken out of the circuit, I A= 1.7 A c. V AB = 10.2 V, Q = C V AB = (5 10-6 F)(10.2 V)=51 µc d. R L = 0, point A and B can be combined in one and the new resistance is 7.3 Ω, I net =4.12 A, I L = I 3 Ω-I 9 Ω = 2.01 A e. U = ½ LI 2 = 0.5(4 10-3 H)(2.01 A) 2 = 8.1 mj a. I= Ɛ/(R1+R2)=60V/30 Ω=2A, V=60V-10V=50V b. L di = V =50V/0.1H=500 A/s dt L c. I = Ɛ/R=60V/5 Ω=12 A d. U = ½ LI 2 =0.5(0.1 H)(12A) 2 = 7.2 J e. V= IR = (12 A)(25 Ω)=300 V f. 9. a. I = 0 b. I = 20 V/400 Ω= 0.05 A
c. d. U= ½ LI 2 = 0.5(0.5 H)(0.05 A) 2 =6.25 10-4 J e. I = 0.05 A f. V = IR=(0.05 A)(250 Ω)=12.5 V 10. a. I = Ɛ/(R1+R2)=40V/40 Ω=1 A, V = (1 A)(15 Ω)=15 V b. V = 0 c. Q = CV = (10 10-6 F)(40 V) = 4 10-4 C d. V = 0 e. I = Ɛ/R net=40v/40 Ω=1 A f. U = ½ LI 2 = 0.5(1 H)(1 A) 2 =0.5 J 11. a. V 40 Ω=60 V, I 40 Ω=1.5 A, V 30 Ω= V C= V L= V 60 Ω =0, I L = 0, I C = 0 b. V 40 Ω=40 V, I 40 Ω=1 A, V 30 Ω=V 60 Ω=V C= 20 V, V L = 0, I 30 Ω=2/3 A, I 60 Ω=1/3 A c. Q = CV = (4 10-6 F)(20 V)= 80 µc d. 12. a. I = 0 b. V = 0 c. I = Ɛ/R net= 18V/9Ω=2 A d. V = IR = (2 A)(6Ω) = 12 V e. I = 2(1-e t 4.4 10 4 ) f. I = 2 A