HW 17 solution

# HW 17 solution - song(shs546 – Homework 17 – weathers...

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Unformatted text preview: song (shs546) – Homework 17 – weathers – (22202) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points An inductor and a resistor are connected with a double pole switch to a battery as shown in the figure. The switch has been in position b for a long period of time. 106 mH 5 . 26 Ω 5 . 8 V S b a If the switch is thrown from position b to position a (connecting the battery), how much time elapses before the current reaches 178 mA? Correct answer: 3 . 54786 ms. Explanation: Let : R = 5 . 26 Ω , L = 106 mH , and E = 5 . 8 V . L R E S b a The time constant of an RL circuit is τ = L R = . 106 H 5 . 26 Ω = 0 . 0201521 s . The final current reached in the circuit is I = E R = 5 . 8 V 5 . 26 Ω = 1 . 10266 A . The switch is in position a in an RL circuit connected to a battery at t = 0 when I = 0. Then the current vs. time is I = I parenleftBig 1 − e − t / τ parenrightBig . Solving the above expression for t , when I = I 1 gives t 1 = − τ ln parenleftbigg 1 − I 1 I parenrightbigg = − (0 . 0201521 s) ln parenleftbigg 1 − . 178 A 1 . 10266 A parenrightbigg = 3 . 54786 ms . 002 (part 2 of 3) 10.0 points What is the maximum current in the inductor a long time after the switch is in position a ? Correct answer: 1 . 10266 A. Explanation: After a long time compared to τ , we have a d.c. circuit with a battery supplying an emf E , which is equal to the voltage drop I R across the resistor. Thus I = E R = 5 . 8 V 5 . 26 Ω = 1 . 10266 A . 003 (part 3 of 3) 10.0 points The switch has brushes within it so that the switch can be thrown from a to b without internal sparking. Now the switch is smoothly thrown from a to b , shorting the inductor and resistor. How much time elapses before the current falls to 98 mA? Correct answer: 48 . 7784 ms. Explanation: The current decay in an RL circuit when there is no voltage source present, and the initial current is I , is I = I e − t /τ . Solving the above equation for t , we obtain t = − τ ln parenleftbigg I I parenrightbigg . song (shs546) – Homework 17 – weathers – (22202) 2 The time t 3 that elapses for the current to fall to I 3 = 0 . 098 A is t 3 = − τ ln parenleftbigg I 3 I parenrightbigg = − (0 . 0201521 s) ln parenleftbigg . 098 A 1 . 10266 A parenrightbigg = 48 . 7784 ms . 004 (part 1 of 3) 10.0 points A long solenoid carries a current I 2 . Another coil (of larger diameter than the solenoid) is coaxial with the center of the solenoid, as in the figure below. ℓ 2 ℓ 1 Outside solenoid has N 1 turns Inside solenoid has N 2 turns A 1 A 2 The current I 2 is held constant....
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HW 17 solution - song(shs546 – Homework 17 – weathers...

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