8.02 Pset 9 Soln

# 8.02 Pset 9 Soln - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2008 Problem Set 9 Solutions Due Wednesday April 16 at

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2008 Problem Set 9 Solutions Due: Wednesday, April 16 at 11 am. Hand in your problem set in your section slot in the boxes outside the door of 32- 082. Make sure you clearly write your name, section, table, group (e.g. L01 Table 3 Group A) Problem 1: Read Experiment 6: Inductance and RL Circuits Pre-Lab Questions (10 points) 1. RL Circuits (3 points) Consider the circuit at left, consisting of a battery (emf ε ), an inductor L , resistor R and switch S . For times t < 0 the switch is open and there is no current in the circuit. At t = 0 the switch is closed. (a) Using Kirchhoff’s loop rules (really Faraday’s law now), write a differential equation relating the emf on the battery, the current in the circuit and the time derivative of the current in the circuit. Walking in the direction of current, starting at the switch 0 dI IR L dt ε −= (b) The solution to your differential equation should look: / () ( ) t It AX e τ =− where A , X , and are constants. Plug this expression into the differential equation you obtained in (a) in order to confirm that it indeed is a solution and to determine what the time constant and the constants A , and X are. What would be a better label for A ? (HINT: You will also need to use the initial condition for current. What is (0 ) = ?)

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() 0 t tt Ae A A X e R L ARX AR L e τ εε ττ −− ⎛⎞ =− = − + ⎜⎟ ⎝⎠ Both the constant and time dependent part must equal zero, giving us two equations. The third (because there are three unknowns) we can get from initial conditions: ( ) 01 0 1 0 0 t It AX X ARX A R XR AL AR L e R ε == = ⇒= −= = = = A better label for A would be I f , the final current. (c) Now that you know the time dependence for the current I in the circuit you can also determine the voltage drop V R across resistor and the EMF generated by the inductor. Do so, and confirm that your expressions match the plots in Fig. 6a or 6b in Experiment 6. We find: 1 (Fig. 6a) 1 (Fig. 6a) (Fig. 6b) t R L AX e e R Vt I R e dI tL L e e dt R =−= Looking at the EMF from the inductor you see that it starts the same as the battery (but in the opposite direction) which explains why no current initially flows. Then as time goes on it relaxes. 2. ‘Discharging’ an Inductor (3 points) After a long time T the current will reach an equilibrium value and inductor will be “fully charged.” At this point we turn off the battery ( =0), allowing the inductor to ‘discharge,’ as pictured at left. Repeat each of the steps a-c in problem 1, noting that instead of exp(- t / ), our expression for current will now contain exp(-( t - T )/ ).
(a) Faraday’s law: Walking in the direction of current, starting at the switch 0 dI IR L dt −= (b) Confirm solution: () ( ) 0 tT Ae A A X e R L ARX AR L e τ ττ −− ⎛⎞ =− = − + ⎜⎟ ⎝⎠ Both the constant and time dependent part must equal zero, giving us two equations. The third (because there are three unknowns) we can get from initial conditions: 00 0 1 t ARX X AL AR L e R It T AX A R R ε = = == = =

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## This note was uploaded on 02/27/2011 for the course PHYSICS 8 taught by Professor Dourmashkin during the Fall '10 term at MIT.

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8.02 Pset 9 Soln - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2008 Problem Set 9 Solutions Due Wednesday April 16 at

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