LR Circuits. . The voltage drop through both resistors will equal. Hence 10 1 A

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1 The diagram shows a classic R circuit, containing both resistors and inductors. The switch shown is initially connected to neither terminal, and is then thrown to position a at time t = 0. R Circuits E a b R 1 R E = 10 V = 15 mh R 1 = 4 R = 6 (a) At t = 0+, just after the switch is thrown to position a, what are the currents 1 and across the two resistors? Just after the switch is thrown, what does the inductor 'look like' to the rest of the circuit? What is the current doing? Do inductors like that? Once you know what the inductor looks like at this point in time, it is highly advisable to redraw the circuit. Just before the switch is thrown, there is no current in the inductor, since the inductor blocks immediate changes in current, the current will be still be zero. Hence the same current will flow through R 1 and R. The voltage drop through both resistors will equal. Hence 10 1 A R R (b) After a very long time, what is the instantaneous power dissipated in the circuit? After a very long time, what will have happened to the current? Now what will the inductor look like to the rest of the circuit? The circuit now takes on a simple form... what current is flowing through the resistors? After a long time, the rate of change of current has stabilized and hence V 0. This means no current flows through R Hence R. Power = R W R (c) Sketch the behavior of the energy stored in the inductor as a function of time. Sure, why not? 15mH Joules What is the final energy stored, after a very long time? Think about the current through the inductor: what is it at time 0? 0A after a very long time? 5/3 E E Next, after a very long time, the 'clock' is reset to 0 and the switch E b R is thrown to position b. t a R 1 E = 10 V = 15 mh R 1 = 4 R = 6

2 We have a new formula available for time constants in R circuits: = /R. But the R in the formula refers to the total resistance in series with the inductor. Redrawing your circuit will help you to determine this R! Essentially the inductor discharges through R and R in parallel. 1 R R The equivalent resistance is RE 4 R1R The time constant is 65. ms R 4. E (e) Sketch the time dependence of the current through the inductor. The current dies away exponentially starting from the computed in part (b) according to = exp t / and resembles the figure. t (f) What is the energy stored in the inductor 8 msec after the switch goes to position b? First you must write down an equation for the time-dependence of the current. Check that your formula is correct: does it produce the right answer at time 0? What about at t =? The stored energy is exp(-t/ t t U exp U exp. 8 ms where U is value calculated in part (c). Thus U mj exp mj 65. ms

3 Physics 1 Week 10 nductors n this question, we will explore the effective inductance of series and parallel combinations of wire coils. n your calculations, you can assume that your coils are all very long compared to the radii of their circular turns which allows you to approximate them as infinite solenoids. The formula is = / just remember, the magnetic flux we re talking about here is that caused by the coil s own field, not some external field as in earlier problems. (That s what we mean by self-inductance). So run some current through the coil and see what flux it causes. The field of the inductor is the field of a solenoid or B= n. This field pierces n loops of area A and hence BnA or na na l n turns per meter coil area A pair single et V be the voltage across the pair, and let V 1 and V be the voltages across the individual inductors. Start with an equation relating these f the two coils are in series, one has the same and / flowing through them. The voltage drops are V1 1 and V. The net voltage drop across both coils is Veq V1V 1 + equiv equiv 1 1 1

4 parallel f the coils are in parallel, the V V V. The voltage drops are equiv 1 1 related to the currents as V1 Vequiv 1 and V Vequiv and Vequiv equiv We now write the rate of change of currents: Vequiv 1 V equiv Vequiv equiv 1 equiv 1 Suppose you have a spool of thin wire of total length D = 5 m and radius R = 0. mm. You can make an inductor out of this wire by wrapping it into tight circular turns, thereby forming a spiral. How should you do the wrapping to achieve the imum possible inductance? Given that you will make circular turns, there are only two parameters you can vary: the radius r of your turns, and the gap g between them (i.e. how tightly you pack your turns.) et s try three designs and see which one gives the greatest self-inductance. (i) Design : Use circular turns of radius r = 1 cm, and pack them as tightly as possible without causing any of the turns to overlap. (ii) Design #: Double the turn radius r to cm. (iii) Design #3: Go back to the 1 cm turn radius, but this time try packing the turns more loosely: leave a gap of 1 wire diameter between each turn. Hint- write in terms of the winding radius r, the total wire length D, and the spacing between the wire centers. We begin by writing in a form that can be used for all three cases. We will write all quantities in the expression = nain terms of the winding radius r, the total length of wire D, and the spacing between the centers of adjacent wires s. A r D of turns one can make with a length D of wire: N = r et N be the total number of number. Write in terms of spacing between sd N 1 adjacent wires and total number of wires: N s and n r s 1 sd rd Substitute these forms for n,, and A to get = na r s r s Continues on next page

5 Design and # have s R since the wires are spaced apart by 1 wire diameter. rd Hence = Thus # Since r# cm and r 1cm 4R 1 Design #3 has r #3=1cm and s# 3 4 R = s Hence #3 =. Hence the order is > rd = = 314 H ; 3 # =68 H ; #3 157H R # #3

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