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RL example L=2 mH (a) (b) In the circuit above, the switch has been in setting (a) for some time. At t = 0, it is moved to (b). 1)Derive an expression.

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Presentation on theme: "RL example L=2 mH (a) (b) In the circuit above, the switch has been in setting (a) for some time. At t = 0, it is moved to (b). 1)Derive an expression."— Presentation transcript:

1 RL example L=2 mH (a) (b) In the circuit above, the switch has been in setting (a) for some time. At t = 0, it is moved to (b). 1)Derive an expression for the voltage across the inductor and the current through it for t > 0. 2)Determine the time t > 0 for which the magnitude of the voltage across the inductor is 3 V. R=10 Ω 1 A VLVL ILIL

2 t= 0 First we need to look at the circuit’s “pre-history” – ie where were all the voltages and currents before flipping the switch at t=0? At t = 0, the circuit is equivalent to that shown above. As the inductor is simply a piece of bent wire (!), no current flows through the resistor, and it can be disregarded when figuring out I start, I finish etc So – the initial current through the inductor is 1 A. If we put the resistor back in, we can see that the initial current I Rpre-start flowing through it is zero Therefore V R =R x I Rpre-start tells us that V Rstart = V Lstart = 0 (a) 1 A I Rstart =0

3 Start … t>0 (just!) Now we get to the actual start of the journey, having figured out where everything war before flipping the switch. For t > 0, the circuit is equivalent to that shown above. The inductor forces I = 1 A through itself and the resistor, although this will reduce with time as P = I 2 R takes energy out of the circuit as heat. Note that the inductor is now forcing current UP through the resistor, so V R is negative and V L = V R – check the direction of the voltage arrow V Rstart = V Lstart = - R x1 A 1 A (a) 1 A

4 t = ∞ The current falls with time as P = I 2 R takes energy out of the circuit as heat. Eventually, it is reduced to I finish = 0, and thus Ohm's Law tells us that V Rfinish is = 0. V Lfinish = V Rfinish = 0 V Rfinish = V Lfinish = A

5 Put it all together I Lstart = 1 A, I Lfinish = 0 A V Lstart = -1 A x 10 Ω = -10 V, V Lfinish = 0 V I L (t) = I start + (I finish – I start ) x ( 1 – e -tR/L ) V L (t) = V start + (V finish – V start ) x ( 1 – e -tR/L ) I L (t) = 1 + (0 – 1) x ( 1 – e -t10/2 ) … inductance in mH V(t) = ( ) x ( 1 – e -t10/2 ) … inductance in mH I L (t) = e -5t V L (t) = -10e -5t V …(NB 5t is actually 5t x … L in mH) 1 A L=2 mH R=10 Ω

6 Some pictures VLVL time V finish V start ILIL time I finish I start -10 V 1 A -3 V t= ? Last bit of question … when is V L = -3V?

7 When is |V L | = -3 V? V L (t) = - 10 e -5t V … inductance in mH … ie 5t is really 5t x So -3 = - 10 e -5t e -5t = 3/10 = 0.3 … x(-1) and log of both sides … -5t = log e (0.3) = -1.2 Turning 5 back into 5 x … and x(-1) again … 5t x = 0.005t = 1.2 t = 241 sec for V L = -6V Check … V L (t) = - 10 e x241 V = -3 V ( ) 1 A L=2 mH R=10 Ω


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