ENGR 210 HW8 Solutions 2008 spring

ENGR 210 HW8 Solutions 2008 spring - ENGR210Spring2008...

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ENGR 210 – Spring 2008 Homework #8 Solutions Problem 1 The three inductors in the circuit shown below are connected across the terminals of a black box at t=0. The resulting voltage for t>0 is measured to be Vo=1250e 25t volts. i 1 (0)=10A and i 2 (0)= 5A, find (a) i o (0) (b) i o (t), t 0 (c) i 1 (t), t 0, (d) i 2 (t), t 0, (e) the initial energy stored in the three inductors, (f) the total energy delivered to the black box, and (g) the energy trapped in the ideal inductors. Solution: a) We need to find i o (0). We can see that i o branches into i 1 and i 2 , and since we are given the initial conditions for i 1 and i 2 : through KCL : i o (0)= i 1 (0)+ i 2 (0)=10 + ( 5) = 5A b) Part b asks us to find i o (t) for t 0. When talking about inductors, we can find the voltage if we are given the value of the inductor and the current. If we need to find the current (this problem) then all we need is the value of the inductor in question and the voltage across it. In this case, our voltage is v o which is the voltage across not one individual inductor, but rather all three inductors in the problem. We do a simple calculation to find the equivalent inductor for the 3 in the problem: (8H||32H)+3.6H. This gives us an equivalent inductance of 10H, and the voltage across this equivalent inductor is v o . Knowing these two things, we can solve the problem with the following equation:
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0 0 00 0 25 0 0 25 0 0 25 0 25 0 1 () ( ) 1 (0 ) 1 ( ) 1250 5 10 ( ) 125 5 25 () 5 ( 1 ) 5 t t t Equiv t x t x t t it Vd t it L Vtd t i L e d x H e e e A =+ ⎛⎞ =− + ⎜⎟ ⎝⎠ + = 0 Some notes about the above solution… Notice when we plug in the value for the inductor its negative . This is because the passive sign convention for inductors states that if the current from the inductor flows into the positive terminal of the voltage source, we are working in the opposite direction of the convention and we need a negative in front of our inductor value. Also notice that we have used the variable ‘x’ to take our integral. This is so we don’t confuse our time with our period of integration (bounds of the integral). See that ‘t’ falls back into the equation when we evaluate the integral over the period. It all works out. c) For part c we are asked to find i 1 (t) for t 0. First we will redraw the circuit: To find the current through the 8H inductor we need 2 things: the value of the inductor (we have that) and the voltage (v c ) across the inductor (we need to find that before we can solve this part of the problem). By inspection we notice that: v c = v a + v b . Since we already have v b , we just need v a to find v c : We know that di vL dt = And the current through the 3.6H inductor is i o , so: v a + + v b _ + v c _
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25 25 () 3 .6 (5 ) ( ) 450 t a t a d vt e dt e V = =− So now that we have v a we can find v c with the equation we found previously: 25 25 25 450 1250 () 800 tt cab t c vvv e e −− =+= + = Finally, now that we know the voltage across the inductor we can find i 1 (t) in the same fashion we did part b: 11 0 8 25 1 0 25 1 25 1 1 () (0 ) 1 ( ) 800 10 8 () 4 4 10 6 t c H t x t t it Vtd t i L e d x e e A =+ + d)
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ENGR 210 HW8 Solutions 2008 spring - ENGR210Spring2008...

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