Download Problem 7.14

Problem 7.14 The switch in the above circuit has been closed for along time before opening at t = 0.
Part A
Find iL(t) at t > = 0
Solutions:
We first find iL at t < 0.
When t < 0, we model the inductor as a short circuit and solve for the current using standard circuit
analysis techniques. Note that at t < 0, the switch is also a short circuit.
Using nodal analysis and assuming all currents flow left to right or top down (except for i which flows as
shown in the above diagram).
Δ
At node A: (-­‐120-­‐Va)/40 = Va/60 + (Va – Vb)/6 At node B: By observation, Vb = 0V At node C: By observation: Vc = 0-­‐20 i = -­‐20 i
Δ
Δ
By observation: -­‐Vc/60 = i
The only possible solution for i is i = 0. And therefore Vc = 0 too. We can now solve for Va: (-­‐120 – Va)/40 = Va/60 + Va/6 -­‐720-­‐6Va = 4Va + 40Va -­‐720 = 50Va Va = -­‐14.4 V From previous results, the only current flowing through the inductor at t < 0 is the current flowing through the 6 Ohm resistor (the current i is zero and the current through the 100 ohm resistor is Vc/100 = 0. Therefore, by KCL iL = i + i (100 ohm resistor) + i (6 Ohm resistor) = i (6 ohm resistor). Thus, at t < 0, the current through the inductor is Va/6 = -­‐2.4A At t = 0+, the current must be the same as at t = 0-­‐, so iL = -­‐2.4A As t approaches infinity, the inductor once again becomes a short circuit: Δ
Δ
Δ
Δ
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By observation, at t = infinity, the current through the inductor is 0A. Now, we need to find the time constant of the circuit. Because of the dependent source, we can’t just simply assume it is L/R, so we return to KVL to find the time constant: Using KVL: vL – 20 iΔ + iL (37.5) = 0 (100 and 60 ohm resistors were combined in parallel). By observation iΔ = iL(37.5)/60 = 0.625 iL Substituting into our KVL equation: LdiL/dt – 20 (0.625 iL) + iL (37.5) = 0 From here, we can see the time constant of this circuit is going to be: L/(-­‐20(0.625) + 37.5) = 10 ms Thus, the current through the inductor as a function of time is: iL (t) = (final value) + (initial – final) exp (-­‐t/6.67msec) iL (t) = 0 + (-­‐2.4 – 0) exp (-­‐150t) = -­‐2.4 exp (-­‐100t) where t is in seconds or iL (t) = -­‐2.4 exp (-­‐0.1t) where t is in milliseconds Part B: Find vL(t) for t > 0 vL(t) = LdiL/dt = 0.250 d/dt (-­‐2.4 exp (-­‐100t)) vL(t) =60 exp (-­‐100t) where t is in seconds or vL(t) =60 exp (-­‐0.1t) where t is in milliseconds Part C: Find iΔ when t > 0 By current divider: iΔ = 100/160 * iL (t) = 0.625* -­‐2.4 exp (-­‐100t) = -­‐1.5 exp (-­‐100t) with t in seconds or iΔ = -­‐1.5 exp (-­‐0.1t) with t in milliseconds