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Unformatted text preview: Solution by Idong Ebong HW08 Extra Credit Problem Solution Part 1: The circuit with a shunt resistor Given: V B = 50 V, R B = 0.5 , R S = 16 , R M = 0.4 . DC steadystate conditions are reached at t = 0 then the switch is open. In the given circuit, at DC steadystate conditions, the power absorbed by the shunt resistor R S is only 2.5% of the power absorbed by the motor resistance R M . Assume that V B and R M are given above, and R B =0 (an ideal battery). If needed, assume L = 1 H. (5 points) Calculate the magnitude V M (t = 0+) in the given circuit. At time t=0, DC steady state conditions are reached. Going from t=0 to t=0+, the inductor current must be continuous, so we will define V M in terms of the current after the switch opens at t=0+. @ t=0 , DC steady state (inductor is a short) @ t>0, after switch 0 0 (Eqn 1) 0 0 0 (Eqn 2) Solution by Idong Ebong KCL @ 1 for t=0 gives: I B = I S + i L 1 1 1 If R B =0 , then V M = V B =50V (you can see this more clearly, if you just replace R B with a wire in the circuit diagram in the previous page) From Eqn 2, 0 0...
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- Winter '07