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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 Ω , R S = 16 Ω , R M = 0.4 Ω . DC steady‐state conditions are reached at t = 0‐ then the switch is open. In the given circuit, at DC steady‐state 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 ܸ ெ ሺݐ ൌ 0ሻ ൌ െ݅ ሺݐ ൌ 0ሻ ൈ ܴ ௌ (Eqn 1) ݅ ሺݐ ൌ 0 ሻ ൌ ݅ ሺݐ ൌ 0 െሻ ൌ ܸ ெ ሺݐ ൌ 0 െሻ ܴ ெ (Eqn 2) Solution by Idong Ebong KCL @ 1 for t=0 ‐ gives: I B = I S + i L ܸ െ ܸ ெ ܴ ൌ ܸ ெ ܴ ௌ ܸ ெ ܴ ெ ሳልልልሰ ܸ ܴ ൌ ܸ ெ ൈ ൬ 1 ܴ ௌ 1 ܴ ெ 1 ܴ ൰ ܸ ெ ൌ ܸ ൈ ൬ ܴ ௌ ܴ ெ ܴ ௌ ܴ ெ ܴ ெ ܴ ܴ ௌ ܴ ൰ If R B =0 Ω , 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, ݅ ሺݐ ൌ 0 ሻ ൌ ݅ ሺݐ ൌ 0 െሻ ൌ...
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This note was uploaded on 10/28/2009 for the course EECS 314 taught by Professor Ganago during the Winter '07 term at University of Michigan.
 Winter '07
 Ganago

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