W09 314 HW08 Extra Credit Solution

W09 314 HW08 Extra Credit Solution - Solution by Idong...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Page1 / 5

W09 314 HW08 Extra Credit Solution - Solution by Idong...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online