ECE2260F09_HW1p3soln

3 ma the switch is also open removing the 23 v source

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Unformatted text preview: ) = 2.3 V = 2.3 mA 1 kΩ At time t = 0+, we treat the C as a voltage source of 0V and the L as a current source of 2.3 mA. (The switch is also open, removing the 2.3 V € source and 1 kΩ resistor from the circuit.) We can solve the circuit for any voltage or current at t = 0+. Using Kirchhoff's laws for voltage loops and current sums at nodes, we have the following results, (with voltages measured with plus on top, and currents measured flowing in the down direction): v R (0+ ) = v L (0+ ) = vC (0+ ) = 0 V € v (0+ ) 0V iR (0+ ) = C = =0 A R 100 mΩ iC (0+ ) = −iR (0+ ) − iL (0+ ) = 0 A − 2.3 mA = −2.3 mA € Now we are ready to find A1 and A2 by matching our symbolic solution to dv ( t ) + . We have already found v(0+) to € circuit values for v(0 ) and dt + t =0 be 0 V. Matching this to the symbolic solution for t = 0 +, we have the following: € A1 = 0V For the value of the derivative in the circuit, we always try to first write our variable, (v in this case), in terms of energy (or state) variables, iL and/or vC. Here...
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This document was uploaded on 03/17/2014 for the course ECE 2260 at University of Utah.

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