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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+ )
iR (0+ ) = C
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
be 0 V. Matching this to the symbolic solution for t = 0 +, we have the
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.
- Fall '08