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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 Fall '08
 Cotter,N
 RLC, RC circuit, parallel rlc circuit

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