Part #6 - Score: 0/10 Next, repeat this for the node voltage, v, of the parallel RLC circuit: 1 = 0 73 w 6V 27 F 00 85 m

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Part 6 Score 0 10 Next Repeat This For The Node Voltage V Of The Parallel Rlc Circuit 1 0 73 W 6v 27 F 00 85 M 1 (43.29 KiB) Viewed 31 times

Part #6 - Score: 0/10 Next, repeat this for the node voltage, v, of the parallel RLC circuit: 1 = 0 73 w 6V 27 F 00 85 m First, what is the final value for the voltage? v(inf): Part #7 - Score: 0/10 Since v is the same as the capacitor voltage (so it stays fixed from time 0 to 0) evaluating it before the switching occurs tells us the initial condition on v. Find this value. vot): Part #8 - Score: 0/10 Since a capacitor's current is proportional to the derivative of its voltage, ic - Cdv/dt, we can use KCL at the top node to find the capacitor current at time of and then evaluate dv/dt(ot). Note that as part of the KCL equation you will need to find the inductor current at time of (but it's the same as the inductor current at time 0+). Evaluate this derivative, entering your value assuming units of volts/second (again, no need to write the words). dv/dt(o); Part #9 - Score: 0/10 Assuming a solution of the form v(t) =B1 exp(- alpha 1) cos (wa t) + B2 exp(- alpha t) sin ( wd 1) + v(inf), find B and B2. Your units should be volts for both (no need to enter). B: Part #10 - Score: 0/10 B2: