Determine Vo for the following network.

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The discussion focuses on determining the output voltage (Vo) for a given electrical network, with an emphasis on applying Kirchhoff's Voltage Law (KVL). A classmate suggested that Vo is 2.45V, but the original poster is struggling to reach this value and is exploring different approaches, including modeling the diode as a voltage source and using superposition. They calculated the capacitor voltage (Vc) to be 25V and derived an output voltage (Vo) of 35V during a specific time period. The conversation highlights the importance of considering changes in input voltage and their impact on the output. Overall, the thread emphasizes the complexity of analyzing time-varying circuits and the need for careful application of circuit analysis techniques.
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Homework Statement



Determine Vo for the following network

[PLAIN]http://img535.imageshack.us/img535/7041/63014601.jpg.

Homework Equations



Too many to list I would think. I'm pretty sure I have to use KVL.

The Attempt at a Solution



My classmate told me the answer was 2.45V. I can't for the life of me get even close to this. I need some fresh ideas.
 
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There is a time-varying input voltage, so the output should also be time-varying. (At least in this case.) It could be useful to model the diode as a voltage source. You could use the principle of superposition to find the voltage over the resistor...
 
Think I got it,

Applying KVL

-20 + Vc - 5 = 0
Vc = 25V

The capacitor is charged up to 25 V.
Thevenin equivalent ckt of that portion of the network which includes Battery and resistor will result in RTH = 0Ω and ETH = V = 5V.

For the period t₂-t₃
Applying KVL
10V + 25 - v₀ = 0
v₀ = 35 V

What do you think?
 
Almost right.

You need to consider what will happen when the input changes.
 
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