- #1

kawaiiRose

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## Homework Statement

Compute V3(t)-V4(t). The capacitors are initially discharged, the voltage source is turned on at t = 0, and the diodes’ forward voltage drop is negligible (Vf = 0). What is the value of V3(10s) – V4(10s)?

Vi=4sin(2∏t)

RC= 0.1s

## Homework Equations

i (over capacitor) = CdV/dt

i (over resistor) = V/R

Assume Diode is either on (id>0) or off (id=0)

## The Attempt at a Solution

I used the node method: node above the top capacitor has voltage V1 and node above the bottom capacitor has voltage V2. Using KCL at those two nodes, I found id1=(CdV1/dt)+(V3/R) and id2=-(CdV2/dt)-(V4/R). Unknowns are id1, id2, V1, V2, V3, and V4.

Based on whether I assume a diode is on/off I can assume V1=Vi or V2=Vi and that id1=0 or id2=0, so that eliminates two unknowns. I'm still left with 4 variables though and can't see any other equations to solve for them. I was thinking use the node between the two diodes and the voltage source, but I don't know the current coming off the source. And there aren't any closed loops for me to use KVL.