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http://www.scribd.com/doc/33828848/Fundamentals-of-Electric-Circuits-Alexander-Sadiku-Chapter-07-Solution-Manual

Problem 67:

I'm working on a homework assignment and have solved all of the problems successfully except this one, so I checked the above link to see if i couldn't work out my error, but I'm missing it.

When I solved, I saw that it was a follower, so set my V(t) to be 5, and v->INF to be 0, since the cap should be drained by the time the sun explodes. Safe bet. Then I said that the Req for the capacitor at t>0 was 20 kohms, so the time constant is $$20 * 10^3 * 1 * 10^{-6} = 20/1000 = 20 mS$$

Given $$v(t) = V(inf) + [V(0) - V(inf)]e^{t/T}$$ I get $$v(t) = 5e^{-50t}$$

The solution is using Kirchoff's Current Law to find ... something? it also multiplies by the unit step function for reasons I fail to comprehend...

What am I missing?

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it also multiplies by the unit step function for reasons I fail to comprehend...

The problem asks that you give the solution for \$t \gt 0\$ but that still doesn't quite explain the presence of \$u(t)\$ in the solution. I think you would be fine if you don't have the \$u(t)\$ term as long as you indicate the solution is for \$t \gt 0\$.

The solution is using Kirchoff's Current Law to find ... something?

To find the voltage at node 1, the common connection of the three resistors. Knowing \$v_1\$ allows you to find the current through the capacitor via the series resistor thus

$$i_C = \frac{v_1 - v_O}{R}$$

You now have the voltage across and the current through the capacitor so, using the capacitor equation

$$C \frac{dv_O}{dt} = i_C = \frac{v_1 - v_0}{R}$$

you can write the differential equation to solve for \$v_O(t)\$

Now, you can go directly to the solution for a 1st order RC circuit if you know the time constant and, for that, you must find the equivalent resistance of the circuit connected to the capacitor.

The twist here is that you have a dependent source to deal with - the output of the opamp - so you must be careful in setting up the calculation for the equivalent resistance.

It's straightforward to show that the equivalent resistance is \$3R\$

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  • \$\begingroup\$ I ... think I see my error, not treating the op amp's output as a dependent source, which it obviously is. Thanks Alfred. \$\endgroup\$ – Daniel B. Jan 26 '14 at 15:10

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