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yesterday my professor introduced us to forced response circuits. I am some what confused on telling the difference when trying to find D.E. for it.

circuit

Determine the D.E. governing V(t), t>0.

This is one of the pratice problems he gave us. Now since t>0 the switch is on B, this means we ignore the 2V and 5omh portion of the circuit.

I found that there is 9V across the capacitor and Tau = 6.

This give me V(t) = 9e^(-t/6).

When I looked up a solution(could be wrong) it was V(t) = 9 - 7e^(-t/6).

Could someone explain this? Is this because the circuit is a forced response?

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    \$\begingroup\$ I don't even know what D.E. is. You haven't accounted for the 2V that the cap is initially charged to before t = 0. \$\endgroup\$ – Andy aka Oct 10 '15 at 12:17
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First step is to analyze the current flow of your circuit. You see that the capacitor is loaded by 2 volts and ends up at 9 volts.

Second step is the convert the current supply into a voltage supply; R stays 3 Ohm and the Voltage becomes 9v. The charging formula of a resistor is:

Uc = U(1-e^(-t/τ)

While

τ = R*C = 3 Ohm * 2 F = 6s

The approach I would choose is with a time offset, just add extra time t0 to t, so that when time t is zero, the formula gives 2v

2V = 9V(1-e^(-t0/6s)) => t0=1.51s

Enter this in the formula:

Uc = 9V*(1-e^(-(t+1.51s)/6s))

And thats your function at the end. You can simplify this formula by splitting the e';

Uc = 9V(1-e^(-t/6s)*e^(-1.51/6))

And you end up by this;

Uc = 9V - 7V*e^(-t/6s)
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  • \$\begingroup\$ Doesn't the left half of the circuit have a different time constant? \$\endgroup\$ – user3482104 Oct 10 '15 at 18:50
  • \$\begingroup\$ Yes but you assume that it has been precharged by the right side. You just slide the time function until the start point is at 2v. Like f(t)->f(t+a) just a linear translation in t direction. \$\endgroup\$ – Sider Oct 10 '15 at 19:02
  • \$\begingroup\$ Is there another approach besides setting it equal to 2v and solving for t? I don't get a calculator on my exams. \$\endgroup\$ – user3482104 Oct 10 '15 at 20:13
  • \$\begingroup\$ The 7V could come from the 9V of the source minus the 2V of the precharged capacitor. Deduce with General params, if this law is true, then you don't need a calculator at your exam for the charging function. \$\endgroup\$ – Sider Oct 12 '15 at 12:23

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