At t = 0, the switch is at Position 1.

We note that `q(0)=0`.




Using SNB to solve this differential equation, we have:


NOTE: By differentiating, this gives us:

`i_1(t)=` `d/(dt)1/100000(1-e^(-4000t))=` `0.04e^(-4000t`

We need to find `tau`:

`tau=RC=500xx0.5xx10^-6=` `0.00025`.

Now, at `t=0.00025`, the charge will be:




Position 2

At `t=tau`, switch at Position 2:

Applying the formula `R(dq_2)/(dt)+1/Cq_2=V` again:


NOTE: The negative voltage is because the current will flow in the opposite direction through the resistor and capacitor.

Once again, we solve using Scientific Notebook:


` q_2(0)=6.3212xx10^-6`

Exact solution is:

`q_2(t)` `=-0.00002+2.6321xx10^-5e^(-4000t)`

So the current transient will be:


`=d/(dt)(-0.00002+2.6321xx10^-` `{:5e^(-4000t))`


This expression assumes that time starts at `t=0`. However, we moved the switch to Position 2 at `t=0.00025`, so we need:

`i_2=-0.10528e^(-4000(t-0.00025))` `=-0.10528e^(1-4000t`

So the complete current transient is:

`i_1(t)=0.04e^(-4000t)` for `0 <= t <= 0.00025`

`i_2(t)=-0.10528e^(1-4000t)` for `t>0.00025`

The graph is quite interesting:

0.000250.00050.000750.0010.020.04-0.02-0.04-0.06-0.08-0.1-0.12tiOpen image in a new page

Graph of current `i` at time `t`.

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