RC circuit (parallel and in series) help please?

Hi,

First at all, u(t) is.a voltage which starts with some previous specified value from t0. Before that time, V1=0 and suddenly, it values 9 times that specified value u(t).

For practical purposes, 9u(t) could be 9 volts.

Lets suppose V in capacitor is equal to 0 too.

Knowing that, you have two choices: To know the permanent response or the transitory one.

Because the capacitor is going to be significantly charged with 9V at approx to 5*R*C, a permanent analysis does not make sense; transitory response is probably the one you could be interested.

If R2>>R1, then consider R2 as an open wire. R1 and Cap are left in a series connection.

Add each voltage and get V1==> IR + (1/C)integral(I.dt) = V1

deviating both sides respect to t: ==> R.dI/dt + I/C = 0

manipulating ==> R.dI/dt = - I/C

Integrating==> ln I = -t / (RC)

Then, I = I0. e power (-t/RC).

Consider initial current I0 = V1/R1, thus I = (V1/R1).e power(-t/RC)

You will see the current is decreasing, from V/R (amps) to zero, which is consistent with reality.

Finally, VI is just V1 - I.R1 ==> VI(t) = V1 - R1.(V1/R1).e power(-t/RC).

VI(t) = V1 (1 - e power(-t/RC))

Hope it helps.

The cap will charge to 97% of Vout in 5 * R * C so that'll be in 5 * 10 * 0.1 = 5 sec. When Vin gets isolated from the circuit the cap will discharge over R2 in 5 * 1000 * 0.1 = 500 sec. The charge/dischargecurve is hyperbolic. Then R1 + R2 = 1010, I through the series R1 and R2 = 9 / 1010 = 0.009A, U over R1 = 0.009 * 10 = 0.09 so Uout 9 - 0.09 = 8.91V