RC circuit (two resitors and 1 cap connected to 10V DC)

[renzoms]

Am I doing something wrong calculating the current and voltages at time t for the RC circuit? I put these equations on google sheets and the voltage across all of the elements passes 10V between 4 and 5 seconds. Then it goes all the way to 35V around 20 seconds!

Super thanks

[bob91343]

It's hard to read.  My approach is to use the Thevenin equivalent, which is to divide the voltage with the resistive divider (ignoring the capacitor) and use that as the source voltage.  Put that as the battery voltage and put the capacitor across the new source in series with a resistor whose value is the parallel combination of your two resistors.

Then it becomes a very simple RC.  One volt source in series with the capacitor and about 91 Ohms.

[rstofer]

At t=0, the capacitor is a short circuit so the entire battery voltage will be across the 1000 Ohm resistor at t=0+ (an infinitesimal period after t=0).
At t=infinity, the voltage across the capacitor will be the same as the voltage drop across the 100 Ohm resistor and this would be
10V*(100/(100+1000)) or about 0.091V.
The time constant (tau) is based on the 100 uF capacitor and the parallel resistance of about 91 Ohms or about 9.09 ms.  Six tau will result in the capacitor being very close to fully charged.

It's always useful to simulate the circuit with something like LTSpice just to be certain you understand what is going on.

The red trace is the voltage across the capacitor as a function of time gained by subtracting the green trace from the blue trace.  The blue trace is 10V and the green trace is the output voltage as a function of time.  It starts at 10V because the capacitor looks like a short circuit and decreases as the capacitor is charged and the 100 Ohm resistor becomes more of a factor in the voltage divider.  It ends up at about 9.1V (9.090909V).