# Current split between resistor and capacitor in parallel

I'm interested in knowing the formula which represents the voltage across the 10 ohm resistor. I know that it's going to be the current multiplied by the resistance, which means I have to find the current passing through the 10 ohm resistor as a function of time. I also know that when the capacitor is fully charged, the voltage across will stop changing (and as i = C dv/dt, the current will go to zero also). This means that from this time on, it's just a series resistor circuit.

However, before this time, how would you find the voltage across the 10 ohm resistor? Is the solution general for non constant input voltage?

So, starting point and ending point are easy to compute for a constant input source. At start the capacitor shunts the resistor and you basically get vo = vi (vo is output voltage and vi is input voltage). At steady state there is no current through the resistor so you get a simple voltage divider vo = 10/110 * vi

You can find the transient behavior by solving a differential equation. Let's take the output node. The current entering the output node has to be the same as the current leaving it so we could write the equation 10e-6*d(vi-vo)/dt + (vi-vo)/100 = vo/10. Simplifying, we have 1e-3*dvo/dt + 11*vo = vi. From the characteristic equation, we know vo has to be of the form vo = A*e^(-11e3*t)+B for this differential equation to be satisfied.

Given the steady state condition, vo = 10/110*vi=A*0+B, then B=10/110*vi and vo = A*e^(-11e3*t)+10/110*vi. If we use the initial condition vo=vi=A+10/110*vi, then A=100/110*vi. Thus, vo = 100/110*vi*e^(-11e3*t)+10/110*vi.

If vi is not constant then dvi/dt is not zero and the output will also be dependent on the time-varying behavior of the input. You will need to solve a non-homogenous differential equation to get the answer depending on vi as a function of time.

You need to solve the differential equation resulting from KCL and KVL:

$$i = i_{R1} + i_c$$

and

$$V = V_{R2} + V_c$$

where R1 = 100 Ohm and R2 = 10 Ohm.

With some algebra you get

$$V = R2\left(\frac{V_c}{R1}+C\frac{dV_C}{dt}\right)+V_c$$

with initial condition

$$V_c(t=0)=V_0$$

Then the current on R1 follows from Ohm's law and the current on the capacitor is

$$i_c(t) = C\frac{dVc}{dt} + i_0$$

The result should be an exponential function plus a constant.

Figured it out. You just sum the currents and use kvl (voltage law). You then see that:

//Let the capacitor == C for simpler writing. Same for R1 (in parallel with C) and R2. Let the voltage across R2 be V2, and of the source V1.

(R1+R2)/(R1*R2*C)*(V2)+V2'=V1'+V1/(R1C)