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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?

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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.

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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.

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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)

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