# Voltage across capacitor in charging RC circuit with constant current source

simulate this circuit – Schematic created using CircuitLab

I couldn't really figure out how to get the switch that I wanted but this is functionally the same.

Before the switch flips, there's a short and no current flows into the other parts of the circuit. Therefore Vc = 0

In order to solve for Vc I set up the Kirchoff's Current Law

$$C\frac{dV}{dt} + \frac{V_c}{R} = I_s$$

This is the step I get stuck at. If this were a voltage source then I could use separation of variables and integrate. But I don't know how to integrate this with a current source.

• Try $V(t) = ke^{-\frac{t}{RC}}$ as a solution for the homogeneous equation (i.e. when the RHS is 0). Nov 7, 2018 at 23:56
• @AndyWalls: Put a forward slash in front of your dollar signs: \$x\$ renders as $x$ (if my rendering is right). Nov 8, 2018 at 0:15
• You could do a thevenin transform of the source, if your homework allows. Nov 8, 2018 at 0:17
• Are you familiar with the usual "integrating factor" method? Change of variable works, too, as mentioned in an answer below. What's your preference?
– jonk
Nov 8, 2018 at 1:33
• @TimWescott I think that's the way it's supposed to be done since my class hasn't doesn't really have a focus on the math. Even though diff eqs are a prereq, the professor always emphasizes that it's an EE class and the math will be as simple as allowed. Thanks!
– mnal
Nov 8, 2018 at 1:59

simulate this circuit – Schematic created using CircuitLab

As @TimWescot said, I solved it by transforming the circuit to it's Thevenin Equivalent which has a voltage source that allows for integration using separation of variables.

$$\frac{V_c - V_s}{R} + C\frac{dV}{dt} = 0$$

Excuse the values since I'm only working with variables and not concerned about the actual values.

• Your original equation was fine without Thevenizing it, as user287001 pointed out.\begin{align*} C\frac{\text{d} V}{\text{d}t}+\frac{V}{R}&=I\\\\ R\:C\:\text{d} V&=\left(R\:I-V\right)\text{d}t\\\\\text{set: } h=R\:I-V,\quad\:\therefore \text{d}h=-\text{d} V\\\\ -R\:C\:\text{d} h&=h\:\text{d}t\\\\ \frac{\text{d} h}{h}&=\frac{-1}{R\:C}\:\text{d}t\\\\ \int \frac{\text{d} h}{h}&=\frac{-1}{R\:C}\:\int\text{d}t\\\\ \operatorname{ln}\left(h\right)&=\frac{-t}{R\:C}+A_0\\\\\text{at }t=0, V=0\quad \therefore A_0=\operatorname{ln}\left(R\:I\right) \end{align*}From there, the rest is easy.
– jonk
Nov 8, 2018 at 5:11

Your equation is ok. Write it dV/dt = (IR-V)/RC. Change variable, let h=(IR-V). See that formally dV = -dh. Invert. You have dt/dh = -RC/h which you should be able to integrate. Do not forget the integration constant. You solve it with condition V=0 when t=0. Do it after integration and returning to original variables.