Solve Parallel RC Circuit Diff Eq | Current Flow

[cpuwildman]

I have an RC circuit driven by a voltage source. The resister and capacitor are in parallel. I'm having a problem figuring out the governing differential equation for the current of the circuit. I could figure it out easily if it were a series circuit as the current is the same for each branch, but since the current splits between the resister and capacitor branches, I cannot figure it out. I would appreciate any help.

 

[chroot]

For any two impedances z₁ and z₂ in parallel, the total impedance is (z₁^-1 + z₂^-1)^-1.

- Warren

 

[cpuwildman]

Do impedances apply for both AC and DC voltage sources?

 

[SGT]

The two currents are independent. You have:
i_R = V/R
i_C = C dV/dt

If The initial voltage of the capacitor is different from the initial value of the voltage source you will have an impulse of current in your capacitor.

Impedances only have meaning with time varying voltages and currents. Notice that a step is a time varying function. So if you have a constant voltage source that is not initially connected to the circuit and you switch it on, this corresponds to a step of voltage. The impedance of the capacitor will be 1/sC, where s is the Laplace transform variable, but you really don't need it.

 

[cpuwildman]

I know that a capacitor's impedance is given by $$Z_c=\frac{1}{\omega C}$$. If the circuit in question is driven by a DC voltage and a switch closes at t=0 to deliver the voltage, what should I use for $$\omega$$?

 

[SGT]

I know that a capacitor's impedance is given by $$Z_c=\frac{1}{\omega C}$$. If the circuit in question is driven by a DC voltage and a switch closes at t=0 to deliver the voltage, what should I use for $$\omega$$?

No, this expression is valid only for a senoidal wave of frequency ω
For a step excitation (DC source switched at t = 0, you should use $$Z_c=\frac{1}{s C}$$, where s = σ + jω is a complex frequency associated with the Laplace transform variable.
In your case you don't need to use this concept. Use directly $$i_c=\frac{C dv}{dt}$$, where v(t) = V.u(t).
The derivative of the step function u(t) is δ(t), the unit impulse.
For the current in the resistor, it is simply $$i_r=\frac{v(t)}{R}$$.
The total current driven from the source is $$i_s=i_c + i_r$$

 

[cpuwildman]

So then the governing differential equation would be $$i(t)=\frac{v(t)}{R}+C\frac{dv(t)}{dt}$$?

 

[SGT]

So then the governing differential equation would be $$i(t)=\frac{v(t)}{R}+C\frac{dv(t)}{dt}$$?

You are right.

 

[cpuwildman]

Thanks for the help.

Now, if I had the same circuit substituting an inductor for the capacitor, I would get $$i(t)=\frac{v(t)}{R}+\frac{1}{L}\int^{t}_{0}v(\tau)d\tau$$. Is this the governing differential equation for $$i(t)$$ in the new circuit? I'm not sure because of the integral.

 

[SGT]

Thanks for the help.

Now, if I had the same circuit substituting an inductor for the capacitor, I would get $$i(t)=\frac{v(t)}{R}+\frac{1}{L}\int^{t}_{0}v(\tau)d\tau$$. Is this the governing differential equation for $$i(t)$$ in the new circuit? I'm not sure because of the integral.

Yes, this would be an integral equation. The solution for a step of amplitude V is:
$$i(t)=\frac{V.u(t)}{R}+\frac{V}{L}t.u(t)$$.