Odd results solving this very simple first-order RC circuit

capacitor

Here's the circuit:

enter image description here

I'm trying to find the differential equation for the current through the capacitor (i_c(t)).

  • Using KCL at the top node, I know that 10u(t) = i_r(t) + i_c(t).
  • Using KVL around the right loop, I know that i_r(t) = v_c(t) seeing as the resistance is 1 Ohm and i_r(t) = v_r(t).

I can now rewrite my KCL equation to incorporate the equalities I've found:

10u(t) = v_c(t) + i_c(t)
because i_r(t) = v_c(t)

However, I need this in first-order differential form. So I take the derivative of both sides of the equation:

10u'(t) = v_c'(t) + i_c'(t)

And using the V-I characteristics of a capacitor, I replace v_c'(t):

10u'(t) = (i_c(t))/C + i_c'(t)

However, the derivative of the step function is the delta function – which is defined as 0 for t < 0 and t > 0. So we can replace it with 0. This makes my final differential equation:

i_c'(t) + i_c(t)/C  = 0

Is this correct? I've tried to illustrated my methodology so that my steps are clear. The reason I'm skeptical this is the actual answer is because when I try to find a value for i_c(0+) (leveraging the continuity condition on the capacitor's voltage), I get zeros all across the board.

Best Answer

Laplace Transform:

Solving in the s-domain will be easier. Writing the current division formula,

$$i_c (s) = \frac{R}{R+1/Cs}\times I(s)$$ $$i_c (s) = \frac{1}{1+1/5s}\times \frac{10}{s} =\frac{10}{s+0.2}$$

Taking the Laplace inverse,

$$i_c(t) = 10e^{-0.2t}u(t)$$ $$\therefore i_c(0^+) = 10A$$

$$------------------------------------$$

Another view to solve this problem: Capacitor acts as a short circuit for high frequency currents. At t = 0, a sudden switching from 0 to 10A (high frequency) occurs and capacitor acts as short circuit and hence the entire current will be flowing through capacitor. Or, \$i_c(0^+) = 10A\$.

$$------------------------------------$$ Using continuity condition on the capacitor's voltage:

Capacitor at \$t = 0^+\$ can be replaced with a voltage source with voltage = \$v_c(t=0^-)\$. But \$v_c(t=0^-) = 0\$. Voltage source with voltage = 0 is short circuit. Hence the entire current will be flowing through capacitor. Or, \$i_c(0^+) = 10A\$.

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