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.

  • \$\begingroup\$ What about t = 0? \$\endgroup\$
    – caveman
    Commented Jan 24, 2015 at 23:44
  • \$\begingroup\$ @caveman At t = 0, u'(t) is undefined. As far as I understand. \$\endgroup\$
    – nopcorn
    Commented Jan 24, 2015 at 23:48
  • \$\begingroup\$ It's actually infinite at that point. This means that you have a non-continuous differential equation. I believe this is why you are having math issues. Another way to look at it is that the current source has to have infinite power at time 0 to create a step on this circuit. \$\endgroup\$
    – caveman
    Commented Jan 25, 2015 at 0:01

2 Answers 2


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\$.


In your final equation, the right hand side is not 0. Just as you said, it is 10delta(t). If you then try to solve the equation (using integrating factor method for example), there is an integration that turns the delta function back to a step function (which is simply a constant for t>0)

Now, what would be the initial condition? Let's look at this another way first.

This problem is easier and less confusing if you choose v_c(t) as the variable. Because for that, the forcing function is a multiple of 10u(t), and for t>=0, 10, a simple constant. And the initial condition is derived from v_c must be continuous, therefore v_c(0) = 0.

Now what is the initial condition when using i_c(t) as the variable? You need to translate v_c(0) = 0 to a equation that specifies i_c(0) = ?.


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