Here's the circuit:
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 andi_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.
t = 0
,u'(t)
is undefined. As far as I understand. \$\endgroup\$