I have the next circuit:
The textbook says that the capacitor is initially charged, but the current directions was selected as in schematic, why? If capacitor is charged and "+" is on upper node then the capacitor must be a voltage source in this circuit and "iC" must go up to node.
The textbook solution is:
Here an author derivate the result equation. But stop at a second and look. He use hard-coded iC direction. What will be if I try to derivate result myself and select the iC to "up"? The KCL will be: $$ C*\frac {\partial v_{C}}{\partial t} = \frac {v}{R} $$.
Solving, get:
$$v(t) = A*e^{\frac{t}{R*C}} $$
And there is no "-" in "e" power. The other current direction give another tau. I know that one task may be solved with different solutions, but anyone can't take in mind not general solutions in mind for every task. I want to see where an error and how to changing directions of currents I can get proper results with method of that textbook.
For more clarification of my problem I redraw the schematic:
Now I think so: as the capacitor is charged and the external voltage source is turned off then I can think about capacitor as a voltage source with it's own stored charge and the "iC" current begin going through the circuit in one direction with "iR" and the capacitor is discharging through the resistor. The textbook recommends to write KVL with components sign equal to first achieved through loop, in my case it would be:
$$ -\frac{1}{C}*\int {i_C}{\partial t} + i_R*R = 0$$
solving it I get:
$$v(t) = A*e^{\frac{t}{R*C}} $$
Here ther is no "-" sign in exponent and that equation shows that the voltage on "R" will be increasing and stay on max. values. But it's incorrect! Where is an error? It seems to be I can't interpret a capacitor with stored voltage as an active component.