A circuit RC it's described by the next equation: $\frac{1}{c} \int i(dt)+Ri=V$ Where the value of resistance is $R=10 k\omega $ , the value of the capacitor is $C=2.5 \mu F$, and the voltage of the circuit is $V(t)=5V$. The switch that controls the circuit is closed in $t=0$ with $V_c=3V$. Calculate $i(t)$ of the circuit.
Now… I know how to calculate the $i(t)$ through laplace, I just don't know how to express $V$ on the equation with the values of $V(t)=5V$ and $V_c=3V$ in $t=0$. Hope someone can help me.
Let the capacitor charge be $q(t)$, then $i(t)=\frac{d}{dt}q(t)$. The equation then writes $$ c q(t)+R\frac{d}{dt}q(t)=V$$with initial data $q(0) = 3V/c$. In other words, we need to solve the Cauchy problem $$\begin{cases} \dot q = -\frac cRq+\frac VR,\\ q(0)=3V/c. \end{cases}$$ The solution is $q(t) = \frac Vc\left(2\exp(-\frac cR t) +1\right) $, $i(t) = -2\frac VR \exp(-\frac cR t) $.
If it's unclear how to find the solution to this Cauchy problem, ask in comments.