Background: The equations are derived from a Physics 2 Lab circuit that has a resistor and a capacitor
Problem: Show that the integral of equation 5 yields equation 2.
I'm given:
$I(t) = \frac{Ve^{-t/RC}}{R}$ Equation 5
$V - \frac{dq}{dt} R - \frac{q}{c} = 0$ Equation 2
My attempt:
I'm just a bit confused on if I should be solving for the integral with respects to time or not. What I was thinking was something like this:
$\int \frac{Ve^{-t/RC}}{R} \frac{dq}{dt}$
though I am not entirely sure. The multiple variables are causing slight confusion for me. Any advice or reassurance would be great.
$$I(t) = \frac{Ve^{-t/RC}}{R}$$ The desired equation does not contain $t$ explicitly except in $\frac{dq}{dt}$, but it does contain $q$. This suggests us to write, $$\tag1I(t)=\frac{dq}{dt}=\frac{Ve^{-t/RC}}{R}$$ Now, integrate with respect to $t$, $$\int_{0}^q\frac{dq}{dt}dt=\int_{0}^t\frac{Ve^{-t/RC}}{R}dt$$ $$q=\left[\frac{Ve^{-t/RC}}{R}(-RC)\right]_0^t=\frac{Ve^{-t/RC}}{R}(-RC)+CV$$ $$q={CV(1-e^{-t/RC})}$$ Thus, $$\frac qC=V(1-e^{-t/RC})$$ $$\tag2V-\frac qC=Ve^{-t/RC}$$ From $(1)$ and $(2)$, we get the desired result.