I need to find the intercept point between a function $V_C=325.6 e^{t/0.22}$ and $V=325.6 \sin(2\pi\cdot 50t)$. I tried solving for $t$ in both equations and then solving when $V=V_C$ but i couldn't find a way to solve it. If it helps, the first equation, $V_C$, is a capacitor discharging, and the second one is a $230$V rms AC voltage source. They both start at the peak, $325.6$V, and I need to find out at what time will the voltage of the capacitor will be the same as the voltage source again. Thanks a lot.
2026-04-07 10:30:15.1775557815
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Intercept between an Exponential and Sinusoidal Function
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The equation $$ 325.6\ e^{t/0.22}= 325.6 \sin(2\pi \cdot 50t) $$ has no 'nice' exact solution. However, there are numeric techniques to finding the values $t$ that satisfy the equation. There are plenty of free calculators that will find intersections for you. For instance, try this link here. A few solutions are: $$ \begin{aligned} t &\approx -0.0138811 \\ t &\approx -0.016207 \\ t &\approx -0.0332926 \\ t &\approx -0.0367900 \end{aligned} $$
You are looking for the zero's of function $$f(t)=\exp \left(\frac{50}{11} t\right)-\sin (100 \pi t)$$ which has an infinite number of solutions, all of them being negative. Because of the exponential term, they will closer and closer to $t_k=-\frac k {100}$.
Building the $[1,2]$ Padé approximant of $f(t)$ around $t=-\frac k {100}$, we can obtain the approximation $$t_k=-\frac k {100}+\frac{33 \left(1-88 \pi e^{k/22} \left((-1)^k-11 \pi e^{k/22}\right)\right)}{50 \left(88 \pi e^{k/22} \left(-99 \pi e^{k/22}+726 \pi ^2 (-1)^k e^{k/11}+\left(3-121 \pi ^2\right) (-1)^k\right)-1\right)}$$
For the first roots, this gives
$$\left( \begin{array}{ccc} k & \text{aproximation} & \text{exact} \\ 1 & -0.0135137 & -0.0138811 \\ 2 & -0.0165562 & -0.0162069 \\ 3 & -0.0331248 & -0.0332926 \\ 4 & -0.0369465 & -0.0367900 \\ 5 & -0.0527927 & -0.0528803 \\ 6 & -0.0572777 & -0.0571973 \\ 7 & -0.0725058 & -0.0725543 \\ 8 & -0.0775623 & -0.0775183 \\ 9 & -0.0922555 & -0.0922834 \\ 10 & -0.0978095 & -0.0977845 \end{array} \right)$$