Solving a function problem

[SamFeatherstone]

1. In a capacitive circuit used for research purposes, the voltage, v and the time, t are related to each other by the following relationship:

v = 210 ( 1 - e⁻ ᴿ / ᴸ ᵀ ) Volts
Given that R = 3 Ω and L is 200 mH, calculate the time when the voltage is expected to be set.

No	1	2	3	4	5	6	7	8	9
Voltage (V)	110	120	130	140	150	160	170	180	190

Can Anybody make any sense of this please? struggling to write it but its r over l to power t

 

[HallsofIvy]

The wording "the voltage is expected to be set" is a little peculiar but I think it is reasonably clear. You are given the formula $\displaystyle V= 210(1- e^{-\frac{R}{LT}})$. You are told that L= 3 ohms and R= 200 micro-Hertz so the formula becomes $\displaystyle V= 210(1- e^{-\frac{200}{3T}})$.

You are then given a list of 9 voltages and asked to find the time, T, when V is each of those. So you are asked to solve 9 equations for T:
$\displaystyle 210(1- e^{-\frac{200}{3T}})= 110$
$\displaystyle 210(1- e^{-\frac{200}{3T}})= 120$
etc.

For the first one, $\displaystyle 210(1- e^{-\frac{200}{3T}})= 110$, an obvious first step is to divide both sides by 210:
$\displaystyle 1- e^{-\frac{200}{3T}}= \frac{110}{210}= 0.5238$ (rounded to four decimal places). Then subtract 1 from both sides:
$\displaystyle -e^{-\frac{200}{3T}}=0.5238- 1= -0.4762$.
Multiply both sides by -1:
$\displaystyle e^{-\frac{200}{3T}}= 0.4762$.
Take the natural logarithm of both sides:
$\displaystyle - \frac{200}{3T}= ln(0.4762)= -0.7419$
and continue to solve for T.

 

[Cubist]

The formula might actually be one of the following...

\(\displaystyle V= 210\left(1- e^{- \left( \frac{R}{L} \right)^T}\right) \quad\enclose{box}{A}\)
\(\displaystyle V= 210\left(1- e^{- \frac{R}{L^T}}\right) \quad\enclose{box}{B}\)
\(\displaystyle V= 210\left(1- e^{- \left( \frac{R}{L} T\right)}\right) = 210\left(1- e^{- \left( \frac{R}{L} \times T\right)}\right)\quad\enclose{box}{C}\)

...since the OP said, "struggling to write it but its r over l to power t", at the very end of the OP

I added option (C) because it seems a more "usual" placement for T in electronic charging/ discharging types of equations. Luckily this doesn't negate most of @HallsofIvy 's post, since the work in post#2 doesn't "change" that part of the formula. We just need to do a search/replace to make it fully correct when we know the proper formula.

OP please verify which one is correct, if any, before we go further