# Finding times in first cycle with given voltage

I have the question "The instantaneous value of voltage in an a.c. Circuit at any time t seconds is given by: $$V = 100\sin(50\pi{}t - 0.523)\ \rm{V}$$

Find:

The times in the first cycle when the voltage is -40V."

Here is my attempt:

My final answer is t = 5.949 ms however the solutions say that the answer should be 25.95 ms.

Where have I gone wrong ?

• did you mean to drop the minus sign on the sin^-1(-0.4) (second bottom line)? Feb 18, 2017 at 18:41
• You have changed the sign of the value inside the arcsin() function in your last two steps.
– jonk
Feb 18, 2017 at 18:45
• Oh no that was a mistake sorry :) Feb 18, 2017 at 18:45
• But even changing this still does not give me the correct answer. Feb 18, 2017 at 18:46
• Remember the sinus goes through -0.4 twice per cycle, but the arcsin function only tells you the location of one of those points. I'd draw a graph of the function to be sure of picking the correct crossing. Feb 18, 2017 at 18:47

You have: \begin{align*} V_t &= 100\: \textrm{V}\cdot\operatorname{sin}\left(2 \pi\:\operatorname{rad}\cdot25\operatorname{Hz}\cdot\: t - 0.523 \operatorname{rad}\right) \end{align*}

And you want to solve for $t$ where $t\ge 0$ and $V_t=-40\:\textrm{V}$. So, let's set $x$ as follows:

$$x= 2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}\cdot\: t - 0.523 \operatorname{rad}$$

Then we have:

\begin{align*} -40\:\textrm{V} &= 100\: \textrm{V}\cdot\operatorname{sin}\left(x\right)\\\\ \frac{-40\:\textrm{V}}{100\: \textrm{V}} &= \operatorname{sin}\left(x\right)\\\\ -0.4 &= \operatorname{sin}\left(x\right) \end{align*}

Before going any further, the value of $x$ has multiple solutions. The solutions are:

\begin{align*} x&=2\pi\cdot n + \operatorname{sin}^{-1}\left(-0.4\right)= 2\pi\cdot n - 0.411516846 \operatorname{rad}\\&=2\pi\cdot n +\pi- \operatorname{sin}^{-1}\left(-0.4\right)= 2\pi\cdot n +3.5531095 \operatorname{rad} \end{align*}

Combining this information, we have:

\begin{align*} 2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}\cdot\: t - 0.523 \operatorname{rad}&= 2\pi\cdot n - 0.411516846 \operatorname{rad}\\2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}\cdot\: t - 0.523 \operatorname{rad}&= 2\pi\cdot n +3.5531095\operatorname{rad} \end{align*}

These solve out as:

\begin{align*} t&= \frac{2\pi\cdot n - 0.411516846 \operatorname{rad} + 0.523 \operatorname{rad}}{2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}}=\frac{2\pi\cdot n +0.111483154\operatorname{rad}}{2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}}\\\\ t&= \frac{2\pi\cdot n +3.5531095 \operatorname{rad} + 0.523 \operatorname{rad}}{2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}}=\frac{2\pi\cdot n +4.0761095\operatorname{rad}}{2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}} \end{align*}

That's the full answer. But for values of $t\ge 0$, you find the following for the first two answers with $n=0$:

\begin{align*} t&=709.723801\:\mu\textrm{s}\\\\ t&=25.949319\:\textrm{ms} \end{align*}

I hope that helps out. As you can see, the trick is mostly in taking very careful steps and not to move too rapidly towards a quick "calculator" solution, which would find the first answer, perhaps, but not the second (which appears to be the desired one.)

• The solutions say that the answers should be 25.95 ms and 40.71 ms so where does the 40.71 ms come from ? Thanks. Feb 24, 2017 at 20:27
• @DanKhan I provided TWO equations and gave you the results for $n=0$. Try using those same equations with $n=1$. I think you will find your answer there. Also, please keep in mind that there are MANY answers. An infinite number of them. In short, the second equation with $n=0$ and the first equation with $n=1$ provides your two answers. You should sketch on paper, by hand, too. Just so you can see what the equations give you, as a picture.
– jonk
Feb 24, 2017 at 20:34
• Oh okay I fully understand why there are many answers now because n can be any value. Thanks :) Feb 24, 2017 at 20:43
• @DanKhan Yes! :) n is like an index, of sorts. Or, which "cycle" you are looking at. Not complicated, really, once you see a picture.
– jonk
Feb 24, 2017 at 21:41