# Electrical – Finding times in first cycle with given voltage

acwaveform

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 ?

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.)