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 ?
There have been a lot of comments but I think it's still valuable to sketch the solution:
The impedances of \$L\$ and \$C\$ are given at an angular frequency \$\omega = 2000\$ rad/s. This means that
$$\omega L = 6\Omega\text{ and } \frac{1}{\omega C}=8\Omega$$ which gives the following values for \$L\$ and \$C\$: $$L=3mH\quad C=62.5\mu F$$
If you don't know the formula for the resonance frequency by heart, it's very easy to derive it (if you know that the impedance must be purely real-valued at resonance):
$$Z = R + j\left (\omega L -\frac{1}{\omega C} \right)$$ The imaginary part of the impedance \$Z\$ disappears for \$\omega L = \frac{1}{\omega C}\$, i.e.
$$\omega_0 = \frac{1}{\sqrt{LC}} = 2309.4\text{ rad/s}$$
Since at resonance the impedance of the capacitor and the inductor are equal, the voltages across them must be equal. So the current through the inductor at resonance must be
$$\frac{4mV}{\omega_0 L} = 0.577\text{ mA}$$
The problem is your current measurement method and assumptions for load current.
I get I(R1) = 177 mV dc /150 Ω = 1.18mA with a 1K load at 9V out on 470uF cap and 9.7V Zener with Iz=17mA pk using Si diodes.
Since nominal Zener Vz= 9.4~10.6 @ 5mA and Zt is around 15 Ω at 17mA pk
I used 1k load for 1.18mA average. Assumptions for load on mine were different than yours.
Your measurements were:
Your assumption on how multimeters work is wrong for AC current.
To make accurate complex current measurements with cheap multimeters, use a shunt resistor and 10K R with 100uF cap and measure DC voltage.
Best Answer
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.)