Question

An alternating current varying sinusoidally at $50Hz$ has the RMS value of $10A$. Write down an equation for instantaneous value of current. Find the value of current at
(A) $0.0025$seconds after passing through the positive maximum value.
(B) $0.0075$seconds after passing through 0 value and increasing negatively.

Answer 1

Hint
We can find the value of current from the RMS value given by multiplying $\sqrt{2}$ with it and then write the instantaneous equation from in the form, $i=I\sin \omega t$ where the $\omega$can be found from the given value of frequency. By substituting the value of time in the equation, we can find the current at various times.
In this solution we will be using the following formula,
$\Rightarrow I_{RMS}={\displaystyle \frac{\mathrm{I}}{\sqrt{2}}}$ where $I_{RMS}$ is RMS value of current and $I$ is the maximum value of current.
$\Rightarrow \omega =2\pi \upsilon$ where $\upsilon$ is the given frequency and $\omega$ is the angular frequency.
$\Rightarrow i=I\cos \omega t$ where $i$ is the instantaneous current at time $t$.

Complete step by step answer
In this problem, we are provided with the RMS value of the current. Now the RMS value is given by the formula, $I_{RMS}={\displaystyle \frac{\mathrm{I}}{\sqrt{2}}}$
So from here we can find the maximum value of current as,
$\Rightarrow I=I_{RMS}\sqrt{2}$
By substituting the values we get the maximum current as,
$\Rightarrow I=10\times \sqrt{2}$
On calculating this gives us,
$\Rightarrow I=14.14A$
The frequency given in the question is, $50Hz$. From this value we can find the angular frequency of the sinusoidal wave by the formula, $\omega =2\pi \upsilon$
So by substituting the value of $\upsilon$ we have,
$\Rightarrow \omega =2\pi \times 50$
This gives us, $\omega =100\pi rad/s$
The equation of the instantaneous current can be written of the form,
$\Rightarrow i=I\sin \omega t$
So using these values of current and angular frequency we can write the equation for instantaneous current as,
$\Rightarrow i=14.14\sin \left(100\pi t\right)$
So this is the equation for the instantaneous value of current.
For the next part of the question, the time values are given from the positive maximum. So in that case we can write the equation in the cosine form, since the cosine wave starts from the positive maximum, as
$i=I\cos \omega t$
Substituting values,
$\Rightarrow i=14.14\cos \left(100\pi t\right)$
For the first part we need to find the instantaneous current at $t=0.0025s$
So substituting the value of time in the equation we have,
$\Rightarrow i=14.14\cos \left(100\pi \times 0.0025\right)$
Here $100\pi \times 0.0025$ is in radian. So to convert it to degrees we multiply, ${\displaystyle \frac{{180}^{\circ }}{\pi }}$
Therefore, we get the current as,
$\Rightarrow i=14.14\cos \left(100\pi \times {\displaystyle \frac{{180}^{\circ }}{\pi }}\times 0.0025\right)$
This gives us a value of,
$\Rightarrow i=14.14\cos \left(100\times {180}^{\circ }\times 0.0025\right)$
Hence we get the value of the angle as
$\Rightarrow i=14.14\cos \left(45\right)$
Since $\cos \left(45\right)={\displaystyle \frac{1}{\sqrt{2}}}$
So we get the value of instantaneous current at $t=0.0025s$ is,
$\Rightarrow i=10\sqrt{2}\times {\displaystyle \frac{1}{\sqrt{2}}}A$, that is,
$\Rightarrow i=10A$
For the second part we need to find the instantaneous current at $t=0.0075s$
So substituting the value of time in the equation we have,
$\Rightarrow i=14.14\cos \left(100\pi \times 0.0075\right)$
Here $100\pi \times 0.0075$ is in radian. So to convert it to degrees we multiply, ${\displaystyle \frac{{180}^{\circ }}{\pi }}$
Therefore, we get the current as,
$\Rightarrow i=14.14\cos \left(100\pi \times {\displaystyle \frac{{180}^{\circ }}{\pi }}\times 0.0075\right)$
This gives us a value of,
$\Rightarrow i=14.14\cos \left(100\times {180}^{\circ }\times 0.0075\right)$
Hence we get the value of the angle as
$\Rightarrow i=14.14\cos \left(135\right)$
Since $\cos \left(135\right)=-{\displaystyle \frac{1}{\sqrt{2}}}$
So we get the value of instantaneous current at $t=0.0075s$ is,
$\Rightarrow i=10\sqrt{2}\times \left(-{\displaystyle \frac{1}{\sqrt{2}}}\right)A$, that is,
$\Rightarrow i=-10A$

Note
The root mean square or the RMS is a statistical measure of a varying quantity. It can be used in AC circuits to measure the average values of current and voltage. It can be found by dividing the peak current or the peak voltage by $\sqrt{2}$.