Question

If the rms current in a $50\;Hz$ ac circuit is $5\;A$, the value of the current 1/300 seconds after its value becomes zero is
(A) $5\sqrt 2 \;{\rm{A}}$
(B) $5\sqrt {\dfrac{3}{2}} \;{\rm{A}}$
(C) $\dfrac{5}{6}$
(D) $\dfrac{5}{{\sqrt 2 }}A$

Answer 1

The expression of the sinusoidal current directly relates to the rms current, time, and frequency. In the question, the value of the rms current frequency and time is given, so we can directly use the expression of the sinusoidal current.

Complete step by step answer:
The alternating current whose amplitude changes with time known as sinusoidal current, or you can say that the alternating current is known as sinusoidal current. Write the expression of the sinusoidal current,

$I = {I_o}\sin \omega t$.... (1)

Here, ${I_o}$ is the peak current, $\omega$ is the angular velocity and $t$ is the time.

We know the expression of the angular velocity is $\omega = 2\pi f$, so put the values in this expression and calculate$\omega$.

Therefore, we get,
$\begin{array}{l} \omega = 2\pi f\\\ \omega = 2\pi \times 50\;\;{\rm{Hz}}\\\ \omega = 100\;{\rm{rd/sec}} \end{array}$

We know the relation of peak current and rms current is ${I_o} = \sqrt 2 {I_{rms}}$, so we will use the value of rms current for the calculation of peak current

Substitute the values in the relation, therefore we get

$\begin{array}{l} {I_o} = \sqrt2 \times 5\;{\rm{A}}\\\ {{\rm{I}}_o} = 5\sqrt 2 \;{\rm{A}} \end{array}$

Now we have the value of $\omega$ and ${I_o}$, by substituting these values in equation (1) we can calculate the current after1/300 seconds, so

I = {I_o}\sin \omega t\\\ I = \left( {5\sqrt 2 \;{\rm{A}}} \right)\sin \left( {100\pi \;{\rm{rad/s}} \times \dfrac{1}{{300}}\;{\rm{s}}} \right)\\\ I = \left( {5\sqrt 2 \;{\rm{A}}} \right)\sin \left( {\dfrac{\pi }{3}} \right)\\\ I = \left( {5\sqrt 2 \;{\rm{A}}} \right) \times \dfrac{{\sqrt 3 }}{2} \end{array}$$ On further solving, we get $$I = 5\sqrt {\dfrac{3}{2}} \;{\rm{A}}$$ Therefore, if the rms current in a $50\;Hz$ ac circuit is $5\;A$, the value of the current 1/300 seconds after its value becomes zero is $$5\sqrt {\dfrac{3}{2}} \;{\rm{A}}$$ **So, the correct answer is “Option B”.** **Note:** Try to remember the expression sinusoidal current and put the correct value of $sin\omega t$, so that we can obtain the correct answer easily. You can revise the concept of trigonometry for the various values of $\sin \theta $ because wrong values may give incorrect answers.