Question

In an AC circuit, current $I=100\sin 200\pi t$ (where I is in ampere and t is in second). The time required for the current to reach its RMS value from zero will be
A. $\dfrac{1}{100}s$
B. $\dfrac{1}{200}s$
C. $\dfrac{1}{400}s$
D. $\dfrac{1}{800}s$

Answer 1

Hint: We know that the standard representation of ac current is given by$I={{I}_{0}}\sin \omega t$, where, ${{I}_{0}}$ is the peak value of current. Comparing the standard expression with the given expression of ac current will give you the peak value. Now, recall that the root mean square value (RMS value)of ac current is $\dfrac{1}{\sqrt{2}}$ times the peak value and then substitute this RMS value in place of I and then find t.

Formula used:
Expression for ac current,
$I={{I}_{0}}\sin \omega t$
Expression for RMS current,
${{I}_{rms}}=\dfrac{{{I}_{0}}}{\sqrt{2}}$

Complete answer:
In the above figure, the source is producing a sinusoidally varying potential difference across the resistor. This sinusoidally varying potential difference is called ac voltage and it is given by,
$V={{V}_{0}}\sin \omega t$
Where, ‘${{V}_{0}}$’ is the amplitude of the oscillating potential difference.
‘$\omega $’ is the angular frequency of the oscillating potential difference.
Now let us find the value of the current through the resistor of resistance R. For that let us apply Kirchhoff’s loop rule that says that the algebraic sum of the potential differences around any closed loop involving resistors and cells is zero. That is,
$\sum{\varepsilon \left( t \right)}=0$
$\Rightarrow {{V}_{0}}\sin \omega t-IR=0$
$\Rightarrow {{V}_{0}}\sin \omega t=IR$
$\Rightarrow I=\dfrac{{{V}_{0}}}{R}\sin \omega t$
Since R here remains constant, this equation can be rewritten as,
$I={{I}_{0}}\sin \omega t$ ……………………………. (1)
Where ‘${{I}_{0}}$’ is the peak current or the current amplitude given by,
${{I}_{0}}=\dfrac{{{V}_{0}}}{R}$
In the question ac current is given as,
$I=100\sin 200\pi t$………………………….. (2)
Comparing (2) and (1),
${{I}_{0}}=100A$
$\omega =200\pi $
But we know that,
$\omega =2\pi f$
$\Rightarrow 2\pi f=200\pi $
$\Rightarrow f=100$
For expressing ac power in the same form as dc power, a special value of current is defined and used. It is the root mean square (rms) or effective current and is denoted by${{I}_{rms}}$ .
By definition,
${{I}_{rms}}=\sqrt{\overline{{{I}^{2}}}}$
$\Rightarrow {{I}_{rms}}=\sqrt{\dfrac{1}{2}{{I}_{0}}^{2}}=\dfrac{{{I}_{0}}}{\sqrt{2}}$ ……………………….. (3)
$\Rightarrow {{I}_{rms}}=0.707{{I}_{0}}$
From the question,
${{I}_{0}}=100A$
$\Rightarrow {{I}_{rms}}=\dfrac{\left( 100 \right)}{\sqrt{2}}A$
Remember that, at time $t=0$ , ac current $I=0$ .
In the question we are asked to find the time taken for the current I to reach its ${{I}_{rms}}$ value.
That is, in equation (1) we are supposed to find t when $I={{I}_{rms}}$
$I=100\sin 200\pi t$
$\Rightarrow {{I}_{rms}}=100\sin 200\pi t$
$\Rightarrow \dfrac{100}{\sqrt{2}}=100\sin 200\pi t$
$\Rightarrow {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)=200\pi t$
$\Rightarrow 200\pi t=\dfrac{\pi }{4}$
$\Rightarrow t=\dfrac{1}{800}s$
Therefore, the time required for the current to reach its RMS value from zero is$\dfrac{1}{800}s$ .

So, the correct answer is “Option D”.

Note:
If you know the basic expression of ac current then all you have to do is compare the given expression in the question and also the standard expression. Then from there you will get almost all the quantities that are required about an ac current like its peak value, frequency, angular frequency, etc. You could memorize these standard forms so that it consumes less time in solving these type questions.