Question

Alternating current in a circuit is given by $I={{I}_{o}}\sin 2\pi nt$. Then the time taken by the current to rise from zero to r.m.s. value equal to
A. $\dfrac{1}{2n}$
B. $\dfrac{1}{n}$
C. $\dfrac{1}{4n}$
D. $\dfrac{1}{8n}$

Answer 1

Hint: AC's Root Mean Square (RMS) value is that steady current that creates the same amount of heat as passing through resistance for a given time as the alternating current emits at the same time in the same resistance. Value of r.m.s. or abstract value or real value of a.c. Range 0.707 times or $\dfrac{1}{\sqrt{2}}$ times the maximum value of a.c.

Formula Used: For solving this question, we will be using the formula for the root mean square value of current, i.e.,
${{I}_{rms}}=\dfrac{{{I}_{o}}}{\sqrt{2}}$

Complete step-by-step solution:
Before we start solving the question, let us take a look at the given parameters
Alternating current = $I={{I}_{o}}\sin 2\pi nt$
Now, as we discussed above.
${{I}_{rms}}=\dfrac{{{I}_{o}}}{\sqrt{2}}$
So, we have
Time taken by the current to rise from zero to r.m.s.
$\dfrac{{{I}_{o}}}{\sqrt{2}}={{I}_{o}}\sin 2\pi nt$
Now, as we know that $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
We have,
$\sin \dfrac{\pi }{4}=\sin 2\pi nt$
$\dfrac{\pi }{4}=2\pi nt$
So, we have on solving,
$t=\dfrac{1}{8n}$
So the time taken by the current to rise from zero to r.m.s. value equal to $t=\dfrac{1}{8n}$, i.e., Option D.

Note: For the same power, the RMS gives you the corresponding DC voltage. If you calculate the temperature of the resistor as a dissipated energy calculation, you may find that it is the same as for a 0.71 V DC voltage, not 0.64 V. Nevertheless, calculating average voltage is cheaper than calculating RMS voltage, and that's what cheaper DMMs do.