Question

In an AC circuit the r.m.s value of the current, ${I_{r.m.s}}$, is related to the peak current${I_ \circ }$ as:
A) ${I_{r.m.s}} = \dfrac{1}{\Pi }{I_0}$
B) ${I_{r.m.s}} = \dfrac{1}{{\sqrt 2 }}{I_0}$
C) ${I_{r.m.s}} = \sqrt 2 {I_0}$
D) ${I_{r.m.s}} = \Pi {I_0}$

Answer 1

The alternating current varies with time according to Sin(wt) or Cos(wt) and reverses periodically. But the mean value of the current does not change with time.

Formula used:
$I = {I_0}\sin wt$

\overline {{I^2}} = \dfrac{1}{T}\int_0^T {{I^2}} dt \\\ \\\ \end{gathered} $$ **Complete step by step answer:** The full form of RMS is the root-mean-square of some instances of values. The alternating current has a root mean square value that is given by direct current that flows through a resistance. The alternating current may be represented as, $$\begin{gathered} I = {I_0}\sin wt \\\ and,I = {I_0}\cos wt \\\ \end{gathered} $$ Let, the alternating current is represented as, $$I = {I_0}\sin wt................(1)$$ So, the mean of the square root of $$I$$is, $$\overline {{I^2}} = \dfrac{1}{T}\int_0^T {{I^2}} dt$$ To integrate we need to substitute the limits. Substituting the limits we get: $$ \rightarrow \overline {{I^2}} = \int_0^T {{I_0}^2{{\sin }^2}wtdt} $$ We can now convert the sin into cos. $$ \rightarrow \overline {{I^2}} = \dfrac{{{I_0}^2}}{{2T}}\int_0^T {(1 - \cos 2wt)dt} $$ Now let us integrate the given equation $$ \rightarrow \overline {{I^2}} = \dfrac{{{I_0}^2}}{{2T}}[t - \dfrac{{\sin 2wt}}{{2w}}]_0^T$$ By Substituting the limits $$ \rightarrow \overline {{I^2}} = \dfrac{{{I_0}^2}}{{2T}}[T - \dfrac{{\sin 4\Pi }}{{2w}} + \dfrac{{\sin 0}}{{2w}}][\because Tw = 2\Pi ]$$ $$ \rightarrow \overline {{I^2}} = \dfrac{{{I_0}^2}}{{2T}} \times T[\because \sin 4\Pi = 0]$$ $$ \rightarrow \overline {{I^2}} = \dfrac{{{I_0}^2}}{2}$$ The answer is; $$\therefore {I_{r.m.s}} = \sqrt {\overline {{I^2}} } = \sqrt {\dfrac{{{I_0}^2}}{2}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$$ **Hence the correct option is, option (B).** **Additional information:** In a conductor if the current passes according to Joules heat law it creates heat which is proportional to voltage square(${V^2}$) and current’s square(${I^2}$). This heat does not depend upon the direction of the current. If any changes happen in the alternating current then the heat of the conductor will change from zero to its maximum value. **Notes:** The alternating current (AC) generates heat in the conductor. If we apply a constant amount of direct current (DC) in that conductor and get the same amount of heat then this value of DC will be known as the effective value of AC. In a full cycle of AC with various time intervals, the mean root of ${V^2}$ or ${I^2}$is the same as the effectictive value. This is known as the RMS value of voltage or current.