Question

The root mean square value of alternating current is equal to
(a) Twice the peak value
(b) Half the peak value
(c) $\dfrac{1}{\sqrt{2}}$ Times the peak value
(d) Peak value

Answer 1

Hint: We can get this by the help of derivation of heat in a particular time dt. This will give the right amount of root mean square value of alternating current. $I={{I}_{\text{o}}}\sin \left( \omega t \right)$. This procedure will result in the direct equation for the root mean square value of alternating current.
Formula used:
$I={{I}_{\text{o}}}\sin \left( \omega t \right),H={{I}^{2}}Rt,\int{tdt=t,\int{\cos \left( \omega t \right)dt}=\sin \left( \omega t \right)},\sin \left( 2\pi \right)=0$

Complete answer:
The root mean square value can be abbreviated to rms value. It is found out in an alternating current in a steady form. As soon as the steady current passes through some opposition in the form of resistance, it will create heat energy similar to alternating current. The rms value of alternating current is represented by ${{I}_{rms}}$.

The value of alternating when it comes across resistance is measured as $I={{I}_{\text{o}}}\sin \left( \omega t \right)$. Let this be with the condition of constant current passing with a resistance at some point of time dt. This will produce very less heat. By the formula of heat $H={{I}^{2}}Rt$ which gives the value of heat for time dt as $dH={{I}^{2}}Rdt$. Since, $I={{I}_{\text{o}}}\sin \left( \omega t \right)$ therefore,

$\begin{align}
  & dH={{\left( {{I}_{\text{o}}}\sin \left( \omega t \right) \right)}^{2}}Rdt \\
 & \Rightarrow dH={{I}_{\text{o}}}^{2}{{\sin }^{2}}\left( \omega t \right)Rdt \\
\end{align}$

The measurement of small total amount of heat in time $\dfrac{T}{2}$ is calculated as,

$\begin{align}
  & \int\limits_{0}^{\dfrac{T}{2}}{dH}=\int\limits_{0}^{\dfrac{T}{2}}{{{I}_{\text{o}}}^{2}{{\sin }^{2}}\left( \omega t \right)Rdt} \\
 & \Rightarrow H={{I}_{\text{o}}}^{2}R\int\limits_{0}^{\dfrac{T}{2}}{{{\sin }^{2}}\left( \omega t \right)dt} \\
\end{align}$

As we know that $\int{{{\sin }^{2}}\left( \omega t \right)}=\dfrac{1-\cos \left( 2\omega t \right)}{2}$ so,

\[\begin{align}
  & H={{I}_{\text{o}}}^{2}R\int\limits_{0}^{\dfrac{T}{2}}{\dfrac{1-\cos \left( 2\omega t \right)}{2}dt} \\
 & \Rightarrow H=\dfrac{{{I}_{\text{o}}}^{2}R}{2}\int\limits_{0}^{\dfrac{T}{2}}{\left( 1-\cos \left( 2\omega t \right) \right)dt} \\
 & \Rightarrow H=\dfrac{{{I}_{\text{o}}}^{2}R}{2}\left( \int\limits_{0}^{\dfrac{T}{2}}{1dt}-\int\limits_{0}^{\dfrac{T}{2}}{\cos \left( 2\omega t \right)dt} \right) \\
\end{align}\]

As we know that $\int{tdt=t,\int{\cos \left( \omega t \right)dt}=\sin \left( \omega t \right)}$ so,

\[\begin{align}
  & H=\dfrac{{{I}_{\text{o}}}^{2}R}{2}\left( \left[ t \right]_{0}^{\dfrac{T}{2}}-\left[ \cos \left( 2\omega t \right) \right]_{0}^{\dfrac{T}{2}} \right) \\
 & \Rightarrow H=\dfrac{{{I}_{\text{o}}}^{2}R}{2}\left( \dfrac{T}{2}-\dfrac{\sin \left( 2\omega t \right)\dfrac{T}{2}}{2\omega }-0 \right) \\
\end{align}\]

As the value of $\omega =\dfrac{2\pi }{T}$ therefore,

\[\begin{align}
  & H=\dfrac{{{I}_{\text{o}}}^{2}R}{2}\left( \dfrac{T}{2}-\dfrac{\sin \left( \left( 2\dfrac{2\pi }{T}t \right)\dfrac{T}{2} \right)}{2\omega }-0 \right) \\
 & \Rightarrow H=\dfrac{{{I}_{\text{o}}}^{2}R}{2}\left( \dfrac{T}{2}-\dfrac{\sin \left( 2\pi \right)}{2\omega } \right) \\
\end{align}\]

Since, $\sin \left( 2\pi \right)=0$ thus, \[H=\dfrac{{{I}_{\text{o}}}^{2}R}{2}\left( \dfrac{T}{2} \right)\] ….(i)
At this time by the rms value of alternating current then, ${{I}_{rms}}=\dfrac{{{I}_{\text{o}}}^{2}}{2}$ therefore, \[H={{I}_{rms}}^{2}R\left( \dfrac{T}{2} \right)\] …(ii).

By using comparison between (i) and (II), we will get that
\[\begin{align}
  & \dfrac{{{I}_{\text{o}}}^{2}R}{2}\left( \dfrac{T}{2} \right)={{I}_{rms}}^{2}R\left( \dfrac{T}{2} \right) \\
 & \Rightarrow {{I}_{rms}}^{2}=\dfrac{{{I}_{\text{o}}}^{2}}{2} \\
 & \Rightarrow {{I}_{rms}}=\dfrac{{{I}_{\text{o}}}}{\sqrt{2}} \\
\end{align}\]

So, the correct answer is “Option c”.

Note:
In order to solve this equation we will apply derivation for the amount of heat present in a small time interval. By this we will be able to get to the equation of root mean square value of alternating current. We can also use a direct formula for the condition if it is known to us. In the equation \[{{I}_{rms}}=\dfrac{{{I}_{\text{o}}}}{\sqrt{2}}\] the symbol \[{{I}_{\text{o}}}\] is working as a peak value. We should solve such derivations by keeping complete focus on it otherwise problems will arise. By taking a small amount of heat from $H={{I}^{2}}Rt$ equation as $dH={{I}^{2}}Rdt$ in the starting of derivation will be helpful here.