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What is the r.m.s. value of the current for A.C. current \[I=100cos\left( 200t+{{45}^{\circ }} \right)\] .
$\left( a \right)50\sqrt{2}A$
$\left( b \right)100A$
$\left( c \right)100\sqrt{2}A$
$\left( d \right)0A$

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Hint: The rms value is also called the effective value of a sinusoidal waveform that gives the same heating effect of an equivalent DC supply. So, the rms value of AC current is calculated by maximum value of current divided by square root of $2$.

Formula used: ${{I}_{rms}}=\dfrac{{{I}_{m}}}{\sqrt{2}}$
Where: ${{I}_{rms}}$ - rms value of sinusoidal AC current
${{I}_{m}}$ - maximum or peak value of current

Complete step by step answer:
As per given equation of AC current:
\[I=100cos\left( 200t+{{45}^{\circ }} \right)A\]
So, the maximum current ${{I}_{m}}$ according to the equation
${{I}_{m}}=100A$
RMS value of current
${{I}_{rms}}=\dfrac{{{I}_{m}}}{\sqrt{2}}$
$\begin{align}
  & \Rightarrow {{I}_{rms}}=\dfrac{100}{\sqrt{2}} \\
 & \Rightarrow {{I}_{rms}}=50\sqrt{2}A \\
\end{align}$
So, for above given equation the ${{I}_{rms}}=50\sqrt{2}A$

So, the correct answer is “Option A”.

Additional Information: The term “RMS” stands for “Root-Mean-Squared”. Mostly define this because the “amount of AC power that produces the same heating effect as an equivalent DC power”, or something similar along these lines, but an RMS value is more than just that. The RMS value is that the root of the mean (average) value of the squared function of the instantaneous values. If I be the basis mean square (rms) value of the complex current, then the equivalent heating effect is going to be ${{I}^{2}}RT$. But from the definition, this must be equal to the heating actually produced by the complex current.

Note: The higher the Mean values of current with same RMS values, the more the absorbed heat, the effect of the Mean value of current depends on the RMS values. It is important to recall that by representing current either by Mean value or by RMS value it does not mean that a similar Mean and RMS value would produce the same amount of heat. In addition, another point must be raised to stop misunderstandings, i.e., the difference between Mean and RMS values of a sign (such as current) and therefore the decomposition of the signal into DC and AC components. The summation of the Mean and RMS values of the DC and AC components, respectively, doesn't describe the present signal.