The rms value of current in a \[50\;{\text{Hz}}\] AC circuit is \[{\text{6}}\;{\text{A}}\]. The average value of AC current over a cycle is
A. \[{\text{6}}\sqrt 2 \]
B. \[\dfrac{3}{{\pi \sqrt 2 }}\]
C. Zero
D. \[\dfrac{6}{{\pi \sqrt 2 }}\]

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Hint: The formula for rms current is given by,
\[{I_{{\text{rms}}}} = \dfrac{{{I_{\text{o}}}}}{{\sqrt 2 }}\].
The formula for angular frequency is given by,
\[\omega = 2\pi f\]

Complete step by step solution:
Given, the rms value of the current,
\[{I_{{\text{rms}}}} = 6\;{\text{A}}\]
\[f = 50\;{\text{Hz}}\]

rms current: The root mean square is defined in mathematics and its applications as the square root of the medium square. The rms is also called the quadratic mean and is a specific case of the generalised mean with exponent 2. The formula for rms current is given by,
\[{I_{{\text{rms}}}} = \dfrac{{{I_{\text{o}}}}}{{\sqrt 2 }}\], where \[{I_{\text{o}}}\] is the peak value of an alternating current.

Rewrite the above equation in order to calculate the value of \[{I_{\text{o}}}\].
\[{I_{\text{o}}} = {I_{{\text{rms}}}}\sqrt 2 \] …… (i)
Substitute the value of \[{I_{{\text{rms}}}}\] in equation (i).
\[{I_{\text{o}}} = 6\sqrt 2 \;{\text{A}}\]

The angular frequency in physics is a scalar function of the amount of rotation. It relates to the angular displacement per unit time or the rate of change in a sinusoidal waveform phase, or as the rate of change in the sine function argument.
The formula for angular frequency is given by,
\[\omega = 2\pi f\]
Substitute the value of frequency \[f\] in the above equation.
  \omega = 2\pi f \\
   = 2 \times \pi \times 50 \\
   = 100\pi \\

The instantaneous value of an alternating voltage or current at a given moment is the value of voltage or current. If the particular instant is the time in the cycle at which the voltage polarity changes, the value may be zero.
The formula for instantaneous current is obtained by using the formula,
\[I = {I_{\text{o}}}\sin \omega t\] …… (ii)

Substitute the values of \[{I_{\text{o}}}\] and \[\omega \] in equation (ii),
\[I = 6\sqrt 2 \sin 100\pi t\]
Average current relates to the average of each and every instantaneous current value from zero to peak and back on a sine wave; a sine wave represents alternating or AC current.
The formula for average current is given by,
\[{I_{avg}} = \dfrac{{\int\limits_0^T {I\left( t \right)dt} }}{{\int\limits_0^T {dt} }}\] …… (iii)
Substitute the values of \[I\left( t \right)\] in equation (iii) and solve.
  {I_{avg}} = \dfrac{{\int\limits_0^{\text{T}} {6\sqrt 2 \sin 100\pi tdt} }}{{\left[ t \right]_0^{\text{T}}}} \\
   = \dfrac{{6\sqrt 2 \int\limits_0^T {\sin \left( {100\pi t} \right)dt} }}{{\text{T}}} \\

Now, we know that the integration of area under the sinusoidal curve over a complete cycle is always zero.
\[{I_{avg}} = \dfrac{{6\sqrt 2 \times 0}}{{\text{T}}} = 0\]

Hence, option C is correct.

Note: In this problem we need to calculate the average current which is given by,
\[{I_{avg}} = \dfrac{{\int\limits_0^T {I\left( t \right)dt} }}{{\int\limits_0^T {dt} }}\]. In order to calculate the average current calculate the value of instantaneous current using the formula, \[I = {I_{\text{o}}}\sin \omega t\] where \[\omega \] is angular frequency and is calculated using the formula, \[\omega = 2\pi f\].