Question

The r.m.s. value of a current given by \( i = (i_1 \cos \omega t + i_2 \sin \omega t) \) is:

• A. \( \frac{1}{\sqrt{2}} (i_1 + i_2) \)
• B. \( \frac{1}{\sqrt{2}} (i_1 - i_2) \)
• C. \( \frac{1}{\sqrt{2}} \sqrt{i_1^2 + i_2^2} \)
• D. \( \frac{1}{\sqrt{2}} (i_1^2 + i_2^2) \)

Hint:

To calculate the r.m.s. value for a combined sinusoidal current, square each term, take the mean, and then the square root.

Answer 1

The Correct Option is C

Calculating the R.M.S. Value of a Current

1: R.M.S. Value FormulaThe root mean square (r.m.s.) value of a current \( i(t) \) is defined as:\[i_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T i^2(t) \, dt}\]where \( T \) represents the oscillation period.
2: Squaring the CurrentGiven the current expression:\[i = i_1 \cos \omega t + i_2 \sin \omega t\]The square of the current is:\[i^2 = (i_1 \cos \omega t + i_2 \sin \omega t)^2 = i_1^2 \cos^2 \omega t + i_2^2 \sin^2 \omega t + 2 i_1 i_2 \cos \omega t \sin \omega t\]
3: Averaging Over One PeriodOver a period \( T \), the average values are \( \langle \cos^2 \omega t \rangle = \frac{1}{2} \), \( \langle \sin^2 \omega t \rangle = \frac{1}{2} \), and \( \langle \cos \omega t \sin \omega t \rangle = 0 \). Therefore:\[\langle i^2 \rangle = \frac{1}{2} i_1^2 + \frac{1}{2} i_2^2\]
4: Final R.M.S. Value CalculationThe r.m.s. value of the current is then:\[i_{\text{rms}} = \sqrt{\frac{1}{2} i_1^2 + \frac{1}{2} i_2^2} = \frac{1}{\sqrt{2}} \sqrt{i_1^2 + i_2^2}\]The final answer is:\[\boxed{(C) \, \frac{1}{\sqrt{2}} \sqrt{i_1^2 + i_2^2}}\]