Question

An alternating current is given by \[ I = I_A \sin \omega t + I_B \cos \omega t. \] The r.m.s. current will be:

• A. \( \sqrt{I_A^2 + I_B^2} \)
• B. \( \frac{\sqrt{I_A^2 + I_B^2}}{2} \)
• C. \( \sqrt{\frac{I_A^2 + I_B^2}{2}} \)
• D. \( \frac{|I_A + I_B|}{\sqrt{2}} \)

Hint:

The r.m.s. current is found by squaring the current expression, averaging over one period, and then taking the square root. When the current is a sum of sine and cosine functions, this simplifies to the formula given above.

Answer 1

The Correct Option is C

The provided alternating current is represented as:

\(I = I_A \sin \omega t + I_B \cos \omega t\)

To determine the root mean square (r.m.s.) value of the total current, we apply the general formula for the r.m.s. value of a sum of two sinusoidal functions:

\(I_{rms} = \sqrt{\text{mean of the square of the total current}}\)

We need to calculate the r.m.s. value of the given current. Each sinusoidal component can be analyzed independently. For a sinusoidal current of the form \(I = I_m \sin \omega t\) or \(I = I_m \cos \omega t\), the r.m.s. value is \(\frac{I_m}{\sqrt{2}}\).

Consequently, we compute the squares of the r.m.s. values of the individual components:

• For \(I_A \sin \omega t\): The r.m.s. value is \(\frac{I_A}{\sqrt{2}}\), and its square is \(\left(\frac{I_A}{\sqrt{2}}\right)^2 = \frac{I_A^2}{2}\).
• For \(I_B \cos \omega t\): The r.m.s. value is \(\frac{I_B}{\sqrt{2}}\), and its square is \(\left(\frac{I_B}{\sqrt{2}}\right)^2 = \frac{I_B^2}{2}\).

The r.m.s. value of the sum of these two sinusoidal components is then calculated as:

\(I_{rms} = \sqrt{\frac{I_A^2}{2} + \frac{I_B^2}{2}}\)

This expression simplifies to:

\(I_{rms} = \sqrt{\frac{I_A^2 + I_B^2}{2}}\)

This result corresponds to option 3. Therefore, the correct answer is:

\( \sqrt{\frac{I_A^2 + I_B^2}{2}} \)