An alternating current is given by
\[
I = I_A \sin \omega t + I_B \cos \omega t.
\]
The r.m.s. current will be:
Show Hint
- \( \sqrt{I_A^2 + I_B^2} \)
- \( \frac{\sqrt{I_A^2 + I_B^2}}{2} \)
- \( \sqrt{\frac{I_A^2 + I_B^2}{2}} \)
- \( \frac{|I_A + I_B|}{\sqrt{2}} \)
The Correct Option is C
Approach Solution - 1
The given alternating current can be expressed as:
\(I = I_A \sin \omega t + I_B \cos \omega t\)
To find the root mean square (r.m.s.) value of the total current, we use the general formula for r.m.s. value of a sum of two sinusoidal functions:
\(I_{rms} = \sqrt{\text{mean of square of total current}}\)
We need to calculate the r.m.s. of the given current. Each sinusoidal component can be considered separately. For a sinusoidal current:
- The r.m.s. value of a sinusoidal component of the form \(I = I_m \sin \omega t\) or \(I = I_m \cos \omega t\) is \(\frac{I_m}{\sqrt{2}}\).
Thus, we calculate the squares of the r.m.s. values of the components:
- For \(I_A \sin \omega t\): \(\left(\frac{I_A}{\sqrt{2}}\right)^2 = \frac{I_A^2}{2}\)
- For \(I_B \cos \omega t\): \(\left(\frac{I_B}{\sqrt{2}}\right)^2 = \frac{I_B^2}{2}\)
The r.m.s. value of the sum of these two sinusoidal components is then given by:
\(I_{rms} = \sqrt{\frac{I_A^2}{2} + \frac{I_B^2}{2}}\)
This simplifies to:
\(I_{rms} = \sqrt{\frac{I_A^2 + I_B^2}{2}}\)
This matches with option 3. Therefore, the correct answer is:
\( \sqrt{\frac{I_A^2 + I_B^2}{2}} \)
Approach Solution -2
Given the alternating current expression:
\[ I = I_A \sin \omega t + I_B \cos \omega t, \] where \( I_A \) and \( I_B \) are constants, and \( \omega \) is the angular frequency of the current.
Step 1: Understanding the RMS Current
The root mean square (RMS) value of an alternating current is given by: \[ I_{\text{rms}} = \sqrt{\langle I^2 \rangle}, \] where \( \langle I^2 \rangle \) is the time average of the square of the current.
Step 2: Calculate \( \langle I^2 \rangle \)
The square of the current \( I \) is: \[ I^2 = \left( I_A \sin \omega t + I_B \cos \omega t \right)^2. \] Expanding the square: \[ I^2 = I_A^2 \sin^2 \omega t + I_B^2 \cos^2 \omega t + 2 I_A I_B \sin \omega t \cos \omega t. \] Taking the time average, we use the following identities: - \( \langle \sin^2 \omega t \rangle = \frac{1}{2} \), - \( \langle \cos^2 \omega t \rangle = \frac{1}{2} \), - \( \langle \sin \omega t \cos \omega t \rangle = 0 \). So, the time average of \( I^2 \) is: \[ \langle I^2 \rangle = I_A^2 \cdot \frac{1}{2} + I_B^2 \cdot \frac{1}{2} = \frac{I_A^2 + I_B^2}{2}. \]
Step 3: Calculate the RMS Current
The RMS current is the square root of the time average: \[ I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}. \]
Final Answer:
The RMS current is \( \boxed{\sqrt{\frac{I_A^2 + I_B^2}{2}}} \).
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