Question

An alternating current is given by i = i₁ sin ωt + i₂ cos ωt. The r.m.s. current is given by 

• A. √ (i₁² + i₂²)/ 2
• B. (i₁ + i₂)/ √2
• C. √ (i₁² + i₂²)/ √2
• D. (i₁ - i₂)/ √2

Answer 1

The Correct Option is A

Given: \( i = i_1 \sin \omega t + i_2 \cos \omega t \)

Step 1: Square the expression 

\( i^2 = (i_1 \sin \omega t + i_2 \cos \omega t)^2\)

\(= i_1^2 \sin^2 \omega t + i_2^2 \cos^2 \omega t + 2i_1 i_2 \sin \omega t \cos \omega t\)

Step 2: Average over one time period

The average of \( \sin^2 \omega t \) and \( \cos^2 \omega t \) over one time period is \( \frac{1}{2} \).

The average of \( \sin \omega t \cos \omega t \) over one time period is 0.

Therefore, the average of \( i^2 \) over one time period is:

\( \frac{(i_1^2 \sin^2 \omega t + i_2^2 \cos^2 \omega t + 2i_1 i_2 \sin \omega t \cos \omega t)}{2} = \frac{i_1^2 + i_2^2}{2}\)

Step 3: Take the square root

\( \text{r.m.s. current} = \sqrt{\frac{i_1^2 + i_2^2}{2}}\)

Comparing this result to the given options:

• (1) \( \sqrt{\frac{i_1^2 + i_2^2}{2}} \) – This matches our calculated expression.
• (2) \( \frac{i_1 + i_2}{\sqrt{2}} \) – This is not equivalent to our expression.
• (3) \( \sqrt{\frac{i_1^2 + i_2^2}{\sqrt{2}}} \) – This is not equivalent to our expression.
• (4) \( \frac{i_1 - i_2}{\sqrt{2}} \) – This is not equivalent to our expression.

Therefore, the correct answer is: (A) \( \sqrt{\frac{i_1^2 + i_2^2}{2}} \).