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) \( \sqrt{\frac{i_1^2 + i_2^2}{2}} \).

Answer 2

Given the alternating current \(i = i_1 \sin(\omega t) + i_2 \cos(\omega t)\), we want to find the RMS current, \(i_{rms}\).

The RMS value of a current \(i(t)\) is defined as:

\(i_{rms} = \sqrt{\frac{1}{T} \int_0^T i^2(t) \, dt}\), where \(T\) is the period.

In this case, \(i(t) = i_1 \sin(\omega t) + i_2 \cos(\omega t)\). Thus,

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

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

We need to integrate this over one period \(T = \frac{2\pi}{\omega}\). The integral becomes:

\(\int_0^T i^2(t) \, dt = \int_0^T (i_1^2 \sin^2(\omega t) + 2 i_1 i_2 \sin(\omega t) \cos(\omega t) + i_2^2 \cos^2(\omega t)) \, dt\)

We know the following integrals over one period:

• \(\int_0^T \sin^2(\omega t) \, dt = \frac{T}{2}\)
• \(\int_0^T \cos^2(\omega t) \, dt = \frac{T}{2}\)
• \(\int_0^T \sin(\omega t) \cos(\omega t) \, dt = 0\)

Thus, our integral simplifies to:

\(\int_0^T i^2(t) \, dt = i_1^2 \frac{T}{2} + 2 i_1 i_2 (0) + i_2^2 \frac{T}{2} = \frac{T}{2}(i_1^2 + i_2^2)\)

Now, we can find \(i_{rms}\):

\(i_{rms} = \sqrt{\frac{1}{T} \int_0^T i^2(t) \, dt} = \sqrt{\frac{1}{T} \frac{T}{2}(i_1^2 + i_2^2)}\)

\(i_{rms} = \sqrt{\frac{i_1^2 + i_2^2}{2}}\)

Therefore, the RMS current is \(\sqrt{\frac{i_1^2 + i_2^2}{2}}\).