Find the effective value of current.
`i=2 sin 100 (pi)t + 2 cos (100 pi t + 30^(@))`.

Explanation

To find the effective (RMS) value of the current given by the equation:

\[ i = 2 \sin(100 \pi t) + 2 \cos(100 \pi t + 30^\circ) \]

we will follow these steps:

### Step 1: Rewrite the Current Equation
The given current can be rewritten as:
\[ i = 2 \sin(100 \pi t) + 2 \cos(100 \pi t + 30^\circ) \]

### Step 2: Expand the Cosine Term
Using the cosine addition formula:
\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]
we can expand the cosine term:
\[ \cos(100 \pi t + 30^\circ) = \cos(100 \pi t) \cos(30^\circ) - \sin(100 \pi t) \sin(30^\circ) \]

Substituting the known values:
- \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)
- \(\sin(30^\circ) = \frac{1}{2}\)

We have:
\[ 2 \cos(100 \pi t + 30^\circ) = 2 \left( \cos(100 \pi t) \cdot \frac{\sqrt{3}}{2} - \sin(100 \pi t) \cdot \frac{1}{2} \right) \]
\[ = \sqrt{3} \cos(100 \pi t) - \sin(100 \pi t) \]

### Step 3: Combine the Terms
Now substituting back into the equation for \(i\):
\[ i = 2 \sin(100 \pi t) + \sqrt{3} \cos(100 \pi t) - \sin(100 \pi t) \]
\[ = (2 - 1) \sin(100 \pi t) + \sqrt{3} \cos(100 \pi t) \]
\[ = \sin(100 \pi t) + \sqrt{3} \cos(100 \pi t) \]

### Step 4: Express in the Form \(R \sin(100 \pi t + \phi)\)
To express this in the form \(R \sin(100 \pi t + \phi)\), we identify:
- \(a = 1\) (coefficient of \(\sin\))
- \(b = \sqrt{3}\) (coefficient of \(\cos\))

Now, calculate \(R\):
\[ R = \sqrt{a^2 + b^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 5: Calculate the Phase Angle \(\phi\)
The phase angle \(\phi\) can be calculated using:
\[ \tan \phi = \frac{b}{a} = \frac{\sqrt{3}}{1} = \sqrt{3} \]
Thus,
\[ \phi = \tan^{-1}(\sqrt{3}) = 60^\circ \text{ or } \frac{\pi}{3} \text{ radians} \]

### Step 6: Final Expression for Current
So we can express the current as:
\[ i = 2 \sin(100 \pi t + 60^\circ) \]

### Step 7: Calculate the RMS Value
The effective (RMS) value of the current is given by:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]
Where \(I_0\) is the peak current:
\[ I_0 = 2 \]
Thus,
\[ I_{\text{rms}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ A} \]

### Final Answer
The effective value of the current is:
\[ \boxed{\sqrt{2} \text{ A}} \]

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