If instantaneous current in a circuit is given by `I=(2+3 sin omegat)` A, then the effective value of resulting current in the circuit is

Options

A. `sqrt((17)/(2))` A
B. `sqrt((2)/(17))` A
C. `sqrt((3)/sqrt(2))`
D. `3sqrt(2)` A

Answer

To find the effective (RMS) value of the instantaneous current given by \( I(t) = 2 + 3 \sin(\omega t) \) A, we will follow these steps:

### Step 1: Understand the formula for RMS current
The effective (RMS) value of a current \( I(t) \) over one complete cycle is given by:
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T I^2(t) \, dt} \]
where \( T \) is the period of the function.

### Step 2: Determine the period \( T \)
The function \( \sin(\omega t) \) has a period of \( T = \frac{2\pi}{\omega} \).

### Step 3: Substitute \( I(t) \) into the formula
We need to calculate \( I^2(t) \):
\[ I(t) = 2 + 3 \sin(\omega t) \]
\[ I^2(t) = (2 + 3 \sin(\omega t))^2 = 4 + 12 \sin(\omega t) + 9 \sin^2(\omega t) \]

### Step 4: Integrate \( I^2(t) \) over one period
Now we can calculate the integral:
\[ \int_0^T I^2(t) \, dt = \int_0^T \left(4 + 12 \sin(\omega t) + 9 \sin^2(\omega t)\right) dt \]
This can be split into three separate integrals:
\[ \int_0^T 4 \, dt + \int_0^T 12 \sin(\omega t) \, dt + \int_0^T 9 \sin^2(\omega t) \, dt \]

### Step 5: Calculate each integral
1. **First integral**:
\[ \int_0^T 4 \, dt = 4T = 4 \cdot \frac{2\pi}{\omega} = \frac{8\pi}{\omega} \]

2. **Second integral**:
\[ \int_0^T 12 \sin(\omega t) \, dt = 0 \quad (\text{since the integral of } \sin \text{ over a full period is zero}) \]

3. **Third integral**:
Using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
\[ \int_0^T 9 \sin^2(\omega t) \, dt = 9 \int_0^T \frac{1 - \cos(2\omega t)}{2} \, dt = \frac{9}{2} \left( T - 0 \right) = \frac{9}{2} \cdot \frac{2\pi}{\omega} = \frac{9\pi}{\omega} \]

### Step 6: Combine the results
Now we can combine the results of the integrals:
\[ \int_0^T I^2(t) \, dt = \frac{8\pi}{\omega} + 0 + \frac{9\pi}{\omega} = \frac{17\pi}{\omega} \]

### Step 7: Substitute back into the RMS formula
Now we substitute this back into the RMS formula:
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T I^2(t) \, dt} = \sqrt{\frac{1}{\frac{2\pi}{\omega}} \cdot \frac{17\pi}{\omega}} = \sqrt{\frac{17\omega}{2}} \]

### Step 8: Final result
Thus, the effective value of the resulting current in the circuit is:
\[ I_{\text{rms}} = \sqrt{\frac{17}{2}} \text{ A} \]