A current in circuit is given by i = 3 + 4 `sin omegat`. Then the effective value of current is :

Options

A. 5
B. `sqrt(7)`
C. `sqrt(17)`
D. `sqrt(10)`

Answer

To find the effective value (RMS value) of the current given by the equation \( i = 3 + 4 \sin(\omega t) \), we will follow these steps:

### Step 1: Understand the RMS Value
The RMS (Root Mean Square) value of a function is calculated by squaring the function, finding the mean (average) of the squared function over one complete cycle, and then taking the square root of that mean.

### Step 2: Square the Current Function
First, we square the current function:
\[ i^2 = (3 + 4 \sin(\omega t))^2 \]
Expanding this, we get:
\[ i^2 = 3^2 + 2 \cdot 3 \cdot 4 \sin(\omega t) + (4 \sin(\omega t))^2 \]
\[ i^2 = 9 + 24 \sin(\omega t) + 16 \sin^2(\omega t) \]

### Step 3: Find the Mean of \( i^2 \)
To find the RMS value, we need to calculate the mean of \( i^2 \) over one complete cycle. The mean value \( \langle i^2 \rangle \) is given by:
\[ \langle i^2 \rangle = \frac{1}{T} \int_0^T i^2 \, dt \]
where \( T \) is the time period of the function. For \( \sin(\omega t) \), the period \( T \) is \( \frac{2\pi}{\omega} \).

### Step 4: Integrate Each Term
We will integrate each term in \( i^2 \) from \( 0 \) to \( T \):
1. **Integrate the constant term (9)**:
\[ \int_0^T 9 \, dt = 9T = 9 \cdot \frac{2\pi}{\omega} = \frac{18\pi}{\omega} \]

2. **Integrate the term with \( \sin(\omega t) \)**:
\[ \int_0^T 24 \sin(\omega t) \, dt = 0 \quad \text{(since the integral of sine over one complete cycle is zero)} \]

3. **Integrate the term with \( \sin^2(\omega t) \)**:
Using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
\[ \int_0^T 16 \sin^2(\omega t) \, dt = 16 \int_0^T \frac{1 - \cos(2\omega t)}{2} \, dt = 8 \int_0^T (1 - \cos(2\omega t)) \, dt \]
\[ = 8 \left( T - 0 \right) = 8 \cdot \frac{2\pi}{\omega} = \frac{16\pi}{\omega} \]

### Step 5: Combine the Integrals
Now, we combine the results:
\[ \langle i^2 \rangle = \frac{1}{T} \left( \frac{18\pi}{\omega} + 0 + \frac{16\pi}{\omega} \right) = \frac{1}{\frac{2\pi}{\omega}} \cdot \frac{34\pi}{\omega} = \frac{34\pi}{\frac{2\pi}{\omega}} = \frac{34\omega}{2} = 17\omega \]

### Step 6: Take the Square Root
Finally, the RMS value \( I_{\text{rms}} \) is:
\[ I_{\text{rms}} = \sqrt{\langle i^2 \rangle} = \sqrt{17} \]

### Conclusion
The effective value of the current is:
\[ \boxed{\sqrt{17}} \]