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)))`A
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 can follow these steps:

### Step 1: Define the RMS value
The effective (RMS) value of a current is defined as:
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T I(t)^2 \, dt} \]
where \( T \) is the time period of the current.

### Step 2: Square the instantaneous current
We need to compute \( I(t)^2 \):
\[ I(t) = 2 + 3 \sin(\omega t) \]
Squaring this gives:
\[ I(t)^2 = (2 + 3 \sin(\omega t))^2 = 4 + 12 \sin(\omega t) + 9 \sin^2(\omega t) \]

### Step 3: Substitute into the RMS formula
Now, we substitute \( I(t)^2 \) into the RMS formula:
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T (4 + 12 \sin(\omega t) + 9 \sin^2(\omega t)) \, dt} \]

### Step 4: Evaluate the integral
We can break this integral into three parts:
\[ \int_0^T (4 + 12 \sin(\omega t) + 9 \sin^2(\omega t)) \, dt = \int_0^T 4 \, dt + \int_0^T 12 \sin(\omega t) \, dt + \int_0^T 9 \sin^2(\omega t) \, dt \]

1. **First Integral**:
\[ \int_0^T 4 \, dt = 4T \]

2. **Second Integral**:
The integral of \( \sin(\omega t) \) over one complete cycle is zero:
\[ \int_0^T 12 \sin(\omega t) \, dt = 0 \]

3. **Third Integral**:
For \( \sin^2(\omega t) \), we use the identity:
\[ \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} \]
Thus,
\[ \int_0^T 9 \sin^2(\omega t) \, dt = 9 \int_0^T \frac{1 - \cos(2\omega t)}{2} \, dt = \frac{9}{2} \left( T - 0 \right) = \frac{9T}{2} \]

### Step 5: Combine the results
Now we can combine the results from the integrals:
\[ \int_0^T (4 + 12 \sin(\omega t) + 9 \sin^2(\omega t)) \, dt = 4T + 0 + \frac{9T}{2} = 4T + \frac{9T}{2} = \frac{8T}{2} + \frac{9T}{2} = \frac{17T}{2} \]

### Step 6: Substitute back into the RMS formula
Now substituting back into the RMS formula:
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \cdot \frac{17T}{2}} = \sqrt{\frac{17}{2}} \]

### Step 7: Final result
Thus, the effective value of the resulting current in the circuit is:
\[ I_{\text{rms}} = \sqrt{\frac{17}{2}} \, \text{A} \]