A direct current of `5` amp is superimposed on an alternating current `I=10 sin omega t` flowing through a wire. The effective value of the resulting current will be:

Options

A. (A) `(15//2)A`
B. (B) `5 sqrt(3)A`
C. (C) `5 sqrt(5)A`
D. (D) `15 A`

Explanation

To find the effective value of the resulting current when a direct current (DC) of 5 A is superimposed on an alternating current (AC) given by \( I = 10 \sin(\omega t) \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Current Components**:
The total current \( I \) can be expressed as:
\[ I = I_{DC} + I_{AC} = 5 + 10 \sin(\omega t) \]

2. **Calculate the RMS Value**:
The effective (RMS) value of the total current \( I \) can be calculated using the formula for the RMS value of a combined current:
\[ I_{rms} = \sqrt{I_{DC}^2 + I_{AC,rms}^2} \]

3. **Identify the RMS Value of the AC Component**:
The RMS value of the AC component \( I_{AC} = 10 \sin(\omega t) \) is given by:
\[ I_{AC,rms} = \frac{I_{peak}}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \]

4. **Calculate the RMS Value of the DC Component**:
The RMS value of the DC component \( I_{DC} = 5 \) A is simply:
\[ I_{DC,rms} = 5 \]

5. **Combine the RMS Values**:
Now substitute the values into the RMS formula:
\[ I_{rms} = \sqrt{(5)^2 + (5\sqrt{2})^2} \]
\[ = \sqrt{25 + 50} = \sqrt{75} \]

6. **Simplify the Result**:
The square root of 75 can be simplified:
\[ \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \]

### Final Answer:
The effective value of the resulting current is:
\[ I_{rms} = 5\sqrt{3} \text{ A} \]

---