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

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. **Identify the Components of the Current:**
- The direct current (DC) component is \( I_{DC} = 5 \, \text{A} \).
- The alternating current (AC) component is \( I_{AC} = 10 \sin(\omega t) \).

2. **Determine the RMS Value of the AC Component:**
- The peak value of the AC current is \( I_0 = 10 \, \text{A} \).
- The RMS (Root Mean Square) value of the AC current is given by:
\[ I_{RMS, AC} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{A} \]

3. **Calculate the Effective Value of the Resulting Current:**
- The effective value of the total current \( I_{total} \) when combining both DC and AC components is given by:
\[ I_{RMS, total} = \sqrt{I_{DC}^2 + I_{RMS, AC}^2} \]
- Substituting the values:
\[ I_{RMS, total} = \sqrt{(5)^2 + (5\sqrt{2})^2} \]
- Calculate \( (5\sqrt{2})^2 \):
\[ (5\sqrt{2})^2 = 25 \times 2 = 50 \]
- Now substitute back into the equation:
\[ I_{RMS, total} = \sqrt{25 + 50} = \sqrt{75} \]

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

5. **Final Result:**
- Therefore, the effective value of the resulting current is:
\[ I_{RMS, total} = 5\sqrt{3} \, \text{A} \]

### Conclusion:
The effective value of the resulting current when a direct current of 5 A is superimposed on an alternating current \( I = 10 \sin(\omega t) \) is \( 5\sqrt{3} \, \text{A} \).