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.
B.
C.
D.

Answer

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 DC and AC Components:**
- The DC component is given as \( I_{DC} = 5 \, \text{A} \).
- The AC component is given by \( I = 10 \sin(\omega t) \).

2. **Determine the Peak Value of the AC Current:**
- The peak value \( I_0 \) of the AC current is \( 10 \, \text{A} \).

3. **Calculate the RMS Value of the AC Current:**
- The RMS (Root Mean Square) value of the AC current is calculated using the formula:
\[ I_{AC} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, \text{A} = 5\sqrt{2} \, \text{A} \approx 7.07 \, \text{A} \]

4. **Calculate the Effective Value of the Resulting Current:**
- The effective value of the resulting current when DC and AC are superimposed is given by:
\[ I_{resultant} = \sqrt{I_{AC}^2 + I_{DC}^2} \]
- Substituting the values we have:
\[ I_{resultant} = \sqrt{(5\sqrt{2})^2 + (5)^2} \]
- Simplifying further:
\[ I_{resultant} = \sqrt{(50) + (25)} = \sqrt{75} \]
- This can be simplified to:
\[ I_{resultant} = 5\sqrt{3} \, \text{A} \approx 8.66 \, \text{A} \]

5. **Final Result:**
- The effective value of the resulting current is approximately \( 8.66 \, \text{A} \).