If an AC produces same heat as that produced by a steady current of 4 A , then peak value of current is :

Options

A. 4A
B. `1.56 A`
C. `5.6 A`
D. `1.41 A`

Answer

To solve the problem, we need to find the peak value of the alternating current (AC) that produces the same heat as a steady direct current (DC) of 4 A.

### Step-by-Step Solution:

1. **Understand the Heat Produced by DC Current**:
The heat produced (H) by a steady current (DC) is given by the formula:
\[ H = I^2 R \]
where \( I \) is the current and \( R \) is the resistance.

2. **Heat Produced by AC Current**:
The heat produced by an alternating current (AC) is given by:
\[ H = \frac{I_0^2}{2} R \]
where \( I_0 \) is the peak value of the AC current.

3. **Equate the Heat Produced by AC and DC**:
Since both currents produce the same amount of heat, we can set the two equations equal to each other:
\[ I^2 R = \frac{I_0^2}{2} R \]

4. **Cancel the Resistance (R)**:
Since \( R \) is common in both terms, we can cancel it out:
\[ I^2 = \frac{I_0^2}{2} \]

5. **Substitute the Given DC Current**:
We know that the DC current \( I = 4 \, \text{A} \):
\[ 4^2 = \frac{I_0^2}{2} \]
This simplifies to:
\[ 16 = \frac{I_0^2}{2} \]

6. **Multiply Both Sides by 2**:
To isolate \( I_0^2 \), multiply both sides by 2:
\[ 32 = I_0^2 \]

7. **Take the Square Root**:
Now, take the square root of both sides to find \( I_0 \):
\[ I_0 = \sqrt{32} = 4\sqrt{2} \]

8. **Calculate the Numerical Value**:
The numerical value of \( \sqrt{2} \) is approximately 1.414, so:
\[ I_0 \approx 4 \times 1.414 \approx 5.656 \, \text{A} \]

9. **Final Result**:
Therefore, the peak value of the AC current is approximately:
\[ I_0 \approx 5.65 \, \text{A} \]

### Final Answer:
The peak value of the AC current is approximately **5.65 A**.

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