An alternating current having peak value `14 A` is used to heat a metal wire. To produce the same heating effect, a constant current `i` can be used where `i` is

Options

A. 14A
B. about 20A
C. 7A
D. about 10A

Explanation

To solve the problem, we need to find the constant current \( i \) that produces the same heating effect as an alternating current (AC) with a peak value of \( 14 \, A \). The heating effect in a resistor is related to the power, which can be calculated using the root mean square (RMS) value of the current.

### Step-by-Step Solution:

1. **Understand the relationship between peak current and RMS current**:
The RMS value of an alternating current is given by the formula:
\[ I_{\text{rms}} = \frac{I_{\text{peak}}}{\sqrt{2}} \]
where \( I_{\text{peak}} \) is the peak value of the AC current.

2. **Substitute the given peak current**:
We know that the peak current \( I_{\text{peak}} = 14 \, A \). Substituting this value into the RMS formula:
\[ I_{\text{rms}} = \frac{14 \, A}{\sqrt{2}} \]

3. **Calculate the RMS current**:
To compute \( I_{\text{rms}} \):
\[ I_{\text{rms}} = \frac{14}{\sqrt{2}} \approx \frac{14}{1.414} \approx 9.899 \, A \]

4. **Determine the equivalent constant current**:
The constant current \( i \) that produces the same heating effect as the AC current is equal to the RMS value:
\[ i = I_{\text{rms}} \approx 9.899 \, A \]

5. **Round the result**:
Rounding \( 9.899 \, A \) gives approximately \( 10 \, A \).

### Final Answer:
The constant current \( i \) that can be used to produce the same heating effect is approximately \( 10 \, A \).