In a heating arrangement , an alternating current having a peak value of 28 A is used . To produce the same heat energy , If the constant current is used, its magnitude must be

Options

A. about 1A
B. about 28 A
C. about 20 A
D. Cannot say

Explanation

To solve the problem, we need to find the equivalent constant (DC) current that produces the same heat energy as an alternating current (AC) with a peak value of 28 A.

### Step-by-Step Solution:

1. **Understand the Heat Produced by DC Current:**
The heat produced in a resistor by a DC current \( I \) over a time \( T \) is given by:
\[ H_{DC} = I^2 R T \]
where \( R \) is the resistance.

2. **Understand the Heat Produced by AC Current:**
For an AC current, we use the root mean square (RMS) value of the current. The RMS value \( I_{RMS} \) is related to the peak value \( I_0 \) by:
\[ I_{RMS} = \frac{I_0}{\sqrt{2}} \]
Given \( I_0 = 28 \, \text{A} \), we can calculate \( I_{RMS} \):
\[ I_{RMS} = \frac{28}{\sqrt{2}} = \frac{28}{1.414} \approx 19.8 \, \text{A} \]

3. **Heat Produced by AC Current:**
The heat produced by the AC current over the same time \( T \) is:
\[ H_{AC} = I_{RMS}^2 R T \]
Substituting \( I_{RMS} \):
\[ H_{AC} = \left(\frac{28}{\sqrt{2}}\right)^2 R T = \frac{28^2}{2} R T = 392 R T \]

4. **Equating Heat Produced:**
Since we want the heat produced by the DC current to equal the heat produced by the AC current:
\[ I^2 R T = 392 R T \]
We can cancel \( R T \) from both sides (assuming \( R \) and \( T \) are the same for both cases):
\[ I^2 = 392 \]

5. **Solving for DC Current \( I \):**
Taking the square root of both sides gives:
\[ I = \sqrt{392} \approx 19.8 \, \text{A} \]

Thus, the magnitude of the constant DC current required to produce the same heat energy is approximately **20 A**.

### Final Answer:
The magnitude of the constant current must be approximately **20 A**.