What is the approxiamnate peak value of an alternating current producing four times the heat produced per second by a steady of 2 A in a resistor ?

Options

A. 2.8 A
B. 4.0 A
C. 5.6 A
D. 8.0 A

Answer

To find the approximate peak value of an alternating current (AC) that produces four times the heat produced per second by a steady direct current (DC) of 2 A in a resistor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Heat Produced**:
The heat produced in a resistor by a current can be calculated using the formula:
\[ H = I^2 R T \]
where \( H \) is the heat produced, \( I \) is the current, \( R \) is the resistance, and \( T \) is the time.

2. **Calculate Heat Produced by DC**:
For a steady current of 2 A, the heat produced in time \( T \) is:
\[ H_{DC} = (2)^2 R T = 4 R T \]

3. **Heat Produced by AC**:
The heat produced by an AC current is given by:
\[ H_{AC} = I_{RMS}^2 R T \]
According to the problem, the heat produced by AC is four times that of the DC:
\[ H_{AC} = 4 H_{DC} = 4 \times 4 R T = 16 R T \]

4. **Set the Equations Equal**:
From the above equations, we can set the heat produced by AC equal to 16 R T:
\[ I_{RMS}^2 R T = 16 R T \]
We can cancel \( R T \) from both sides (assuming \( R \) and \( T \) are not zero):
\[ I_{RMS}^2 = 16 \]

5. **Solve for RMS Current**:
Taking the square root of both sides gives:
\[ I_{RMS} = 4 \, \text{A} \]

6. **Calculate Peak Current**:
The relationship between the peak current \( I_0 \) and the RMS current \( I_{RMS} \) is given by:
\[ I_{RMS} = \frac{I_0}{\sqrt{2}} \]
Rearranging this gives:
\[ I_0 = I_{RMS} \times \sqrt{2} \]
Substituting \( I_{RMS} = 4 \, \text{A} \):
\[ I_0 = 4 \times \sqrt{2} \]

7. **Calculate the Numerical Value**:
Since \( \sqrt{2} \approx 1.414 \):
\[ I_0 \approx 4 \times 1.414 \approx 5.656 \, \text{A} \]

8. **Final Answer**:
Thus, the approximate peak value of the alternating current is:
\[ I_0 \approx 5.656 \, \text{A} \quad \text{or} \quad 5.6 \, \text{A} \text{ (to one decimal place)} \]