What is the approximate peak value of an altemating currrent producing four times the heat produced per second by a steady current 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 solve the problem step by step, we need to determine the peak value of an alternating current (AC) that produces four times the heat per second compared to a steady current of 2 A in a resistor.

### Step 1: Calculate the heat produced by the steady current
The heat produced per second (H) by a steady current (I) in a resistor (R) is given by the formula:
\[ H = I^2 R \]
For a steady current of 2 A:
\[ H_1 = (2 \, \text{A})^2 R = 4R \]

### Step 2: Determine the heat produced by the AC
According to the problem, the AC current produces four times the heat of the steady current:
\[ H_2 = 4 \times H_1 = 4 \times 4R = 16R \]

### Step 3: Relate the heat produced by AC to its RMS value
For an alternating current, the heat produced per second can also be expressed in terms of the root mean square (RMS) value of the current (\(I_{RMS}\)):
\[ H_2 = I_{RMS}^2 R \]
Setting this equal to the heat produced by the AC:
\[ I_{RMS}^2 R = 16R \]

### Step 4: Simplify the equation
Since R is common on both sides, we can cancel it out:
\[ I_{RMS}^2 = 16 \]
Taking the square root of both sides gives:
\[ I_{RMS} = 4 \, \text{A} \]

### Step 5: Relate RMS current to peak current
The relationship between the peak current (\(I_0\)) and the RMS current is given by:
\[ I_{RMS} = \frac{I_0}{\sqrt{2}} \]
Rearranging this to find \(I_0\):
\[ I_0 = I_{RMS} \times \sqrt{2} \]
Substituting the value of \(I_{RMS}\):
\[ I_0 = 4 \, \text{A} \times \sqrt{2} \]

### Step 6: Calculate the peak current
Using the approximate value of \(\sqrt{2} \approx 1.414\):
\[ I_0 \approx 4 \, \text{A} \times 1.414 \approx 5.656 \, \text{A} \]

### Final Answer
Thus, the approximate peak value of the alternating current is:
\[ I_0 \approx 5.6 \, \text{A} \]

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