Alternating current with peak value 20`sqrt(2)` A is used to produce required amount of heat across a resistor. What constant current is needed to produce same amount of heat across the same resistance in same time ?

Options

A. 40 A
B. 20 A
C. `10`A
D. `11.14`A

Explanation

To solve the problem of finding the constant current needed to produce the same amount of heat across a resistor as an alternating current (AC) with a peak value of \(20\sqrt{2}\) A, we can follow these steps:

### Step 1: Understand the relationship between AC and DC heating
The heat produced in a resistor due to an AC current is given by:
\[ H_{AC} = I_{rms}^2 \cdot R \cdot t \]
where \(I_{rms}\) is the root mean square (RMS) current, \(R\) is the resistance, and \(t\) is the time.

For a direct current (DC), the heat produced is given by:
\[ H_{DC} = I'^2 \cdot R \cdot t \]
where \(I'\) is the constant DC current.

### Step 2: Set the heats equal
Since we want the heat produced by both the AC and DC to be the same, we can set the equations equal to each other:
\[ I_{rms}^2 \cdot R \cdot t = I'^2 \cdot R \cdot t \]
We can cancel \(R\) and \(t\) from both sides (assuming they are the same for both cases):
\[ I_{rms}^2 = I'^2 \]

### Step 3: Find the RMS value of the AC current
The RMS value of an AC current is related to its peak value (\(I_0\)) by the formula:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
Given that the peak value \(I_0 = 20\sqrt{2}\) A, we can substitute this into the equation:
\[ I_{rms} = \frac{20\sqrt{2}}{\sqrt{2}} = 20 \text{ A} \]

### Step 4: Find the constant current
From the previous step, we have:
\[ I' = I_{rms} \]
Thus, the required constant current \(I'\) to produce the same amount of heat is:
\[ I' = 20 \text{ A} \]

### Final Answer
The constant current needed to produce the same amount of heat across the same resistance in the same time is \(20\) A.

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