The RMS value of AC which when passed through a resistor produces heat, which is twice that produced by a steady current of 1.414A in the same resistor is

Options

A. `2A`
B. 3.46 A
C. 2.818 A
D. 1.732 A

Answer

To solve the problem, we need to find the RMS value of an alternating current (AC) that produces heat in a resistor, which is twice the heat produced by a steady current of 1.414 A in the same resistor.

### Step-by-Step Solution:

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

2. **Define the Two Scenarios**:
- Let \( H_1 \) be the heat produced by the AC current with RMS value \( I_{RMS} \).
- Let \( H_2 \) be the heat produced by the steady current of \( 1.414 \, \text{A} \).

3. **Set Up the Heat Equations**:
For the steady current:
\[ H_2 = (1.414)^2 R T \]
For the AC current:
\[ H_1 = I_{RMS}^2 R T \]

4. **Relate the Two Heat Values**:
According to the problem, the heat produced by the AC current is twice that produced by the steady current:
\[ H_1 = 2 H_2 \]
Substituting the expressions for \( H_1 \) and \( H_2 \):
\[ I_{RMS}^2 R T = 2 \times (1.414)^2 R T \]

5. **Cancel Out Common Terms**:
Since \( R \) and \( T \) are common in both sides, we can cancel them out:
\[ I_{RMS}^2 = 2 \times (1.414)^2 \]

6. **Calculate \( (1.414)^2 \)**:
\[ (1.414)^2 = 2 \]
Therefore, substituting this back in:
\[ I_{RMS}^2 = 2 \times 2 = 4 \]

7. **Find \( I_{RMS} \)**:
Taking the square root of both sides:
\[ I_{RMS} = \sqrt{4} = 2 \, \text{A} \]

### Final Answer:
The RMS value of the AC current is \( 2 \, \text{A} \).

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