What is the r.m.s. value of an alternating current which when passed through a resistor produces heat which is thrice of that produced by a direct current of `2` amperes in the same resistor?

Options

A. `6` ampere
B. `2` ampere
C. `2sqrt3` ampere
D. `0.65` ampere

Answer

To solve the problem, we need to find the root mean square (r.m.s.) value of an alternating current (AC) that produces heat in a resistor, which is three times the heat produced by a direct current (DC) of 2 amperes.

### Step-by-Step Solution:

1. **Understanding Heat Produced by DC:**
The heat produced by a direct current (I_dc) in a resistor (R) over time (t) is given by the formula:
\[ H_{DC} = I_{DC}^2 \cdot R \cdot t \]
Given that \( I_{DC} = 2 \, \text{A} \), we can calculate:
\[ H_{DC} = (2 \, \text{A})^2 \cdot R \cdot t = 4R \cdot t \]

2. **Heat Produced by AC:**
The heat produced by an alternating current (I_rms) in the same resistor is given by:
\[ H_{AC} = I_{rms}^2 \cdot R \cdot t \]

3. **Relationship Between Heat Produced:**
According to the problem, the heat produced by the AC is three times that produced by the DC:
\[ H_{AC} = 3 \cdot H_{DC} \]
Substituting the expression for \( H_{DC} \):
\[ H_{AC} = 3 \cdot (4R \cdot t) = 12R \cdot t \]

4. **Setting the Heat Equations Equal:**
Now we can set the equations for heat produced by AC and DC equal to each other:
\[ I_{rms}^2 \cdot R \cdot t = 12R \cdot t \]

5. **Simplifying the Equation:**
Since \( R \) and \( t \) are common in both sides of the equation, we can cancel them out (assuming \( R \) and \( t \) are not zero):
\[ I_{rms}^2 = 12 \]

6. **Finding the r.m.s. Value:**
Taking the square root of both sides gives us:
\[ I_{rms} = \sqrt{12} = 2\sqrt{3} \, \text{A} \]

### Final Answer:
The r.m.s. value of the alternating current is \( 2\sqrt{3} \, \text{A} \).

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