In an electric iron heat producedm is same, whether it is connected across an A.c. source or across 50 V constant voltage. R.M.S. value of the A.C. voltage applied is

Options

A. 50 V
B. Zero
C. `50sqrt(2)V`
D. `(50)/(sqrt(2))V`

Answer

To solve the problem, we need to find the RMS value of the AC voltage applied to the electric iron, given that the heat produced is the same whether it is connected to an AC source or a 50 V constant voltage (DC).

### Step-by-Step Solution:

1. **Understanding the Heat Produced**:
The heat produced (H) in an electric iron can be expressed in terms of power (P) and time (t):
\[ H = P \times t \]
For resistive heating, the power can also be expressed as:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the iron and \( R \) is its resistance.

2. **Heat Produced with AC Voltage**:
For an AC source, the peak voltage is denoted as \( V_{peak} \). The average power (and hence heat produced) over one cycle can be expressed as:
\[ H_1 = \frac{V_{peak}^2}{R} \times t \]
However, since we are dealing with AC, we need to consider the RMS value of the voltage:
\[ H_1 = \frac{V_{rms}^2}{R} \times t \]

3. **Heat Produced with DC Voltage**:
For a DC source of 50 V, the heat produced can be expressed as:
\[ H_2 = \frac{V_{0}^2}{R} \times t \]
where \( V_{0} = 50 \, \text{V} \).

4. **Equating the Heat Produced**:
According to the problem, the heat produced in both cases is the same:
\[ H_1 = H_2 \]
Substituting the expressions for \( H_1 \) and \( H_2 \):
\[ \frac{V_{rms}^2}{R} \times t = \frac{50^2}{R} \times t \]
The resistance \( R \) and time \( t \) can be canceled from both sides:
\[ V_{rms}^2 = 50^2 \]

5. **Calculating the RMS Voltage**:
Taking the square root of both sides gives:
\[ V_{rms} = 50 \, \text{V} \]

### Conclusion:
The RMS value of the AC voltage applied is:
\[ \boxed{50 \, \text{V}} \]