Two heaters designed for the same voltage `V` have different power ratings. When connected individually across as source of voltage` V`, they produce `H` amount of heat each in time `t_1` and `t_2` respectively. When used together acros the same source, they produce H amount of heat in time t

Options

A. if they are in series `t=t_1+t_2`
B. if they are in series `t=2(t_1+t_2)`
C. if they are in parallel `t=(t_1t_2)/((t_1+t_2))`
D. if they are in parallel `t=(t_1t_2)/(2(t_1+t_2))`

Explanation

To solve the problem, we need to analyze the situation where two heaters are connected individually and then together, and derive the relationships between the time taken to produce a certain amount of heat in different configurations (series and parallel).

### Step-by-Step Solution:

1. **Understanding Power and Heat Production**:
- The power of the first heater (P1) when connected to voltage V is given by:
\[ P_1 = \frac{V^2}{R_1} \]
- The power of the second heater (P2) is:
\[ P_2 = \frac{V^2}{R_2} \]
- The heat produced by each heater in time \( t_1 \) and \( t_2 \) respectively is given by:
\[ H = P_1 \cdot t_1 \quad \text{and} \quad H = P_2 \cdot t_2 \]

2. **Expressing Resistance in Terms of Heat**:
- Rearranging the heat equations gives:
\[ R_1 = \frac{V^2}{H} \cdot t_1 \quad \text{and} \quad R_2 = \frac{V^2}{H} \cdot t_2 \]

3. **Connecting Heaters in Series**:
- When the heaters are connected in series, the equivalent resistance \( R_{eq} \) is:
\[ R_{eq} = R_1 + R_2 \]
- The heat produced when both heaters are connected in series for time \( t \) is:
\[ H = \frac{V^2}{R_{eq}} \cdot t \]
- Substituting for \( R_{eq} \):
\[ H = \frac{V^2}{R_1 + R_2} \cdot t \]

4. **Substituting for \( R_1 \) and \( R_2 \)**:
- Substitute \( R_1 \) and \( R_2 \) from the earlier equations:
\[ H = \frac{V^2}{\left(\frac{V^2}{H} t_1 + \frac{V^2}{H} t_2\right)} \cdot t \]
- Simplifying gives:
\[ H = \frac{H \cdot t}{t_1 + t_2} \]
- Thus, we find:
\[ t_1 + t_2 = t \]

5. **Connecting Heaters in Parallel**:
- When connected in parallel, the equivalent resistance is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
- The heat produced in time \( t \) is:
\[ H = \frac{V^2}{R_{eq}} \cdot t \]
- Substituting for \( R_1 \) and \( R_2 \):
\[ H = \frac{V^2}{\left(\frac{H}{V^2} \cdot \frac{1}{t_1} + \frac{H}{V^2} \cdot \frac{1}{t_2}\right)} \cdot t \]
- Simplifying gives:
\[ H = \frac{H \cdot t}{\frac{t_1 t_2}{t_1 + t_2}} \]
- Thus, we find:
\[ \frac{1}{t} = \frac{1}{t_1} + \frac{1}{t_2} \quad \text{or} \quad t = \frac{t_1 t_2}{t_1 + t_2} \]

### Final Results:
- When connected in series: \( t_1 + t_2 = t \)
- When connected in parallel: \( t = \frac{t_1 t_2}{t_1 + t_2} \)