A heater boils `1kg` water in time `t_(1)` and another heater bolis the same water in time `t_(2)`. If both the heater are connected in parallel, the combination will bolt the water in time

Options

A. `(t_(1)t_(2))/(t_(1) -t_(2))`
B. `(t_(1)t_(2))/(t_(1) +t_(2))`
C. `(t_(1)t^(2) + t_(2)^(2))/(t_(1) +t_(2))`
D. `(t_(1)t^(2) - t_(2)^(2))/((t_(1) +t_(2)))`

Explanation

To solve the problem of how long it will take for two heaters connected in parallel to boil 1 kg of water, we can follow these steps:

### Step 1: Understand the Problem
We have two heaters:
- Heater 1 takes time \( t_1 \) to boil 1 kg of water.
- Heater 2 takes time \( t_2 \) to boil the same amount of water.

When these heaters are connected in parallel, we need to find the time \( t \) it takes for the combination to boil the same amount of water.

### Step 2: Relate Heat and Time
The heat produced by each heater can be expressed using the formula:
\[ Q = V^2 \cdot \frac{t}{R} \]
where \( Q \) is the heat required to boil the water, \( V \) is the voltage, \( t \) is the time, and \( R \) is the resistance of the heater.

For Heater 1:
\[ Q = V^2 \cdot \frac{t_1}{R_1} \]
For Heater 2:
\[ Q = V^2 \cdot \frac{t_2}{R_2} \]

### Step 3: Set Up the Equation for Parallel Connection
When the heaters are connected in parallel, the total power (or heat) produced by both heaters is the sum of the power produced by each heater:
\[ \frac{Q}{t} = \frac{Q}{t_1} + \frac{Q}{t_2} \]
This can be rearranged to:
\[ \frac{1}{t} = \frac{1}{t_1} + \frac{1}{t_2} \]

### Step 4: Solve for \( t \)
From the equation above, we can find \( t \):
\[ \frac{1}{t} = \frac{1}{t_1} + \frac{1}{t_2} \]
Taking the reciprocal gives us:
\[ t = \frac{t_1 \cdot t_2}{t_1 + t_2} \]

### Final Answer
Thus, the time taken to boil 1 kg of water when both heaters are connected in parallel is:
\[ t = \frac{t_1 \cdot t_2}{t_1 + t_2} \]

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