An electric kettie has two coils. When one coil is switched on it takes 15 minutes and the other takes 30 minutes to boil certain mass of water. The ratio of times taken by them, when connected in series and in parallel to boil the same mass of water is

Explanation

To solve the problem step by step, we will analyze the situation of the electric kettle with two coils and calculate the time taken to boil water when the coils are connected in series and in parallel.

### Step 1: Understand the power and heat relationship
The heat supplied to boil water is given by the formula:
\[ H = P \cdot t \]
where \( H \) is the heat supplied, \( P \) is the power, and \( t \) is the time taken.

### Step 2: Establish the relationship for each coil
For the first coil (taking 15 minutes):
- Let the resistance of the first coil be \( R_1 \).
- The power \( P_1 \) is given by:
\[ P_1 = \frac{V^2}{R_1} \]
- The heat supplied when this coil is used:
\[ H = P_1 \cdot t_1 = \frac{V^2}{R_1} \cdot 15 \]

For the second coil (taking 30 minutes):
- Let the resistance of the second coil be \( R_2 \).
- The power \( P_2 \) is given by:
\[ P_2 = \frac{V^2}{R_2} \]
- The heat supplied when this coil is used:
\[ H = P_2 \cdot t_2 = \frac{V^2}{R_2} \cdot 30 \]

### Step 3: Equate the heat supplied
Since the heat supplied is the same in both cases, we can equate the two equations:
\[ \frac{V^2}{R_1} \cdot 15 = \frac{V^2}{R_2} \cdot 30 \]

### Step 4: Simplify the equation
Cancel \( V^2 \) from both sides:
\[ \frac{15}{R_1} = \frac{30}{R_2} \]

Cross-multiplying gives:
\[ 15R_2 = 30R_1 \]
Thus,
\[ \frac{R_1}{R_2} = \frac{1}{2} \]
This implies:
\[ R_2 = 2R_1 \]

### Step 5: Calculate time when coils are in series
When connected in series, the equivalent resistance \( R_s \) is:
\[ R_s = R_1 + R_2 = R_1 + 2R_1 = 3R_1 \]
The power in series is:
\[ P_s = \frac{V^2}{R_s} = \frac{V^2}{3R_1} \]
The heat supplied when using both coils in series:
\[ H = P_s \cdot t_s = \frac{V^2}{3R_1} \cdot t_s \]

Setting this equal to the heat supplied by the first coil:
\[ \frac{V^2}{3R_1} \cdot t_s = \frac{V^2}{R_1} \cdot 15 \]
Cancel \( V^2 \) and \( R_1 \):
\[ \frac{t_s}{3} = 15 \]
Thus,
\[ t_s = 45 \text{ minutes} \]

### Step 6: Calculate time when coils are in parallel
When connected in parallel, the equivalent resistance \( R_p \) is:
\[ R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{R_1 (2R_1)}{R_1 + 2R_1} = \frac{2R_1^2}{3R_1} = \frac{2R_1}{3} \]
The power in parallel is:
\[ P_p = \frac{V^2}{R_p} = \frac{V^2}{\frac{2R_1}{3}} = \frac{3V^2}{2R_1} \]
The heat supplied when using both coils in parallel:
\[ H = P_p \cdot t_p = \frac{3V^2}{2R_1} \cdot t_p \]

Setting this equal to the heat supplied by the first coil:
\[ \frac{3V^2}{2R_1} \cdot t_p = \frac{V^2}{R_1} \cdot 15 \]
Cancel \( V^2 \) and \( R_1 \):
\[ \frac{3t_p}{2} = 15 \]
Thus,
\[ t_p = 10 \text{ minutes} \]

### Step 7: Find the ratio of times
Now, we can find the ratio of the times taken when connected in series and parallel:
\[ \text{Ratio} = \frac{t_s}{t_p} = \frac{45}{10} = \frac{9}{2} \]

### Final Answer
The ratio of times taken by the coils when connected in series and in parallel is:
\[ \frac{9}{2} \]