Sure, let's go through the detailed solution step-by-step.
### Heating Time When Connected in Series
When two heater coils are connected in series, the total time taken to heat the same amount of water is simply the sum of the time taken by each heater coil individually.
Given values:
- Time taken by the first heater coil: [tex]\( t_1 \)[/tex]
- Time taken by the second heater coil: [tex]\( \theta_2 \)[/tex]
So, the total time taken ([tex]\( T_{\text{series}} \)[/tex]) when the two heater coils are connected in series is:
[tex]\[ T_{\text{series}} = t_1 + \theta_2 \][/tex]
From the given data, we have:
[tex]\[ t_1 = 10 \][/tex]
[tex]\[ \theta_2 = 20 \][/tex]
Thus, substituting the values:
[tex]\[ T_{\text{series}} = 10 + 20 = 30 \][/tex]
So, the time taken to heat the water when the coils are connected in series is 30 units of time.
### Heating Time When Connected in Parallel
When two heater coils are connected in parallel, the combined time taken can be found using the reciprocal formula for parallel connections. The formula to find the total time [tex]\( T_{\text{parallel}} \)[/tex] when the coils are connected in parallel is:
[tex]\[ \frac{1}{T_{\text{parallel}}} = \frac{1}{t_1} + \frac{1}{\theta_2} \][/tex]
Rewriting it to solve for [tex]\( T_{\text{parallel}} \)[/tex]:
[tex]\[ T_{\text{parallel}} = \frac{1}{\left( \frac{1}{t_1} + \frac{1}{\theta_2} \right)} \][/tex]
Substitute the given values for [tex]\( t_1 \)[/tex] and [tex]\( \theta_2 \)[/tex]:
[tex]\[ T_{\text{parallel}} = \frac{1}{\left( \frac{1}{10} + \frac{1}{20} \right)} \][/tex]
First, find the common denominator to add the fractions:
[tex]\[ \frac{1}{T_{\text{parallel}}} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} \][/tex]
Thus:
[tex]\[ T_{\text{parallel}} = \frac{20}{3} = 6.666666666666666 \][/tex]
So, the time taken to heat the water when the coils are connected in parallel is approximately 6.67 units of time.
Summary:
- Series Connection: 30 units of time
- Parallel Connection: 6.67 units of time
