
An electric tea kettle has two heating coils. When the first coil of resistance \[R_1\] is switched on, the kettle begins to boil tea in \[6\] minutes. When the second coil of resistance \[R_2\] is switched on, the boiling begins in \[8\] minutes. The value of \[{R_1}/{R_2}\] Is
A. \[\dfrac{7}{3} \\ \]
B. \[\dfrac{3}{7} \\ \]
C. \[\dfrac{3}{4} \\ \]
D. \[\dfrac{4}{3}\]
Answer
163.5k+ views
Hint: To answer this question, we must apply the concept of heat created because it will be common at the end, so we will evaluate the R from that formula, and then reduce some of the variables, and lastly the equation will be in pure time form.
Formula used:
\[{R_1} = \dfrac{Q}{{{V^2}{t_1}}}\]
Where V is the voltage, R is the resistance, and t is the time required to bring the water to a boil.
Complete step by step solution:
To answer this question, we must use the heat created in the first coil formula:
\[Q = \dfrac{{{V^2}}}{{{R_1}}} \times {t_1}\]
Now, on substituting the values, we get
\[{R_1} = \dfrac{Q}{{{V^2}{t_1}}}\]
Because we are boiling the same amount of water at both times and the source has not changed, our heat produced and voltage will be the same for the second coil as for the first.
As a result, the heat generated when the coils are connected in parallel:
\[Q = \dfrac{{{V^2}}}{R} \times t\]
Now, again on substituting the value we get
\[R = \dfrac{Q}{{{V^2}t}}\]
Now, we obtain a resultant resistance, we have
\[\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}\]
Now, we have to replace the values of \[{R_1},{R_2},{R_3}\] we get
\[\dfrac{Q}{{{V^2}t}} = \dfrac{Q}{{{V^2}{t_1}}} = \dfrac{Q}{{{V^2}{t_2}}}\]
Now, let’s solve further the above expression, we get
\[\dfrac{1}{t} = \dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}}\]
Now, it becomes,
\[\dfrac{{{t_1}}}{{{R_1}}} = \dfrac{{{t_2}}}{{{R_2}}}\]
Now, we have to substitute the values of \[{t_1}\] and \[{t_2}\] we get
\[\dfrac{6}{{{R_1}}} = \dfrac{8}{{{R_2}}}\]
Let us simplify the above expression, we get
\[\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{3}{4}\]
Therefore, the value of \[\dfrac{{{R_1}}}{{{R_2}}}\] is \[\dfrac{3}{4}\].
Hence, the option C is correct.
Note: This is essentially a question of calculating the comparable resistance, but we need to know the relationship between heat produced and resistance or else this subject is unsolvable; therefore, we must be cautious in our thinking and not mix things up.
Formula used:
\[{R_1} = \dfrac{Q}{{{V^2}{t_1}}}\]
Where V is the voltage, R is the resistance, and t is the time required to bring the water to a boil.
Complete step by step solution:
To answer this question, we must use the heat created in the first coil formula:
\[Q = \dfrac{{{V^2}}}{{{R_1}}} \times {t_1}\]
Now, on substituting the values, we get
\[{R_1} = \dfrac{Q}{{{V^2}{t_1}}}\]
Because we are boiling the same amount of water at both times and the source has not changed, our heat produced and voltage will be the same for the second coil as for the first.
As a result, the heat generated when the coils are connected in parallel:
\[Q = \dfrac{{{V^2}}}{R} \times t\]
Now, again on substituting the value we get
\[R = \dfrac{Q}{{{V^2}t}}\]
Now, we obtain a resultant resistance, we have
\[\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}\]
Now, we have to replace the values of \[{R_1},{R_2},{R_3}\] we get
\[\dfrac{Q}{{{V^2}t}} = \dfrac{Q}{{{V^2}{t_1}}} = \dfrac{Q}{{{V^2}{t_2}}}\]
Now, let’s solve further the above expression, we get
\[\dfrac{1}{t} = \dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}}\]
Now, it becomes,
\[\dfrac{{{t_1}}}{{{R_1}}} = \dfrac{{{t_2}}}{{{R_2}}}\]
Now, we have to substitute the values of \[{t_1}\] and \[{t_2}\] we get
\[\dfrac{6}{{{R_1}}} = \dfrac{8}{{{R_2}}}\]
Let us simplify the above expression, we get
\[\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{3}{4}\]
Therefore, the value of \[\dfrac{{{R_1}}}{{{R_2}}}\] is \[\dfrac{3}{4}\].
Hence, the option C is correct.
Note: This is essentially a question of calculating the comparable resistance, but we need to know the relationship between heat produced and resistance or else this subject is unsolvable; therefore, we must be cautious in our thinking and not mix things up.
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