JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2022
- Hard
- Numerical
Two coils require \(20\) minutes and \(60\) minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be________ \(min\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(15\)
Step-by-step Solution
Detailed explanation
\(\frac{d Q}{d t}=i^{2} R=\frac{V^{2}}{R}\) (we know)
\(\Rightarrow \text { In 't' time , } \Delta Q =\left(\frac{ V ^{2}}{ R }\right) t\)
Given that, (for same source, \(v=\) same)
\(Q _{0}=\frac{ v ^{2}}{ R _{1}} \times 20=\frac{ V ^{2}}{ R _{2}} \times 60 \ldots\) \((i)\)
\(\Rightarrow R _{2}=3 R _{1} \ldots.\) \((ii)\)
If they are connected in parallel then
\(\operatorname{Req}=\frac{ R _{2} R _{1}}{ R _{1}+ R _{2}}=\frac{3 R _{1} \cdot R _{1}}{3 R _{1}+ R _{1}}=\left(\frac{3 R _{1}}{4}\right)\)
To produce same heat, using equation ...\((i)\)
\(Q _{0}=\frac{ V ^{2}}{ R _{1}} \times 20=\frac{ v ^{2}}{\left(\frac{3 R _{1}}{4}\right)} \times t\)
\(t =\frac{3 \times 20}{4}=15 \,min\)
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