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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Physics
Chapter
Current Electricity

JEE Advanced 2010 · Paper 1 · Physics · Current Electricity

When two identical batteries of internal resistance 1 each are...

When two identical batteries of internal resistance 1Ω1 \Omega each are connected in series across a resistor RR, the rate of heat produced in RR is J1J_1. When the same batteries are connected in parallel across RR, the rate is J2J_2. If J1=2.25J2J_1=2.25 J_2 then the value of RR in Ω\Omega is

Correct answer

0

Step-by-step solution

In series, i=2E2+Ri=\frac{2 E}{2+R} \therefore J_1=i² R=\left(\frac{2 E}{2+R}\right)^2 ⋅\cdotR In parallel, i=E0.5+Ri=\frac{E}{0.5+R}∴\quadJ2=i2R=(E0.5+R)2\cdotRJ1J2=2.25=4(0.5+R)2(2+R)2\begin{aligned} \therefore \quadJ_2 & =i² R=\left(\frac{E}{0.5+R}\right)^2 \cdotR \\ \frac{J_1}{J_2}& =2.25=\frac{4(0.5+R)^2}{(2+R)^2} \end{aligned}$oror1.5=\frac{2(0.5+R)}{(2+R)}Solvingweget,Solving we get,R=4 \Omega\therefore$ The answer is 4 .
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