JEE Advanced · Physics · 19. Current Electricity
When two identical batteries of internal resistance \(1 \Omega\) each are connected in series across a resistor \(R\), the rate of heat produced in \(R\) is \(J_1\). When the same batteries are connected in parallel across \(R\), the rate is \(J_2\). If \(J_1=2.25 J_2\) then the value of \(R\) in \(\Omega\) is
- A 2
- B 3
- C 5
- D 4
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
In series, \(i=\frac{2 E}{2+R}\)
\(
\therefore J_1=i^2 R=\left(\frac{2 E}{2+R}\right)^2 \cdot R
\)
In parallel, \(i=\frac{E}{0.5+R}\)
\(
\begin{aligned}
\therefore \quad J_2 & =i^2 R=\left(\frac{E}{0.5+R}\right)^2 \cdot R \\
\frac{J_1}{J_2} & =2.25=\frac{4(0.5+R)^2}{(2+R)^2}
\end{aligned}
\)
or \(1.5=\frac{2(0.5+R)}{(2+R)}\)
Solving we get, \(R=4 \Omega\)
\(\therefore\) The answer is 4.
\(
\therefore J_1=i^2 R=\left(\frac{2 E}{2+R}\right)^2 \cdot R
\)
In parallel, \(i=\frac{E}{0.5+R}\)
\(
\begin{aligned}
\therefore \quad J_2 & =i^2 R=\left(\frac{E}{0.5+R}\right)^2 \cdot R \\
\frac{J_1}{J_2} & =2.25=\frac{4(0.5+R)^2}{(2+R)^2}
\end{aligned}
\)
or \(1.5=\frac{2(0.5+R)}{(2+R)}\)
Solving we get, \(R=4 \Omega\)
\(\therefore\) The answer is 4.
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