Current Electricity Ques 109 — JEE PYQ Chapterwise

Answer:

Correct Answer: 109.$(4)$

Solution:

Formula:

Electrical Power:

109. In series, $i=\frac{2 E}{2+R} \Rightarrow J _1=i^{2} R=(\frac{2 E}{2+R})^{2} \cdot R$

In parallel, $\quad i=\frac{E}{0.5+R}$

$\therefore \quad J _2 =i^{2} R=(\frac{E}{0.5+R}){ }^{2} \cdot R $

$ \quad \quad \frac{J _1}{J _2} =2.25 $

$=\frac{4(0.5+R)^{2}}{(2+R)^{2}} \text { or } 1.5=\frac{2(0.5+R)}{(2+R)}$

Solving we get, $R=4 \Omega$

$\therefore \quad $ The answer is $4$ .