Current Electricity Ques 109
- 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
(2010)
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Answer:
Correct Answer: 109.$(4)$
Solution:
Formula:
- 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 $4$ .
BETA
