Resistance are connected in a meter bridge circuit as shown in figure. balancing length l_{1} is 40,

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3.Current Electricity

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Resistance are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is $40\,cm$. Now an unknown resistance $x$ is connected in series with $P$ and new balancing length is found to be $80\,cm$ measured from the same end. Then the value of $x$ will be $.......\Omega$

A

$2.2$

B

$22$

C

$200$

D

$20$

(JEE MAIN-2022)

Solution

Initially, $\frac{ P }{ Q }=\frac{40 cm }{60 cm }=\frac{2}{3}$

Finally, $\frac{ P + x }{ Q }=\frac{80 cm }{20 cm }=\frac{4}{1} \ldots(2)$

Divide $(2)$ by $(1)$

$\frac{ P + x }{ P }=4 \times \frac{3}{2}=6$

$1+\frac{ x }{ P }=6 \Rightarrow \frac{ x }{ P }=5$

$\therefore x =5 P =5 \times 4=20 \Omega$

Standard 12

Physics

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