Question

Help plz___________________​

Answer 1

Answer:

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Answer 2

$\Large\underline{\underline{\mathfrak \orange{ Answer:}}}$

❏ Resistance R is 15 Ω

$\Large\underline{\underline{\mathfrak \orange{ Solution:}}}$

For first case ,

$\implies{\sf \dfrac{5}{l_1}=\dfrac{R}{(100-l_1)}\:\:\:→(1) }$

Now , by shunting resistance R by an equal resistance R , new Resistance in that arm becomes $\sf{\dfrac{R}{2}}$

so,

$\implies{\sf \dfrac{5}{1.6l_1}=\dfrac{R/2}{(100-1.6l_1)}\:\:\:→(2)}$

❍ From Equation 1 and Equation 2

$\implies{\sf \dfrac{1.6}{1}=\dfrac{(100-1.6l_1)}{100-l_1}×2 }$

$\implies{\sf 160-1.6l_2=200-3.2l_1 }$

$\implies{\sf 1.6l_1=40}$

$\implies{\sf l_1=\dfrac{400}{1.6}}$

$\implies{\sf l_1=25\:cm }$

❍ From Equation 1

$\implies{\sf \dfrac{5}{25}=\dfrac{R}{75} }$

$\Large\implies{\sf \red{R=15Ω }}$