Question

please help me solve this​

Answer 1

Answer

Equivalent resistance 2.75 Ω

Concept Used

When resistances are connected in series ,

$\rm R_{eq}=R_1+R_2+R_3+...$

When resistances are connected in parallel ,

$\rm \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...$

Explanation

Name the resistances a/c to the attachment .

R₂ , R₃ , R₄ are connected in series

$\to \rm R_{234}=R_2+R_3+R_4\\\\\to \rm R_{234}=1+1+1\\\\\to \rm R_{234}=3\ \Omega\ \; \bigstar$

R₂ , R₃ , R₄ are connected in parallel with R₄

$\to \rm \dfrac{1}{R_{2345}}=\dfrac{1}{R_5}+\dfrac{1}{R_{234}}\\\\\to \rm \dfrac{1}{R_{2345}}=\dfrac{1}{1}+\dfrac{1}{3}\\\\\to \rm \dfrac{1}{R_{2345}}=\dfrac{4}{3}\\\\\to \rm R_{2345}=\dfrac{3}{4}$

Now , R₂ , R₃ , R₄ are connected in series along with R₁ and R₆

$\rm \to R_{123456}=R_1+R_{2345}+R_6\\\\\rm \to R_{123456}=1+\dfrac{3}{4}+1\\\\\rm \to R_{123456}=2+\dfrac{3}{4}\\\\\rm \to R_{123456}=\dfrac{11}{4}\\\\\rm \to R_{123456}=2.75\ \Omega\ \; \bigstar$

$\rm So,R_{eq}=R_{123456}=2.75\ \Omega$