Question

THE RESULTANT RESISTANCE BETWEEN A AND B IS 2 OHM'S IN THE GIVEN DIAGRAM. THEN THE VALUE OF R=

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Answer 1

${ \blue{ \underline{ \underline{ \sf{ \: \: Answer↓ \: \: }}}}}$

Given,

${ \tt{ \red{R_{AB }= 2Ω}}}$

First add all the resistances which are in series,

${ \tt{ \red{R_{CD} = 2Ω + 1Ω => 3Ω}}}$

${ \tt{ \red{R_{EF} = 4Ω + R }}}$

Now, they are in parallel,

${ \pink{ \implies{\tt \: \dfrac{1}{R_{AB} } = \dfrac{1}{R_{CD}} + \dfrac{1}{R_{EF}} }}}$

${ \pink{ \implies{\tt \: \dfrac{1}{ 2Ω} = \dfrac{1}{3Ω} + \dfrac{1}{ 4Ω + R } }}}$

${ \pink{ \implies{\tt \: \dfrac{1}{ 2Ω} = \dfrac{4Ω + R + 3Ω}{3Ω(4Ω + R )} }}}$

${ \pink{ \implies{\tt \: \dfrac{1}{ 2Ω} = \dfrac{7Ω + R }{12Ω {}^{2} + 3ΩR } }}}$

${ \pink{ \implies{\tt \: {12Ω {}^{2} + 3ΩR}{} = { 2Ω(7Ω + R) } }}}$

${ \pink{ \implies{\tt \: {12Ω {}^{2} + 3ΩR}{} = { 14Ω {}^{2} + 2ΩR} }}}$

${ \pink{ \implies{\tt \: {12Ω {}^{2} - 14Ω {}^{2} }{} = { 2ΩR - 3ΩR} }}}$

${ \pink{ \implies{\tt \: {- 2Ω {}^{2} }{} = { - ΩR} }}}$

${ \pink{ \implies{\tt \: \dfrac{ \cancel{ - } 2Ω ^{ \cancel{2}} }{ \cancel{- } \cancel{Ω}} = R}}}$

${ \boxed{ \pink{ \therefore{\tt \: R = 2Ω }}}}$

Answer 2

Answer:

4+5+2+1 is 12 mark me as brainiest