Equivalent resistance 3

Find the equivalent resistance between the two points A A and B B in the above combination of resistors.

R 2 Ω \frac{R}{2} \text{ } \Omega 3 R Ω 3R \text{ } \Omega R Ω R \text{ } \Omega 2 R Ω 2R \text{ } \Omega

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2 solutions

Chew-Seong Cheong
Aug 11, 2014

The equivalent resistance

R e q = 2 R / / [ R + 2 R / / ( R + R ) ] R_{eq}=2R//[R+2R//(R+R)]

= 2 R / / ( R + 2 R / / 2 R ) =2R//(R+2R//2R)

= 2 R / / ( R + 2 R × 2 R 2 R + 2 R ) =2R//\left( R+\cfrac{2R\times 2R}{2R+2R}\right)

= 2 R / / ( R + R ) =2R//(R+R)

= 2 R / / 2 R =2R//2R

= R Ω =\boxed{R\space \Omega}

Kshitij Johary
Jul 13, 2014

The circuit will have following configurations:

  1. One resistance of 2R is in parallel.

  2. One resistance of R in parallel connected to two parallel resistors of 2R (R+R) and 2R.

Thus the resultant resistance will be the reciprocal of

1 2 R + { 1 R + ( 2 R x 2 R 2 R + 2 R ) } = 1 2 R + 1 2 R = 1 R \frac { 1 }{ 2R } \quad +\quad \{ \frac { 1 }{ R\quad +\quad (\frac { 2R\quad x\quad 2R }{ 2R\quad +\quad 2R } ) } \} \quad =\quad \frac { 1 }{ 2R } +\frac { 1 }{ 2R } \quad=\quad \frac { 1 }{ R } .

The reciprocal of 1/R is the resultant resistance of the circuit which is R.

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