Question

$\underline{\overline{\underline{\huge{\mathfrak{ Challenge}}}}}$

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Answer 1

Answer:

In this kind of circuits ,we need to apply a basic tool (concept) in order to simplify the circuit and find the equivalent resistance.

Follow the steps :

1. First look for a plane of symmetry
2. If there is a valid plane / line of symmetry , then any joint of wires or resistance can be opened up.

Diagram:

Look at the attached photo to understand better.

Calculation:

We get to see a line of symmetry along AB axis.

So the joint region at the centre can be opened up. And the circuit is divided into 3 parts :

• Upper
• Middle
• Lower

And these parts are on parallel to one another

Resistance on the upper part is 8r/3

Resistance in the middle part is 2r

Resistance in the lower part is 8r/3

Considering them to be in parallel , we get R_(eq) as :

$\dfrac{1}{R_{eq}} = \dfrac{1}{ \frac{8r}{3} } + \dfrac{1}{ \frac{8r}{3} } + \dfrac{1}{2r}$

$= > \dfrac{1}{R_{eq}} = \dfrac{3}{8r} + \dfrac{3}{8r} + \dfrac{1}{2r}$

$= > \dfrac{1}{R_{eq}} = \dfrac{3 + 3 + 4}{8r}$

$= > \dfrac{1}{R_{eq}} = \dfrac{10}{8r}$

$= > R_{eq}= \dfrac{4r}{5}$

So final answer :

$\boxed{ \red{ \huge{ R_{eq}= \dfrac{4r}{5} }}}$

Answer 2

$</p><p>\tt{\huge{\red{\implies{R_{AB} = \dfrac{6}{7} \: ohm. }}}}$