Question

4 resistors each of resistance 2ohm are joined end to end form a square ABCD. Calculate the equivalent resistance of the combination between any two adjacent corners ?

Answer 1

The net resistance between any two adjacent corner will be 1.5Ω

Explanation :

If we take any two adjacent corners, it will be a parallel combination of one resistor in that branch and 3 resistor in series in other 3 branches

Series resistance of 3 branches = 2 + 2 + 2 = 6Ω

Hence the net resistance will be a parallel combination between 2Ω and 6Ω

we know that parallel resistance is given by;

R = R₁R₂/R₁+R₂

=> R = 2x6/2+6

= 12/8

=> R = 1.5Ω

Hence the net resistance between any two adjacent corner will be 1.5Ω

Answer 2

Solution:

The given configuration is shown in attached figure.

To Calculate the equivalent resistance of the combination between any two adjacent corners ,let us take AB adjacent corners

So,resistors between other three sides are in series

$R_{s} = R_{bc} + R_{cd} + R_{ab} \\ \\ R_{s} = 2 + 2 + 2 \\ \\ = 6 \: ohm$
Now, 6 ohm is parallel with resistance of 2 ohm

$\frac{1}{R_{eq}} = \frac{1}{R_{s}} + \frac{1}{R_{ab}} \\ \\ = \frac{1}{6} + \frac{1}{2} \\ \\ = \frac{1 + 3 }{6} \\ \\ = \frac{4}{6} \\ \\ = \frac{2}{3} \\ \\ R_{eq} = \frac{3}{2} \\ \\ R_{eq} = 1.5 \: ohm$

hope it helps you