Four resistors, each of `10 Omega` are connected to form a square as shown in the given figure. Find the eqUivalent resistance between the opposite corner A and C and between two points on anyone side (AB).

Explanation

( i ) Resistance in the path
`ADC = 10 Omega + 10 Omega = 20Omega`
Resistance in the path
`ABC = 10 Omega + 10Omega = 20 Omega`
Now,`1/(R_(AC)) = 1/20 + 1/20 = 1/10`
`:. ` EqUivalent resistance,
`R_(AC)= 10 Omega`
(ii) Resistance in the path ADCB
`=10 Omega + 10 Omega + 10 Omega = 30 Omega`
The resistance of `30 Omega` is in parallel with `AB (=10 Omega)`
Now, `1/(R_(AB)) = 1/30 + 1/10`
`=(1 +3)/(30) = 4/30`
`:.` EqUivalent resistance,
`R_(AB) = 30/4 = 7.5 Omega`