(i) Calcuate the equivalent resistance of the given electronical network point A and B of

(ii) Also calculate the current through `ACB` if a `10 V` d.c source is connected between A and B and th e value of `R` is is assumed as `2 Omega`

Answer

`2.5 A`

Explanation

(i) The equivalent circuit will be .According to whearstone bridge principle `(R )/(R ) =(R )/(R )`

so bright is balance NO current flow through arm `CD` it mean resistance `R` of arm `CD` is ineffective .Now resistance of arm `ACB(= R + R = 2 R)` is in paralllel to the resistance of arm `ADC (= R +_ R = 2R)`
Equivalent resistance between `A and B` is
`R = (2R xx 2R)/(2R + 2R) = R`
(ii) Given `R = 2 Omega ,V = 10 V` current through
ACB is `i = (V ) 2R = (10) /(2 xx 2) = 2.5 A`