In the given circuit , calculate the (i) effective resistance between A and B (ii) current through the circuit and (iii) current through `3 Omega` resistor.

Explanation

`R_(1) = 12 Omega " "R_(3) = 4 Omega`
`R_(2) =8Omega " "R _(4) =6 Omega`
` R_(5) =3 Omega " "R_(6) = 2 Omega`
Now ` R _(1) || R_(2)`
`R _(n et1 ) = (R_(1) R_(2))/(R_(1) + R _(2)) = (12xx 8)/( 12 +8)`
Again `R _(n et1) = 4.8 Omega`
`R _(3) || R_(4)`
` R _( n et2) = ( R _(3) R_(4))/(R_(3) + R_(4)) = ( 4xx 6)/( 4 + 6)`
`R _(n et2) =2.4 Omega`
Now `R_(n et1)` is series to `R _( n et2)`
`R _( n et3) = R _(n et1 ) + R _(n et2)`
`R _(n et3) = 7.2Omega`
Here `R _(n et3) || R_(5) ||R_(6)`
`(I)/( R _(eff)) = (1)/(7.2) + 1/3 + 1/2 = (70)/(72)`
`R _(eff) = (72)/(70 ) = (36)/(35)`
Now `I = V/R = (35)/(36) xx 12`
`I = ( 12 xx 35)/(36)`
`I = 11. 66 A`
Current through `3 Omega` resistor is
`I = V/R = (12)/(3), I = 4 A.`