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In the givencircuit , calculate the (i) effective resistance between A and B (ii) current through the circuit and (iii) current through 3 Omega resistor. |
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Answer» SOLUTION :`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.` |
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