In the given circuit diagram, culculate: (i) The main current through the circuit and (i) Also current through `9 Omega ` resistor .

Explanation

Given ` R_(1) = 6 Omega ` ,
` R_(2) = 12 Omega `
` R_(3) = 9 Omega ` ,
` E = 3V ,R = 0.24 Omega `
` (##SPH_ABT_PHY_QB_XII_JUL_18_E04_009_S01.png" width="80%">
` w.kt. (1) /(R_(eq)) = (1) /(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)) `
` i.e., (1)/(R_(eq)) = (1)/(6) + (1)/(12) + (1)/(9) = (13)/(36) `
` therefore R_(eq) = (36)/(13) = 2.76 Omega `
`w.k.t " "I = (E) /(R_(eq) + r,) I = (3)/(2.76+ 0.24) ~~ 1 A `
P.d across resistance `V_(tau) = Ir = 1 xx 0.24 = 0.24 V `
` therefore ` P.d across each resistors = 3 - 0.24 = 2.76 V
` therefore ` Current in ` 9 Omega = (2.76)/(9) = 0.31 A ~~ 0.3 A `
Hence Current in ` 9 Omega ~~ 0.3 A ` .