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In the given circuit diagram, culculate: (i) The main current through the circuit and (i) Also current through 9 Omegaresistor . |
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Answer» Solution :Given ` R_(1) = 6 Omega `, ` R_(2) = 12Omega ` ` R_(3) = 9Omega `, ` E = 3V ,R = 0.24Omega ` ` (##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 ` . |
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