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In the givencircuit diagram ,calculate:The maincurrentthroughthe circuit Also currentthrough9Omegaresistor. |
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Answer» SOLUTION :Given`R_1=6Omega` , `R_2=12Omega` , `R_3=9OMEGA` , `E=3V, R=0.24 Omega`  w.k.t. `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)~~ 1A` P.d. across internalresistance`V_r` =Ir = 1 x 0.24 =0.24 V `therefore` P.d. across each resistors =3-0.24 = 2.76 V `therefore `CURRENT in `9Oemga = 2.76/9=0.31 A ~~ 0.3A` Hence Current in `9Omega ~~ 0.3A` |
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