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Three resistors 1Omega , 2Omega " and " 3Omegaare combined in series. What is the total resistance of thecombination ?(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance,obtain the potential drop across each resistor. |
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Answer» Solution :(a) Here `R_1 = 1Omega , R_2 = 2 Omega` and `R_3 = 3Omega ` TOTAL resistance in series combination`R = R_1 + R_2 + R_3 = 1 + 2 + 3 = 6Omega` (b) As `epsi = 12 V, r = 0 `, hence current `I = (epsi)/(R+r ) = (12)/(6+0) = 12/6= 2.0 A` POTENTIAL drop across the three RESISTORS are respectively `V_1 = IR_1 = 2 xx 1 = 2V, V_2 = IR_2 = 2 xx 2 = 4V " and " V_3 = IR_3 = 2 xx 3 = 6V` |
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