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(a) Three resistors of resistance 2Omega,3Omega and 4Omega are combined in series. What is the total resistance of the combination ? (b) It this combination is connected to a battery of emf 10 V and negligible internal resistance, obtain the potential drop across each resistor. |
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Answer» Solution :`R_(1)=2OMEGA,R_(2)=3omega,R_(3)=4omega,R_(s)=?` (a)In series ,EFFECTIVE resistance `R_(s)=R_(1)+R_(2)+R_(3)=2+3+4=9omega` (b)Current through the circuit is `I=(epsi)/(R_(s)+r)=(10)/(9)(because r=0)` I=1.11A `therefore` POTENTIAL drop across =`1R_(1)=2xx1.11=2.22V` Potential drop across =IR_(2)=3xx1.11=3.33V` Potentia drop across `R_(3)=IR_(3)=4xx1.11=4.44V` |
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