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Two cells of emf 3V and 2V and internal resistances 1.5Omega and 1Omega respectively are connected in parallel across a 3Omega resistor such that they tend to send current through the resistor in the same direction . Calculate the potential differences across the 3Omega resistor . |
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Answer» Solution :`E_1=3V " " E_2=2V , "" r_1=1.5Omega "" r_2=1Omega ""R=3Omega` Equivalent internal resistance of the combination, `r_(eq)=(r_1r_2)/(r_1+r_2)=(1.5xx1)/(1.5+1)=0.6Omega` Equivalent emf of the combination, `E_(eqv)/r_(eq)=E_1/r_1+E_2/r_2 , E_(eqv)/0.6=3/1.5+2/1, E_(eq)`=2.4 V CURRENT through the `3Omega` resistor `I=E_(eq)/(R+r_(eq))=2.4/(3+0.6)` `I=2/3A` Potential DIFFERENCES across `3Omega` resistor `V=IR=(2//3)(3)=2V` |
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