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Two cells E_(1) (with emf 4V and internal resistance 2Omega) and E_(2) (with emf 2V and intenral resistnace 2Omega) are connected in parallel. The combination is connected in parallel with a 8Omega resistance R as shown in figure. Calcualte the currents passing through 2V cell and through the resistance R. |
Answer» Solution : Applying Kirchhoff.s second law to the loop ABEFA. then `-2+2I_(2)-2I_(1)+4=0` `2I_(1)-2I_(2)=2"….(1)"` Applying Kirchhoff.s seocnd law to the loop BCDEB, then `-8(I_(1)+I_(2))-2I_(2)+2=0` `8I_(1)+10I_(2)=2"..........(2)"` Solving equation (1) and (2), we GET `I_(1)=(2)/(3)A,I_(2)=-(1)/(3)A,I_(1)+I_(2)=(1)/(3)A` Thus, the current PASSING through 2V CELLS is `I_(2)=-(1)/(3)A.` That is `(1)/(3)A` current flows though 2V CELL from B to E direction. And `(1)/(3)A` current flows through `8OMEGA` resistance from C to D. |
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