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The resistance of a 10 m long potentiometer wire is 10 Omega. If the current through it is 0.4 A, what are the balancing lenghts when two cells of emfs 1.3 V and 1.1V are connected so as to (i) assist (ii) oppose each other? |
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Answer» Solution :Data: L=10 m, `r=10 Omega, I=0.4 A, E_(1)=1.3 V, E_(2)=1.1 V` Potential gradient, `k=V/L = (Ir)/L = (0.4 xx 10)/10 xx 0.4 V//m` When the CELLS assist each other, the resultant EMF `=E_(1) + E_(2)`, LET `l_(1)` be the balancing length. `therefore E_(1) + E_(2) = kl_(1)` `therefore l_(1) = (E_(1) + E_(2))/(k) = (1.3 + 1.1)/(0.4) = 2.4/0.4 = 6m` When the cells oppose each other, the resultant emf `=E_(1) -E_(2)`. Let `I_(2)` be the balancing length. `therefore E_(1)-E_(2)=kl_(2)` `therefore I_(2) = (E_(1) +E_(2))/k = (1.3-1.1)/0.4 = 2.4/0.4 = 0.5`m |
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