Two cells of e.m.f.s E_1 and E_2 (E_1 E_2) | vedclass

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(B) Let $k$ be the potential gradient of the potentiometer wire.When the potentiometer is connected between $A$ and $B$,the potential difference measu

Vedclass · Accepted Answer

$4:3$

Vedclass · Answer

(B) Let $k$ be the potential gradient of the potentiometer wire.When the potentiometer is connected between $A$ and $B$,the potential difference measured is $E_1$. Thus,$E_1 = k 	imes 3.60$.When the potentiometer is connected between $A$ and $C$,the cells are in series opposition because they are connected with opposite polarities. The net e.m.f. is $E_1 - E_2$. Thus,$E_1 - E_2 = k 	imes 0.90$.Dividing the two equations: $E_1 / (E_1 - E_2) = 3.60 / 0.90 = 4$.This implies $E_1 = 4(E_1 - E_2) = 4E_1 - 4E_2$.Rearranging gives $3E_1 = 4E_2$,so $E_1 / E_2 = 4/3$.

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