Two cells of e.m.f. E_1 and E_2 and of neglig | vedclass

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(B) The circuit consists of a loop with cell $E_1$ and resistors $P$ and $Q$ in series.When the galvanometer shows no deflection,it means no current f

Vedclass · Accepted Answer

$P/(P + Q)$

Vedclass · Answer

(B) The circuit consists of a loop with cell $E_1$ and resistors $P$ and $Q$ in series.When the galvanometer shows no deflection,it means no current flows through the branch containing $E_2$ and the galvanometer.Therefore,the potential difference across the resistor $P$ is equal to the $e.m.f.$ of the cell $E_2$.The current $I$ flowing through the series combination of $P$ and $Q$ due to cell $E_1$ is $I = E_1 / (P + Q)$.The potential difference across resistor $P$ is $V_P = I 	imes P = [E_1 / (P + Q)] 	imes P$.Since $V_P = E_2$,we have $E_2 = E_1 	imes [P / (P + Q)]$.Thus,the ratio $E_2 / E_1 = P / (P + Q)$.

Two cells of $e.m.f.$ $E_1$ and $E_2$ and of negligible internal resistances are connected with two variable resistors as shown in the figure. When the galvanometer shows no deflection,the values of the resistances are $P$ and $Q$. What is the value of the ratio $E_2/E_1$?

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