Question 493: Current Electricity

Question

Two cells of e.m.f.s $E_1$ and $E_2\left(E_1>E_2\right)$ are connected as shown in figure.
When the potentiometer is connected between A and B , the balancing length of the potentiometer wire is 3.60 m . On connecting the potentiometer between A and C , the balancing length is 0.90 m . The ratio $E_1 / E_2$ is

Accepted Answer

$$ \begin{aligned} &	ext { For comparing the e.m.f of the two cells, }\ &\frac{E_1}{E_2}=\frac{l_1+l_2}{l_1-l_2}=\frac{360+90}{360-90}=\frac{450}{270}=\frac{5}{3} \end{aligned} $$

Answer

$5: 3$

Answer

$4: 3$

Answer

$3: 4$

Answer

$4: 5$

Current Electricity

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