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Two cells of emfs $E_{1} a n d E_{2}$ and of negligible internal resistances are connected with two variable resistors as shown in 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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$\frac{P + Q}{P}$

$\frac{P}{Q}$

$\frac{P}{P + Q}$

$\frac{Q}{P + Q}$

answer is B.

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Detailed Solution

Potential difference $V_{P}$ across P as determined from $E_{1}$ is given by $V_{P} = (\frac{E_{1}}{P + Q}) P .$

Potential $V_{P}$ across P as determined from $E_{2}$ is same as $E_{2}$ because no current is drawn, i.e., $V_{P} = E_{2} .$

Therefore,

$E_{2} = E_{1} (\frac{P}{P + Q}) o r \frac{E_{2}}{E_{1}} = \frac{P}{P + Q}$

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