Two cells of e.m.f. s, E_1 and E_2 and of negligible internal resistances are connected with two variable resistors as shown in Fig. When the

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

b Potential difference $\mathrm{V}_{\mathrm{P}}$ across $\mathrm{P}$ as determined from $\mathrm{E}_{1}$ is given by $\mathrm{V}_{\mathrm{P}}=\left(\frac{\mathrm{E}_{1}}{\mathrm{P}+\mathrm{Q}}\right) \mathrm{P}.$ Also, $\mathrm{V}_{\mathrm{P}}=\mathrm{E}_{2}$. Therefore, $\mathrm{E}_{2}=\mathrm{E}_{1}\left(\frac{\mathrm{P}}{\mathrm{P}+\mathrm{Q}}\right) \Rightarrow \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{P}}{\mathrm{P}+\mathrm{Q}}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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

A$\frac{P}{Q}$
B$\frac{P}{{P + Q}}$
C$\frac{Q}{{P + Q}}$
D$\frac{{P + Q}}{P}$

Medium

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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