Question 549: Current Electricity

Question

In potentiometer experiment, the balancing length is $$8 \mathrm{~m}$$ when two cells $$E_1$$ and $$E_2$$ are joined in series. When two cells are connected in opposition the balancing length is $$4 \mathrm{~m}$$. The ratio of the e.m.f. of the two cells $$\left(\frac{E_1}{E_2}\right)$$ is

Accepted Answer

$$\begin{aligned}
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1+l_2}{l_1-l_2}=\frac{8+4}{8-4} \
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{12}{4} \
& 	herefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2}=3 \
&
\end{aligned}$$

Answer

$$1: 2$$

Answer

$$2: 1$$

Answer

$$1: 3$$

Answer

$$3: 1$$

Current Electricity

In potentiometer experiment, the balancing length is $8 \mathrm{~m} when two cells E_1 and E_2 are joined in series. When two cells are connected in opposition the balancing length is 4 \mathrm{~m}. The ratio of the e.m.f. of the two cells \left(\frac{E_1}{E_2}\right)$ is

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