Question 563: Current Electricity

Question

In potentiometer experiments, two cells of e. m. f. '$$E_1$$' and '$$E_2$$' are connected in series $$\left(E_1>E_2\right)$$, the balancing length is $$64 \mathrm{~cm}$$ of the wire. If the polarity of $$E_2$$ is reversed, the balancing length becomes $$32 \mathrm{~cm}$$. The ratio $$\mathrm{E}_1 / \mathrm{E}_2$$ is

Accepted Answer

For potentiometer,
$$\begin{aligned}
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1+l_2}{l_1-l_2} \
	herefore \quad & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{64+32}{64-32} \quad=\frac{96}{32}=\frac{3}{1}=3: 1
\end{aligned}$$

Answer

$$3: 2$$

Answer

$$2: 3$$

Answer

$$1: 3$$

Answer

$$3: 1$$

Current Electricity

In potentiometer experiments, two cells of e. m. f. '$E_1' and 'E_2' are connected in series \left(E_1>E_2\right), the balancing length is 64 \mathrm{~cm} of the wire. If the polarity of E_2 is reversed, the balancing length becomes 32 \mathrm{~cm}. The ratio \mathrm{E}_1 / \mathrm{E}_2$ is

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