Question 542: Current Electricity

Question

When cell of E.M.F. ' $E_1$ ' is connected to potentiometer wire, the balancing length is $l_1$. Another cell of E.M.F. ' $E_2$ ' $\left(E_1>E_2\right)$ is connected so that two cells oppose each other, then the balancing length is $l_2$. The ratio $\mathrm{E}_1: \mathrm{E}_2$ is

Accepted Answer

Assume the potential gradient along the uniform potentiometer wire is $$k$$ volts per unit length. This means that a length $$l$$ along the wire corresponds to a potential drop of $$k \, l$$ volts.
When the cell with EMF $$E_1$$ is connected to the potentiometer, the balancing length is given as $$l_1$$. This tells us that:
$$E_1 = k \, l_1.$$
Next, when the cell with EMF $$E_2$$ is connected in opposition to $$E_1$$, the net EMF acting on the wire becomes $$E_1 - E_2$$. The balancing length in this configuration is given as $$l_2$$. Hence, we write:
$$E_1 - E_2 = k \, l_2.$$
Now substitute the value of $$k$$ from the first equation:
$$ k = \frac{E_1}{l_1}. $$
So, replacing in the second equation:
$$ E_1 - E_2 = \frac{E_1}{l_1} \, l_2. $$
Rearranging to solve for $$E_2$$ gives:
$$E_2 = E_1 - \frac{E_1 \, l_2}{l_1} = E_1\left(1 - \frac{l_2}{l_1}\right).$$
Finally, to find the ratio $$\frac{E_1}{E_2},$$ we have:
$$\frac{E_1}{E_2} = \frac{E_1}{E_1\left(1 - \frac{l_2}{l_1}\right)} = \frac{1}{1 - \frac{l_2}{l_1}} = \frac{l_1}{l_1 - l_2}.$$
So, the ratio $$E_1 : E_2$$ is:
$$\frac{l_1}{l_1 - l_2},$$
which corresponds to Option B.

Answer

$\frac{l_1}{l_1+l_2}$

Answer

$\frac{l_1}{l_1-l_2}$

Answer

$\frac{l_1+l_2}{l_1}$

Answer

$\frac{l_1+l_2}{l_1-l_2}$

Current Electricity

When cell of E.M.F. ' E_1 ' is connected to potentiometer wire, the balancing length is l_1. Another cell of E.M.F. ' E_2 ' \left(E_1>E_2\right) is connected so that two cells oppose each other, then the balancing length is l_2. The ratio \mathrm{E}_1: \mathrm{E}_2 is

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