Question 527: Current Electricity

Question

In potentiometer experiment, cells of e.m.f. $E_1$ and $E_2$ are connected in series $\left(E_1>E_2\right)$ the balancing length is 80 cm of the wire. If the polarity of $E_2$ is reversed, the balancing length becomes 20 cm. The ratio $\mathrm{E}_1 / \mathrm{E}_2$ is

Accepted Answer

In a potentiometer experiment, when cells with electromotive forces (emfs) $ E_1 $ and $ E_2 $ are connected in series $ (E_1 > E_2) $, the balancing length of the wire is 80 cm. If the polarity of $ E_2 $ is reversed, the balancing length changes to 20 cm. To find the ratio $ \frac{E_1}{E_2} $, we use the following formula for potentiometer readings:
$ \frac{E_1}{E_2} = \frac{l_1 + l_2}{l_1 - l_2} $
Plugging in the given values:
$ \frac{E_1}{E_2} = \frac{80 + 20}{80 - 20} = \frac{100}{60} = \frac{5}{3} $
Thus, the ratio $ \frac{E_1}{E_2} $ is $ 5:3 $.

Answer

$1: 2$

Answer

$2: 3$

Answer

$3: 4$

Answer

$5: 3$

Current Electricity

In potentiometer experiment, cells of e.m.f. E_1 and E_2 are connected in series \left(E_1>E_2\right) the balancing length is 80 cm of the wire. If the polarity of E_2 is reversed, the balancing length becomes 20 cm. The ratio \mathrm{E}_1 / \mathrm{E}_2 is

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