AP EAMCET · PHYSICS · Current Electricity
In a potentiometer experiment, when two cells of emfs \(\mathrm{E}_1\) and \(\mathrm{E}_2\left(\mathrm{E}_2>\mathrm{E}_1\right)\) are connected in series, the balancing length is 160 cm. If one of the cells is reversed, the balancing length decreases by \(75 \%\). If \(\mathrm{E}_1=1.2 \mathrm{~V}\), then \(\mathrm{E}_2=\)
- A 2 V
- B 2.4 V
- C 1.8 V
- D 1.5 V
Answer & Solution
Correct Answer
(A) 2 V
Step-by-step Solution
Detailed explanation
\( \frac{E_1 + E_2}{E_2 - E_1} = \frac{L_1}{L_2} \) \( L_2 = L_1 - 0.75 L_1 = 0.25 L_1 \) \( \frac{E_1 + E_2}{E_2 - E_1} = \frac{160 \mathrm{~cm}}{0.25 \times 160 \mathrm{~cm}} \) \( \frac{E_1 + E_2}{E_2 - E_1} = 4 \) \( E_1 + E_2 = 4E_2 - 4E_1 \) \( 5E_1 = 3E_2 \)…
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