AP EAMCET · PHYSICS · Current Electricity
When the terminals of a cell are connected by a wire of resistance \(4 \Omega\), the potential difference across the cell is 1.6 \(\mathrm{V}\). If a wire of the same resistance is connected in parallel with the first, the potential difference becomes \(1.33 \mathrm{~V}\). The emf and internal resistance of the cell are respectively
- A \(1 \mathrm{~V}, 1 \Omega\)
- B \(2 \mathrm{~V}, 1 \Omega\)
- C \(1 \mathrm{~V}, 2 \Omega\)
- D \(2 \mathrm{~V}, 2 \Omega\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{~V}, 1 \Omega\)
Step-by-step Solution
Detailed explanation
\(E=V+I r\) Given, \(V=1.6 \mathrm{~V} ; R=4 \Omega\) \( I_1=\frac{1.6}{4}=0.4 \mathrm{~A} \) \(E=1.6+0.4 r\) ...(i) When a resistance of \(4 \Omega\) is connected in parallel the resultant resistance…
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