JEE Mains · Physics · STD 12 - 3. current electricity
Two cells are connected in opposition as shown. Cell \(\mathrm{E}_1\) is of \(8 \mathrm{~V}\) emf and \(2 \ \Omega\) internal resistance; the cell \(E_2\) is of \(2 \mathrm{~V}\) emf and \(4\ \Omega\) internal resistance. The terminal potential difference of cell \(\mathrm{E}_2\) is _______.

- A \(10\)
- B \(6\)
- C \(7\)
- D \(45\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
\(I=\frac{8-2}{2+4}=\frac{6}{6}=1 \mathrm{~A}\) Applying Kirchhoff from \(C\) to \(B\) \( V_C-2-4 \times 1=V_B \) \( V_C-V_B=6 \mathrm{~V} \) \( =6 \mathrm{~V}\)
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