WBJEE · Physics · Current Electricity
Two cells \(A\) and \(B\) of emf \(2 \mathrm{V}\) and \(1.5 \mathrm{V}\) respectively, are connected as shown in figure through an external resistance \(10 \Omega\) The internal resistance of each cell is \(5 \Omega\). The potential difference \(E_{A}\) and \(E_{B}\) across the terminals of the cells \(A\) and \(B\) respectively are
- A \(E_{A}=20 \mathrm{V}, E_{B}=1.5 \mathrm{V}\)
- B \(E_{A}=2.125 \mathrm{V}, E_{B}=1.375 \mathrm{V}\)
- C \(E_{A}=1.875 \mathrm{V}, E_{B}=1.625 \mathrm{V}\)
- D \(E_{A}=1.875 \mathrm{V}, E_{B}=1.375 \mathrm{V}\)
Answer & Solution
Correct Answer
(C) \(E_{A}=1.875 \mathrm{V}, E_{B}=1.625 \mathrm{V}\)
Step-by-step Solution
Detailed explanation
The figure can be redrawn as, The current through the circuit \[ \begin{aligned} i &=\frac{\text { net emf }}{\text { effective resistance }} \\ &=\frac{2-1.5}{5+5+10}=\frac{0.5}{20} \\ &=\frac{1}{40}=0.025 \mathrm{A} \end{aligned} \] The terminal potential difference of the…
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