JEE Mains · Physics · STD 12 - 3. current electricity
A cell \(E _{1}\) of \(emf 6 V\) and internal resistance \(2 \Omega\) is connected with another cell \(E _{2}\) of \(emf 4 V\) and internal resistance \(8 \Omega\) (as shown in the figure). The potential difference across points \(X\) and \(Y\) is............ \(V\)
- A \(10.0\)
- B \(3.6\)
- C \(5.6\)
- D \(2.0\)
Answer & Solution
Correct Answer
(C) \(5.6\)
Step-by-step Solution
Detailed explanation
\(I =\frac{6-4}{10}=\frac{1}{5} A\) \(V _{ x }+4+8 \times \frac{1}{5}- V _{ y }=0\) \(V _{ x }- V _{ y }=-5.6 \Rightarrow| Vx - Vy |=5.6 V\)
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