JEE Mains · Physics · STD 12 - 3. current electricity
Two batteries with e.m.f \(12\ V\) and \(13\ V\) are connected in parallel across a load resistor of \(10\,\Omega\) . The internal resistances of the two batteries are \(1\,\Omega\) and \(2\,\Omega\) respectively. The voltage across the load lies between
- A \(11.5\) \( V\) and \(11.6\) \( V\)
- B \(11.4\) \( V\) and \(11.5\) \( V\)
- C \(11.7\) \(V\) and \(11.8\) \( V\)
- D \(\;\) \(11.6\) \( V\) and \(11.7\) \( V\)
Answer & Solution
Correct Answer
(A) \(11.5\) \( V\) and \(11.6\) \( V\)
Step-by-step Solution
Detailed explanation
Using Kirchhoff's law at \(P\) we get \(\frac{V-12}{1}+\frac{V-13}{2}+\frac{V-0}{10}=0\) [Let potential at \(\mathrm{P}, \mathrm{Q}, \mathrm{U}=0\) and at \(\mathrm{R}=\mathrm{V}\)…
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