JEE Mains · Physics · STD 12 - 3. current electricity
The series combination of two batteries, both of the same emf \(10 \mathrm{\;V},\) but different internal resistance of \(20\; \Omega\) and \(5\; \Omega,\) is connected to the parallel combination of two resistors \(30\; \Omega\) and \(\mathrm{R}\; \Omega .\) The voltage difference across the battery of internal resistance \(20\; \Omega\) is zero, the value of \(\mathrm{R}(\text { in } \Omega)\) is
- A \(30\)
- B \(26\)
- C \(36\)
- D \(15\)
Answer & Solution
Correct Answer
(A) \(30\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}_{1} =\mathrm{E}-\mathrm{ir}\) \(=10-\mathrm{i} \times20=0\) \(\mathrm{i}=0.5 \mathrm{A}\) \(\mathrm{E}_{2}=\mathrm{E}-\mathrm{ir}\) \(=10-0.5 \times 5\) \(=7.5 \mathrm{V}\) \(\mathrm{E}_{\mathrm{net}}=\mathrm{E}_{1}+\mathrm{E}_{2}=7.5 \mathrm{V}\)…
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