JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2020
- Hard
- Numerical
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
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Answer & Solution
Correct Answer
(A) \(30\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}_{1} =\mathrm{E}-\mathrm{ir}\) \(=10-\mathrm{i} \times20=0\)
\(\mathrm{i}=0.5 \mathrm{A}\)
\(\mathrm{E}_{2}=\mathrm{E}-\mathrm{ir}\)
\(=10-0.5 \times 5\) \(=7.5 \mathrm{V}\)
\(\mathrm{E}_{\mathrm{net}}=\mathrm{E}_{1}+\mathrm{E}_{2}=7.5 \mathrm{V}\)
\(\mathrm{i}=\mathrm{i}_{1}+\mathrm{i}_{2}\)
\(0.5=\frac{7.5}{x}+\frac{7.5}{30}\)
\(\mathrm{x}=30 \Omega\)
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