Two sources of equal emf are connected to an external resistance R. The internal…
Physics · JEE Advanced · NTA Exams — Current Electricity
Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R1 and R2 (R2 > R1). If the potential difference across the source having internal resistance R2, is zero, then
- \(\mathrm{R}=\frac{\mathrm{R}_{2} \times\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)}{\left(\mathrm{R}_{2}-\mathrm{R}_{1}\right)}\)
- R = R2 – R1
- \(\mathrm{R}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)}\)
- \(\mathrm{R}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\left(\mathrm{R}_{2}-\mathrm{R}_{1}\right)}\)
Answer
(B) R = R 2 – R 1
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