Two sources of equal emf are connected to an external resistance R. The internal…

Physics · JEE Advanced · NTA ExamsCurrent 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
  1. \(\mathrm{R}=\frac{\mathrm{R}_{2} \times\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)}{\left(\mathrm{R}_{2}-\mathrm{R}_{1}\right)}\)
  2. R = R2 – R1
  3. \(\mathrm{R}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)}\)
  4. \(\mathrm{R}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\left(\mathrm{R}_{2}-\mathrm{R}_{1}\right)}\)

Answer

(B) R = R 2 – R 1

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

More Current Electricity questions · Browse all practice questions