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Physics · Internal resistance, potential difference and emf of a cell, a combination of cells in series and parallel

A cell sends a current through a resistance for time next the same cell sends cu

A cell sends a current through a resistance \( R \) for time \( t ; \) next the same cell sends current through another resistance \( r \) for the heat is developed in both the resistance, then the internal resistance of the cell is:

  • A. \( (R+r) / 2 \)
  • B. \( (R-r) / 2 \)
  • C. \( \sqrt{R r} \)
  • D. \( \sqrt{R r} / 2 \)

Step-by-step solution

Let E be emf and r_i be internal resistance. For resistance R, current I1 = E/(R+r_i). Heat in time t: H_R = I1^2 R t = (E^2 R t)/(R+r_i)^2. Similarly for resistance r: H_r = (E^2 r t)/(r+r_i)^2. Equating H_R = H_r and simplifying gives R/(R+r_i)^2 = r/(r+r_i)^2, leading to R(r+r_i)^2 = r(R+r_i)^2. Expanding and canceling 2R r r_i yields R r^2 + R r_i^2 = r R^2 + r r_i^2, or R r(r - R) = r_i^2 (r - R). Assuming r ≠ R, we get r_i^2 = R r, so r_i = √(R r).
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