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n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R is
  • a)
    nE/R + nr
  • b)
    nE/nR + r
  • c)
    E/R + nr
  • d)
    nE/R + r
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
n identical cells each of e.m.f. E and internal resistance r are conne...
Explanation:

Given:
n identical cells with e.m.f. E and internal resistance r connected in series
External resistance R connected in series to this combination

Formula:
The total e.m.f. in a series combination is the sum of the e.m.f. of each cell, which is nE.
The total internal resistance in a series combination is the sum of the internal resistance of each cell, which is nr.

Calculation:
- The total e.m.f. in the circuit = nE
- The total internal resistance in the circuit = nr
- The total resistance in the circuit = R + nr

Using Ohm's Law:
- The total current in the circuit is given by I = V / (R + nr), where V is the total e.m.f.
- Substituting V = nE, the current through R is I = nE / (R + nr)
- Simplifying, we get I = nE / R + nr
Therefore, the current through the external resistance R is nE / R + nr, which corresponds to option A.
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n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R isa)nE/R + nrb)nE/nR + rc)E/R + nrd)nE/R + rCorrect answer is option 'A'. Can you explain this answer?
Question Description
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