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n identical cells of emf ε and internal resistance r are connected in parallel with resistor R. The current flowing through resistor R is _____.

Asked in GSEB Board July 2015 · Combination of cells

• (1) (nε)/(R+nr)
• (2) (nε)/(nR+r)
• (3) ε/(R+r)
• (4) ε/(nR+r)

Step-by-step solution

Why the other options are wrong

• (1) (nε)/(R+nr) is the current for n cells in series, where the emfs add to nε and the internal resistances add to nr. In parallel the emf stays ε and the internal resistance falls to r/n.
• (3) ε/(R+r) is the current from a single cell. Putting n cells in parallel keeps the emf at ε but cuts the internal resistance to r/n, so the current is larger than this.
• (4) ε/(nR+r) comes from clearing the fraction in ε/(R+r/n) by multiplying only the denominator by n. The numerator must be multiplied by n as well, giving (nε)/(nR+r).

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