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Asked in GSEB Board March 2020 (old course) · Terminal voltage and internal resistance
Given: terminal voltage V=2 V, internal resistance r=0.2 Ω, current I=0.5 A.
Idea: while a battery drives current, part of its emf is used up inside it, so V=ε-Ir.
Internal drop: Ir=0.5×0.2=0.1 V.
ε=V+Ir=2+0.1.
ε=2.1 V.
Only a battery being charged has a terminal voltage above its emf (then ε=V-Ir=1.9 V); nothing in the question says it is being charged.
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