A cell of emf 4 V and internal resistance 0.5 Ω is connected to a 7.5 Ω external resistance. The terminal potential difference of the cell is (a) 0.375 V (b) 3.75 V (c) 4.25 V (d) 4 V

Answer: (b). I = (4)/(7.5 + 0.5) = 0.5 A V = IR = 0.5 × 7.5 = 3.75 V So the answer is (b) 3.75 V.

A cell of emf 4 V and internal resistance 0.5 Ω is connected to a 7.5… | NEET 2022 (Phase 2)

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Cells: EMF, Internal Resistance & Combinations

A cell of emf 4 V and internal resistance 0.5 $\Omega$ is connected to a 7.5 $\Omega$ external resistance. The terminal potential difference of the cell is

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Answer: (b)

$I = \frac{4}{7.5 + 0.5} = 0.5\text{ A}$
$V = IR = 0.5 \times 7.5 = 3.75\text{ V}$
So the answer is (b) 3.75 V.

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