The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is: (a) 4 V (b) 6 V (c) 8 V (d) 10 V

Answer: (c). I = (E)/(R + r) = (10)/(4 + 1) = 2 A V = E - Ir = 10 - 2 × 1 = 8 V (same as IR = 2 × 4) So the answer is (c) 8 V.

The terminal voltage of the battery, whose emf is 10 V and internal… | NEET 2024

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Cells: EMF, Internal Resistance & Combinations

The terminal voltage of the battery, whose emf is $10\text{ V}$ and internal resistance $1\ \Omega$ , when connected through an external resistance of $4\ \Omega$ as shown in the figure is:

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Answer: (c)

$I = \frac{E}{R + r} = \frac{10}{4 + 1} = 2\text{ A}$
$V = E - Ir = 10 - 2 \times 1 = 8\text{ V}$ (same as $IR = 2 \times 4$ )
So the answer is (c) 8 V.

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