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A 50 V battery is connected across a $10 Ω$ resistor. The current is 4.5 ampere. The internal resistance of the battery is :

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$1.5 Ω$

$0.5 Ω$

$1.1 Ω$

$5.0 Ω$

answer is C.

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Detailed Solution

current $I = \frac{E}{R + r}$

$R + r = \frac{E}{I} = \frac{50}{4.5} r = \frac{50}{4.5} - 10 = 1.1 Ω$

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