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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityCircuit AnalysisMedium2 minPYQ_2004

PhysicsMediumsingle choice

When a resistance of $9Ω$ is connected at the ends of a battery, its potential difference decreases from 40 V to 30 V. The internal resistance of the battery is

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Solution:

$E = 40 V, V = 30 V, R = 9Ω$ Internal resistance $r = (\frac{E}{V} - 1) R = (\frac{40}{30} - 1) 9$ $= \frac{1}{3} \times 9 = 3Ω$

Stream:JIPMERSubject:PhysicsTopic:Current ElectricitySubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2004

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