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PhysicsCurrent ElectricityBatteryEasy2 minQB

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A resistance $R$ whose value is varied from $1 Ω$ to $5 Ω$ is connected to a cell of internal resistance $3 Ω$ . The power consumed by $R$

Answer:

Solution:

$P = {(\frac{E}{R + 3})}^{2} R$

$\frac{dP}{dR} = 0 at R = 3 Ω$ so power is maximum at

Therefore increase continuously till $R = 3 Ω$ and then decreases

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Battery

⏱ 2mℹ️ Source: QB

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