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PhysicsCurrent ElectricityElectrical PowerEasy2 minPYQ_2024

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The maximum power rating of a $20Ω$ resistor is 2 kW . Then, it cannot be used with 300 V DC source because the

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Maximum power rating of $20Ω$ resistor, $P_{r} = 2 kW$ Power dissipated through $20Ω$ resistor when it is connected with 300 VD , $P_{1} \Rightarrow P_{1} = \frac{V ^{2}}{R} = \frac{30 0 ^{2}}{20} = \frac{300 \times 300}{20} = 4500 W = 4.5 kW$ $∴ P_{1} > P_{r}$ Hence, resistor may be damaged due to production of excessive heat.

Stream:VITEEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical Power

⏱ 2mℹ️ Source: PYQ_2024

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