The maximum power rating of a `20 Omega` resistor is 2.0 kW. [That is, this is the maximum power the resistor can dissipate (as heat) without melting or changing in some other undersirable way]. Would you connect this resistor directly across a 300 V d.c. source of negligible internal resistance ? Explain your answer.
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Given, Max. Power rating of a `20 Omega` resistor P'= 2.0 kW. Current in resistance, `I = 2000W/20 Omega = 10A` When this resistor is connected to 300 V dc then rate of production of heat, `P = V^(2)/R = (300)^(2)/20 W = 45000 W = 4.5 kW` As `4.5 kW gt 2.0 kW`, hence `20 Omega` resistor should not be connected directly across the 300 V d.c. source. However, the `20 Omega` resistor can be connected to 300 V d.c. source by using some resistance `R_(1)` say in series with it so that current in the circuit remains within the safety limit `(= 10A)`. Thus current, `10 = 300/20+R_(1) or R_(1) = 10 Omega`
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