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How much energy is dissipated as heat during a two-minute time interval by a 1.5- kΩ resistor which has a constant 20- V potential difference across its leads?

Respuesta :

Answer:

32 J

Explanation:

Power is V²/R = 20²/1500 = 4/15 . . . watts

In 120 seconds, the energy is ...

(4/15 J/s)×(120 s) = 32 J

In a 2-minute period 32 joules are dissipated as heat.