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A 6.0 Ω and a 12. Ω resistor are connected in series to a 36. V battery, what power is dissipated by the 12.0 Ω resistor? A) 24. W B) 486. W C)216. W D) 12. W E) 48. W

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Answer:

Is dissipated E) P= 48 W in the 12 Ω resistor.

Step-by-step explanation:

V= 36v

Req= (12 + 6)Ω

Req= 18Ω

I = V/ Req

I= 2 A

P= I² * R(12Ω)

P= 48 W

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