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A flashlight uses two 1.5 volt batteries in series. The flashlight bulb draws a current of 0.5 A. What is the power rating of the bulb?


A.
6 W


B.
0.75 W


C.
3 W


D.
1.5 W

Respuesta :

  • V=1.5V
  • I=0.5A

P=VI

  • P=0.5(1.5)
  • P=0.75W

Option B

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Answer:

b) 0.75 W

Explanation:

Given that,

→ Potential difference (V) = 1.5 V

→ Current (I) = 0.5 A

→ Power (P) = ?

Formula we use,

→ P = V × I

Then the answer will be,

→ P = V × I

→ P = 1.5 × 0.5

→ [ P = 0.75 ]

Hence, the power is 0.75 W.

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