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NARA [144]
2 years ago
14

A flashlight bulb with a 6.00 resistor uses 18.0W of power. What is the current through the bulb

Physics
2 answers:
marishachu [46]2 years ago
8 0

Answer:

<u>c. I = 1.73A</u>

Explanation:

The formula for power :

  • P = VI

The formula for potential difference :

  • V = IR

Calculating P by inputting the formula of V :

  • P = (IR)(I)
  • P = I²R

Solving using the given values :

  • 18 = I² x 6
  • I² = 3
  • I = √3
  • <u>I = 1.73A</u>
devlian [24]2 years ago
6 0

We need voltage first

  • P=V²/R
  • V²=PR
  • V²=18(6)
  • V²=108
  • V=√108
  • V=10.4V

Apply oHMS law

  • V/I=R
  • I=V/R
  • I=10.4/6
  • I=1.73A
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