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:

c. I = 1.73A

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
I = 1.73A

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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