Question

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

a. I= 2.00A
b. I=3.00A
c. I=1.73A
d. I =0.333A

Answer 1

c. I = 1.73A

Step-by-step 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

Answer 2

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