The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. What is the resistance of (a) a 150-W bulb and (b) a 60-W bulb? (c) How much current does each bulb draw in normal use?

a) 225 ohms

b) 240 ohms

c) I(150)= 0.82A

I(60)= 0.5A

Explanation:

Using the equation: R= V^2/P

Where R = resistance in ohms

V= voltage in volts

P= power in watts

a) R = 150v^2/100watts

R= 22500/100 = 225 ohms

b) R = V^2/ P = 120^2v / 60watts

R = 14400/60 = 240 ohms

c) Using the equation: P = I^2R

I = Sqrt(P/R)

I (150) = Sqrt(150/225)

I = Sqrt(0.6667)

I = 0.82A

I(60) = Sqrt(P/R)

I= Sqrt(60/240)

I = Sqrt(0.25)

I = 0.5A

(a) 96 Ω

(b) 240 Ω

(c) for the 150 W bulb, I = 1.25 A and for the 60 W bulb, I = 0.5 A

Explanation:

(a)

From electrical power,

P = V²/R Equation 1

Where P = power, V = Voltage, R = resistance.

make R the subject of the equation

R = V²/P Equation 2

Given: V = 120 V, P = 150 W

Substitute into equation 2

R = 120²/150

R = 14400/150

R = 96 Ω

(b)

Also, using equation 2 above,

R = V²/P

Given: V = 120 V, P = 60 W

Substitute into equation 2

R = 120²/60

R = 14400/60

R = 240 Ω

(c)

Power = Voltage×current

P = VI

make I the subject of the equation,

I = P/V Equation 3

For the 150 W bulb,

I = 150/120

I = 1.25 A,

For the 60 W bulb,

I = 60/120

I = 0.5 A.