Two lightbulbs are each connected to a voltage of 120 V. One has a power of 25 W, the other 100 W. (a) Which lightbulb has the higher resistance? (b) Which lightbulb carries more current?

(a) The 25 W bulb

(b) The 100 W bulb

Explanation:

(a)

Using,

P = V²/R Equation 1

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

make R the subject of the equation

R = V²/P Equation 2

For the 25 W bulb,

Given: P = 25 W, V = 120 V

Substitute into equation 2

R = 120²/25

R = 576 Ω

For the 100 W bulb,

Given: P = 100 W, V = 120 V

Substitute into equation 2

R = 120²/100

R = 14400/100

R = 144 Ω

Hence the the 25 W bulb has the highest resistance.

(b)

Using,

P = VI Equation 3,

Where I = current

Make I the subject of the equation

I = P/V Equation 4

For the 25 W bulb,

Given: P = 25 W, V = 120 V

Substitute into equation 4

I = 25/120

I = 0.208 A.

For the 100 W bulb,

Given: P = 100 W, V = 120 V

Substitute into equation 4

I = 100/120

I = 0.833 A.

Hence the 100 W bulb carries more current.

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