If you put two 60 W bulbs in series across a 120 V outlet, how much power would each consume if its resistance were constant?

Answer:15 watts

Explanation:

To get the current

Power P=iV

(Where P is power of bulb 60watts, i is current, and V is the voltage 120V

i=P/V

i = 60/120

i = 1/2A

But

V=iR where R is the resiance

Therefore R=V/i

=120:1/2

=120*2

=240ohms

For series connection, each bulb draws the same current since,

Since the two bulbs are 60-watt bulbs, they will have the same resistance, therefore the voltage across each bulb is the same and equals halve of the applied voltage, 120/2 = 60 volts.

Taken the resistance 240 ohm is constant across the series, we can roughly estimate the current flow and calculate power dissipation in the series connection.

We have 120 V / (240 + 240) ohms = 1/4 A.

The power dissipated in each bulb is (1/4) * 60 = 15 watts.

Each would use 15 W power.

Together they would use 30 W.

Explanation:

For the 60 W bulb R = ΔV²/P = (120) ²/60=14400/60 = 240 OHMS

We can use Ohms Law to find the current

I = ΔV/R = 120/240 = 0.5 Amperes

When they are placed in series the total resistance is about 480 ohms, so each bulb sees a current of 120/480 = 0.25 A.

Using the power equation to find the power of the bulbs in series

P = I²R = 0.25² * 240 = 0.0625 * 240 = 15 W

Each would use 15 W power.

Together they would use 30 W.