Question

Two light bulbs have resistances of 400 Ω and 800 Ω

The two light bulbs are connected in series across a 120- V line. Find the current through each bulb.
Enter your answers numerically separated by a comma.
I 400, I 800 =______A

Part B
Find the power dissipated in each bulb.
P 400, P 800 =________W


Part C
Find the total power dissipated in both bulbs.
P =______W

Answer 1

The current through the 400 Ω bulb is 0.3 A, and the current through the 800 Ω bulb is 0.15 A. The power dissipated by the 400 Ω bulb is 36 W, and the power dissipated by the 800 Ω bulb is 18 W. The total power dissipated in both bulbs is 54 W.

Step-by-step explanation:

To find the current through each bulb, we can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R).

For the bulb with resistance 400 Ω, the current is I400 = 120 V / 400 Ω = 0.3 A.

For the bulb with resistance 800 Ω, the current is I800 = 120 V / 800 Ω = 0.15 A.

To find the power dissipated by each bulb, we can use the formula P = IV, where P is power, I is current, and V is voltage.

Using the calculated currents, the power dissipated by each bulb is P400 = I400 * 120 V = 0.3 A * 120 V = 36 W and P800 = I800 * 120 V = 0.15 A * 120 V = 18 W.

The total power dissipated in both bulbs can be found by adding the power dissipated by each bulb. The total power is P = P400 + P800 = 36 W + 18 W = 54 W.