Two light bulbs have resistances of 400 Ω and 800 Ω. Part APart complete 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. I400, I800 = 0.100,0.100 A Previous Answers Correct Part BPart complete Find the power dissipated in each bulb. Enter your answers numerically separated by a comma. P400, P800 = 4.00,8.00 W Previous Answers Correct Part CPart complete Find the total power dissipated in both bulbs. P = 12.0 W Previous Answers Correct Part D The two light bulbs are now connected in parallel across the 120-V line. Find the current through each bulb.

A) Rt = 1200 Ω B) I = 0.100 A the current in the two bulbs is the same

Explanation:

Part A

As the bulbs are connected in series the equivalent resistance is

Rt = R1 + R2

Rt = 400 + 800

Rt = 1200 Ω

With this value we can find the current

V = I Rt

I = V / Rt

I = 120/1200

I = 0.100 A

In a circuit, the current is constant, so the current in the two bulbs is the same

Part B

The equation for power is

P = I V = I² R

In the bulb of R = 400 Ω

P₄₀₀ = 0.1² 400

P₄₀₀ = 4.00 W

In the bulb of R = 800 Ω

P₈₀₀ = 0.1² 800

P₈₀₀ = 8.00 W

Part C

Pt = I² Rt

Pt = 0.1² 12

Pt = 12.0 W

Part D.

The bulbs are in parallel. In this case the voltage in each bulb is constant,

V = I R

i = V/R

I₄₀₀ = V/R₄₀₀

I₄₀₀ = 120 / 400

I₄₀₀ = 0.300 A

I₈₀₀ = V/R800

I₈₀₀ = 120/800

I₈₀₀ = 0.150 A