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