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.

Answer:

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

The current in the bulb with the resistance of 400 Ω is 0.3 A, while the current in the bulb with the resistance of 800 Ω is 0.15 A.

What is power?

Power is the rate at which the energy is been transformed or transferred per unit time.

A.) As we know that the bulbs are connected in series, therefore, the equivalent resistance of both the bulbs can be written as,

R_e = R_1 + R_2\\\\R_e = 400 +800\\\\R_e = 1200\rm\ \Omega

As the voltage is known to be 120 V, while the equivalent resistance is 1200 Ω. therefore, the current can be written as,

V = I \times R_e\\\\I = \dfrac{V}{R_e}\\\\I = \dfrac{120}{1200}\\\\I = 0.100\rm\ A

Since the current in the circuit is constant, the current in the two bulbs is the same.

B.) We know that the equation of power for the circuit can be written as

\rm P = I \times V = I^2 \times R

Since the resistance in the first bulb is of R = 400 Ω, therefore, the power in the first bulb can be written as,

P_{400} = 0.1^2 \times 400\\\\P_{400}= 4.00\rm\ W

The power in the second bulb with resistance, R = 800Ω can be written as,

P_{800} = 0.1^2 \times 800\\\\P_{800}= 8.00\rm\ W

C.) The total power dissipated in both bulbs can be written as,

P_t = I^2 \times R_e\\\\P_t = 0.1^2 \times 12\\\\P_t = 12.0\rm\ W

D.) We know when the bulbs are connected in parallel. therefore, the voltage in each bulb is constant,

I = \dfrac{V}{R}

The current in the bulb with the resistance of 400 Ω,

I_{400} = \dfrac{V}{R_{400}}\\\\I_{400} = \dfrac{120}{400}\\\\I_{400} = 0.300\rm\ A

The current in the bulb with the resistance of 800 Ω,

I_{800} = \dfrac{V}{R_{800}}\\\\I_{400} = \dfrac{120}{800}\\\\I_{400} = 0.15\rm\ A

Hence, the current in the bulb with the resistance of 400 Ω is 0.3 A, while the current in the bulb with the resistance of 800 Ω is 0.15 A.