Light bulb packages will have at least two numbers on them; a power and what the potential difference across the bulb must be so that the bulb has the listed power. You go to the store and purchase two light bulbs. The package for Bulb A says 40 W in a large font with 120 V in a smaller font; the package for Bulb B says 60 W in a large font with 120 V in a smaller font. You then connect the light bulbs as shown below (where the emf is 120 V):

To determine the current flowing through each bulb, we can use Ohm's Law, which states that the current (I) through a device is equal to the voltage (V) across the device divided by its resistance (R):

I = V/R

For each bulb, we know the power (P) and the voltage (V), and we can use these values to calculate the resistance (R).

First, let's calculate the resistance for Bulb A. The power of Bulb A is 40 W, and the voltage is 120 V.

We can use the formula for power in terms of voltage and resistance:

P = V^2 / R

Rearranging the equation to solve for resistance:

R = V^2 / P

Plugging in the values for Bulb A:

R_A = (120 V)^2 / 40 W

R_A = 3600 V^2 / 40 W

Simplifying:

R_A = 90 Ω

Similarly, for Bulb B, the power is 60 W, and the voltage is 120 V.

Using the same formula:

R_B = (120 V)^2 / 60 W

R_B = 3600 V^2 / 60 W

Simplifying:

R_B = 60 Ω

Now, let's analyze the circuit:

In this circuit, the bulbs are connected in parallel, meaning they share the same voltage (120 V).

For parallel resistors, the total resistance (R_total) is calculated using the formula:

1/R_total = 1/R_A + 1/R_B

Plugging in the values:

1/R_total = 1/90 Ω + 1/60 Ω

Simplifying:

1/R_total = (1/90 + 1/60) Ω^ -1

1/R_total = (2/180 + 3/180) Ω^ -1

1/R_total = (5/180) Ω^ -1

R_total = 180/5 Ω

R_total = 36 Ω

Finally, we can determine the current flowing through each bulb using Ohm's Law:

I_A = V/R_A

I_A = 120 V / 90 Ω

I_A ≈ 1.33 A

I_B = V/R_B

I_B = 120 V / 60 Ω

I_B = 2 A

Therefore, in this circuit configuration, Bulb A will have a current of approximately 1.33 A flowing through it, and Bulb B will have a current of 2 A flowing through it.