You have two light bulbs: One with resistance R1 and one resistance R2. Assume R1 > R2.

A) Which bulb is brighter if connected in series. Explain.

B)Which bulb is brighter if connected in parallel. Explain.

I will gladly critique your thinking, when any is shown.

Well I was thinking:

A) Since the bulbs are in series, the R1 is going to be brighter than R2, because it has less resistance.

B) The bulbs will have the same brightness, because when the bulbs are parallel, they both receive the same voltage.

Also, I don't fully understand this resistance thing. Can you please explain?

Brightness means power.

Power= V^2/R= I^2R
in series, they have the same current. Which then has generates more power?
In parallel, they have the same voltate, which has more power?

so from what I understood, in parallel, R2 will be the brightest. and in series R1 will be the brightest?

A) When the two light bulbs are connected in series, the total resistance of the circuit is the sum of the resistances of the individual bulbs (R_total = R1 + R2). According to Ohm's Law (V = I * R), the current flowing through the circuit will be the same for both bulbs since they are in a series connection.

Now, the brightness of a light bulb is directly proportional to the power it consumes, which is given by the formula P = V^2 / R, where P is the power, V is the voltage, and R is the resistance. Since the voltage across the bulbs (V_total) is divided between them in a series connection, with each bulb having half the voltage (V_total / 2), the power consumed by each bulb can be calculated.

For bulb 1, the power is P1 = (V_total / 2)^2 / R1 = V_total^2 / (4 * R1), and for bulb 2, the power is P2 = (V_total / 2)^2 / R2 = V_total^2 / (4 * R2).

Comparing these two equations, we can see that the brightness of a bulb is inversely proportional to its resistance. Since R1 > R2, it means that bulb 1 has a higher resistance compared to bulb 2. Thus, according to the power equation, bulb 1 will consume less power and offer less brightness compared to bulb 2 when connected in series. Therefore, bulb 2 will be brighter.

B) When the two light bulbs are connected in parallel, the voltage across them is the same (V_total). However, the current passing through each bulb will be different due to their different resistances. The total current in a parallel connection is the sum of the currents passing through each branch (I_total = I1 + I2).

Using Ohm's Law again, we know that the current passing through individual bulbs is given by I1 = V_total / R1 and I2 = V_total / R2. Comparing these two equations, we can see that the current passing through bulb 1 is larger than the current passing through bulb 2 when connected in parallel.

Since the brightness of a bulb is proportional to the current passing through it, bulb 1 will consume more power and, therefore, be brighter than bulb 2 when connected in parallel.

In summary, when connected in series, bulb 2 is brighter, but when connected in parallel, bulb 1 is brighter.