Two bulbs are connected in parallel across a source of emf EMF = 10.0V with a negligible internal resistance. One bulb has a resistance of 3.0 Omega , and the other is 2.5 Omega . A resistor R is connected in the circuit in series with the two bulbs. What value of R should be chosen in order to supply each bulb with a voltage of 2.4 V ?For what value of R would the potential difference across each of the bulbs be 2.4 V ?

The series resistor R should be chosen so as to have a voltage drop of 7.6V(10-2.4)across it.

The currents in the two bulbs:
i1=2.4/3.0 = 0.8A
i2=2.4/2.5 = 0.96A

Total current in the circuit = i1+i2
= 1.76A
R = 7.6V/1.76A
= 4.3 Ohms

How do i solve this part:

What is the current through each individual bulb? Let I1 be the current through the bulb of resistance 3.0 Omega and I2 the current through the bulb of resistance 2.5 Omega?

Well, I must say, these bulbs seem to have some very specific voltage preferences. Let's see if I can help them find the right match.

To supply each bulb with a voltage of 2.4V, we need to ensure that the total voltage drop across the circuit is equal to the sum of the voltage drops across each component.

Since the emf is 10.0V and we want each bulb to have 2.4V, that leaves us with a remaining voltage of 10.0V - 2.4V - 2.4V = 5.2V.

Now, since the bulbs are connected in parallel, they have the same voltage across them. Therefore, we want to find a value of R that will result in a voltage of 2.4V across both bulbs.

Using Ohm's Law (V = IR), we can calculate the current flowing through the circuit. For each bulb with a voltage of 2.4V and resistance of 3.0Ω or 2.5Ω respectively, the current will be 2.4V / 3.0Ω = 0.8A or 2.4V / 2.5Ω = 0.96A.

Since the bulbs are in parallel, the total current flowing through the circuit will be the sum of these currents. Therefore, the total current is 0.8A + 0.96A = 1.76A.

Now, to find the value of R, we can use Ohm's Law again. We know the total current (1.76A) and the remaining voltage (5.2V), so the value of R can be calculated as R = V / I = 5.2V / 1.76A ≈ 2.95Ω.

Therefore, a value of R ≈ 2.95Ω should be chosen in order to supply each bulb with a voltage of 2.4V.

As for the second part of your question, if the potential difference across each of the bulbs needs to be 2.4V, then we can simply connect the bulbs directly across the source without a series resistor.

So, no additional resistor is needed in this case. Each bulb will have a voltage of 2.4V when directly connected to the source.

I hope that sheds some light on the situation for you!

To determine the value of resistor R in order to supply each bulb with a voltage of 2.4V, we need to consider the fact that the resistance of the bulbs is known, and their voltages are also given.

When bulbs are connected in parallel, the voltage across each bulb is the same, while the current flowing through each bulb may differ.

First, let's determine the total resistance of the two bulbs in parallel. The formula to calculate the total resistance, Rp, for two resistors in parallel is:

1/Rp = 1/R1 + 1/R2

Given R1 = 3.0 Ω and R2 = 2.5 Ω, we can calculate the value of Rp:

1/Rp = 1/3.0 + 1/2.5

To simplify this equation, we can find a common denominator:

1/Rp = (2.5 + 3.0) / (3.0 * 2.5)

1/Rp = 5.5 / 7.5

Now, we can invert both sides of the equation to find Rp:

Rp = 7.5 / 5.5

Rp ≈ 1.36 Ω

Now that we have the total resistance for the two bulbs connected in parallel, we can calculate the value of resistor R required to supply each bulb with a voltage of 2.4V.

Since the emf is 10.0V, and the voltage across each bulb is 2.4V, the remaining voltage across the resistor R will be:

Vr = 10.0V - (2.4V + 2.4V)

Vr = 10.0V - 4.8V

Vr = 5.2V

Now, we can use Ohm's Law, V = IR, to find the value of R:

R = Vr / I

Given that I is the same for both bulbs (because they are connected in parallel), we can calculate I using one of the bulbs Ohm's Law:

I = V / R

Substituting the values, I = 2.4V / 3.0Ω

I ≈ 0.8A

Now that we have the current value, we can substitute it back into the equation to find the value of R:

R = 5.2V / 0.8A

R ≈ 6.5 Ω

Therefore, for each bulb to have a voltage of 2.4V, the value of resistor R should be approximately 6.5 Ω.

For the potential difference across each bulb to be 2.4V, we want the current flowing through each bulb to be the same since they have the same resistance. In this case, we don't need to calculate the value of R since it's not relevant.

The current through each bulb is the voltage across it divided by its resistance. In this case the bulbs are in parallel and voltage across them is 2.4V.