Two bulbs are connected in parallel across a source of emf E = 11.0 V with negligible internal resistance. One bulb has a resistance of 3.0 Ω, and the other is 2.5 Ω. 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?

Question

Two bulbs are connected in parallel across a source of emf E = 11.0 V with negligible internal resistance. One bulb has a resistance of 3.0 Ω, and the other is 2.5 Ω. 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?

Answer 1

I am not up on my physics, last time I studied physics was 63 years ago.

Answer 2

The two bulbs in parallel produce an equivalent resistance X such that

1/X = 1/3 + 1/2.5
So, X = 1.36 Ω
Now you just have an equivalent series circuit with two resistors, R and 1.36Ω. So now you can probably figure out how to find R so that you get a 2.4V drop across 1.36Ω, yeah?

Answer 3

To find the value of R that will supply each bulb with a voltage of 2.4 V, we can use Ohm's Law:

V = IR

In this case, the voltage across each bulb is given as 2.4 V, and the resistance of the bulbs are 3.0 Ω and 2.5 Ω. We need to find the current flowing through each bulb.

For the first bulb:
2.4 V = (I1)(3.0 Ω)

Solving for I1:

I1 = 2.4 V / 3.0 Ω = 0.8 A

Similarly, for the second bulb:
2.4 V = (I2)(2.5 Ω)

Solving for I2:

I2 = 2.4 V / 2.5 Ω = 0.96 A

Since the bulbs are connected in parallel, the total current, Itotal, is the sum of I1 and I2:

Itotal = I1 + I2 = 0.8 A + 0.96 A = 1.76 A

Now, we can find the value of R needed to supply this total current.

Using Ohm's Law again:

V = IR

11.0 V = (1.76 A)(R)

Solving for R:

R = 11.0 V / 1.76 A ≈ 6.25 Ω

Therefore, in order to supply each bulb with a voltage of 2.4 V, a resistor R with a value of 6.25 Ω should be chosen.

To find the value of R that would result in a potential difference of 2.4 V across each bulb, we can approach it differently.

In this case, the potential difference across each of the bulbs is given as 2.4 V. Since they are connected in parallel, the potential difference across each bulb is the same.

Hence, no additional resistance is needed (R = 0), as the potential difference across each of the bulbs will remain 2.4 V.