A 5.0 Ω resistor, a 9.0 Ω resistor, and a 12.0 Ω resistor are connected in parallel across a 24.0 V battery.

a) What is the equivalent resistance of the circuit? Answer in units of Ω

b) What is the current in the 5.0 Ω resistor?
Answer in units of A

E = emf = 24V.

R1 = 5 Ohms
R2 = 9 Ohms
R3 = 12 Ohms

a. 1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/5 + 1/9 + 1/12 = 0.3944444
Req = 2.535 Ohms

b. It = I1+I2+I3 = E/Req = 24/2.535 = 9.47A

I2 = 12/9 * I3 = 1.33I3

I1 = 12/5 * I3 = 2.4I3

I3 + 1.33I3 + 2.4I3 = 9.47
4.733I3 = 9.47
I3 = 2.0A

I1 = 2.4*I3 = 2.4 * 2.0 = 4.8A. = Current through the 5 ohm resistor.

a) Well, if you put three resistors in parallel, it's like trying to untangle a knot in your headphones – it can be quite a challenge! Anyway, to find the equivalent resistance of the circuit, you can use the formula:

1/Req = 1/R1 + 1/R2 + 1/R3

Plugging in the given values, we get:

1/Req = 1/5.0 + 1/9.0 + 1/12.0

Now, let me grab my calculator, don't tell anyone - the equivalent resistance of the circuit is approximately 2.071 Ω.

b) Ah, the current flowing through the 5.0 Ω resistor. Well, according to Ohm's Law, the current (I) can be calculated using the formula:

I = V / R

Plugging in the voltage (V = 24.0 V) and the resistance (R = 5.0 Ω), we find that the current through the 5.0 Ω resistor is approximately 4.8 A.

So, there you have it! The equivalent resistance is around 2.071 Ω, and the current flowing through the 5.0 Ω resistor is approximately 4.8 A. I hope my clownish explanation brought a smile to your face!

To find the equivalent resistance of resistors connected in parallel, we can use the formula:

1/Req = 1/R1 + 1/R2 + 1/R3 + ...

a) Given resistors R1 = 5.0 Ω, R2 = 9.0 Ω, and R3 = 12.0 Ω, we can calculate the equivalent resistance as follows:

1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/5.0 + 1/9.0 + 1/12.0

Taking the reciprocal of both sides:

Req = 1 / (1/5.0 + 1/9.0 + 1/12.0)
Req = 1 / (0.2 + 0.111 + 0.083)

Calculating the sum in the denominator:

Req = 1 / 0.394
Req ≈ 2.538 Ω

Therefore, the equivalent resistance of the circuit is approximately 2.538 Ω.

b) The current in the 5.0 Ω resistor can be found using Ohm's Law:

I = V/R

Where I is the current, V is the voltage across the resistor, and R is the resistance of the resistor.

Given V = 24.0 V and R = 5.0 Ω:

I = 24.0 V / 5.0 Ω
I ≈ 4.8 A

Therefore, the current in the 5.0 Ω resistor is approximately 4.8 A.

To find the equivalent resistance of a set of resistors connected in parallel, you can use the formula:

1/Req = 1/R1 + 1/R2 + 1/R3 + ...

where Req is the equivalent resistance and R1, R2, R3, etc. are the resistances of the individual resistors.

Let's use this formula to solve the problem:

a) First, let's calculate the equivalent resistance of the circuit:
1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/5.0 Ω + 1/9.0 Ω + 1/12.0 Ω

To simplify this, we can get the common denominators:
1/Req = (12/12) / 5.0 Ω + (20/20) / 9.0 Ω + (15/15) / 12.0 Ω
1/Req = 12/60 + 20/180 + 15/180
1/Req = (36 + 20 + 15) / 180
1/Req = 71 / 180

Now, let's take the reciprocal of both sides to find Req:
Req = 180 / 71
Req ≈ 2.54 Ω

So, the equivalent resistance of the circuit is approximately 2.54 Ω.

b) To find the current in the 5.0 Ω resistor, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R).

In this case, the voltage across all resistors is the same, which is the voltage of the battery, 24.0 V. So, we can use the formula:

I = V / R

where I is the current, V is the voltage, and R is the resistance.

Let's calculate the current:

I = 24.0 V / 5.0 Ω
I = 4.8 A

So, the current flowing through the 5.0 Ω resistor is 4.8 A.