A 2.00 resistor, and a 12.0 resistor are connected in parallel across a 20.0-V battery. What is the current flowing through the 2.00 resistor?

A)1.67 A
B)1.40 A
C)11.6 A
D)10.0 A

This is all I understand. Can you explain step by step and show how to get the answer?

To find current, C = vsource/R
C = 20.0/2.00 they're only asking for the current in the 2.00 resistor, not the whole thing. I REALLY don't understand where ohm's are comming into the picture, maybe they have something to do with it but the question still asks "what is the current for the 2.00 resistor" and either way you do the math, the only answer you get is 10.0

Arrrgggg. Where did I read a ten ohm resistor was added in series. You are right. Sorry.

Well, my electrifying friend, let me help you solve this puzzling problem with a dash of humor. Here's a step-by-step breakdown of how to find the current flowing through the 2.00-ohm resistor:

1. Start by using Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across it divided by its resistance (R). In this case, we're interested in finding the current through the 2.00-ohm resistor.

2. Given that the voltage (V) across the parallel combination of resistors is 20.0 V, we can use this value to calculate the total current (I_total) flowing through the circuit. This current will be the same for both resistors.

3. To find the total current (I_total), divide the voltage (V) by the equivalent resistance (R_eq) of the combination of 2.00 ohms and 12.0 ohms. In parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances. Mathematically, 1/R_eq = 1/R_1 + 1/R_2.

4. Plugging in the values, we get 1/R_eq = 1/2.00 + 1/12.0, which simplifies to 1/R_eq = 0.5 + 0.0833.

5. Adding the reciprocals, we find that 1/R_eq = 0.5833. To get R_eq, we take the reciprocal of both sides: R_eq = 1/0.5833.

6. Calculating this out, we find that R_eq is approximately 1.714 ohms (rounded to three decimal places).

7. Now that we have the equivalent resistance (R_eq) and the voltage (V), we can determine the total current (I_total) flowing through the circuit using Ohm's Law: I_total = V/R_eq. Plugging in the values, we get I_total = 20.0/1.714, which is approximately 11.657 amperes (rounded to three decimal places).

8. However, remember that we're interested in finding the current through the 2.00-ohm resistor alone, not the total current. Since the two resistors are in parallel, they have the same voltage (V) across them, but different currents (I_1 and I_2) flowing through them.

9. To find the current (I_1) through the 2.00-ohm resistor, we can use the fact that the current through each resistor in a parallel circuit is inversely proportional to its resistance. Mathematically, I_1/I_2 = R_2/R_1.

10. Plugging in the values, we have I_1/I_2 = 12.0/2.00, which simplifies to I_1/I_2 = 6.

11. Solving for I_1, we find that it is six times smaller than I_2. In other words, I_1 = I_total/6.

12. Substituting the value of I_total we found earlier (11.657 A), we get I_1 = 11.657/6, which is approximately 1.943 amperes (rounded to three decimal places).

So, my electrified friend, the current flowing through the 2.00-ohm resistor is approximately 1.94 amperes. I hope this explanation shocked you with its clarity!

To find the current flowing through the 2.00 resistor, you can use Ohm's Law and the concept of parallel resistors.

1. Start by calculating the total resistance (Rt) of the parallel circuit:

1/Rt = 1/R1 + 1/R2

Where R1 and R2 are the resistances of the two resistors.

Plugging in the values:
1/Rt = 1/2.00 + 1/12.0

Calculating the reciprocals:
1/Rt = 0.5 + 0.0833

Adding the fractions:
1/Rt = 0.5833

Taking the reciprocal:
Rt = 1/0.5833
≈ 1.714 Ω

2. Now that you have the total resistance, you can use Ohm's Law to find the current (I) flowing through the circuit:

I = V/Rt

Where V is the voltage across the circuit.

Plugging in the values:
I = 20.0 V / 1.714 Ω
≈ 11.65 A

So, the current flowing through the 2.00 resistor is approximately 11.65 A. The closest option from the given choices is C) 11.6 A.

To find the current flowing through the 2.00 resistor, you can use Ohm's Law, which states that the current (I) in a circuit is equal to the voltage (V) divided by the resistance (R).

In this case, the voltage across the resistors is 20.0 V, and you are trying to find the current through the 2.00 resistor.

The first step is to find the equivalent resistance of the parallel combination of the two resistors. When resistors are connected in parallel, the reciprocal of the equivalent resistance (1/Req) is equal to the sum of the reciprocals of the individual resistances (1/R1 + 1/R2).

For the given resistors, the equivalent resistance (Req) is:

1/Req = 1/2.00 + 1/12.0
1/Req = (12 + 2)/24
1/Req = 14/24
Req = 24/14
Req ≈ 1.714

Now, you can use Ohm's Law to find the current flowing through the parallel combination of resistors:

C = V / Req
C = 20.0 / 1.714
C ≈ 11.66 A

The current flowing through the parallel combination of resistors is approximately 11.66 A. However, we need to find the current through the 2.00 resistor specifically.

To find the current through the 2.00 resistor, we can use the concept that when resistors are connected in parallel, they share the same voltage. Therefore, the voltage across the 2.00 resistor is still 20.0 V.

Now, you can apply Ohm's Law again using the voltage across the 2.00 resistor (V = 20.0 V) and the resistance of the 2.00 resistor (R = 2.00 Ω):

C = V / R
C = 20.0 / 2.00
C = 10.0 A

Therefore, the current flowing through the 2.00 resistor is 10.0 A.

The correct answer is D) 10.0 A.