What is the total resistance in a parallel circuit with resistors of 3 Ω, 6 Ω, a

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Question: What is the total resistance in a parallel circuit with resistors of 3 Ω, 6 Ω, and 12 Ω?

Options:

Correct Answer: 2 Ω

Solution:

1/R_eq = 1/R1 + 1/R2 + 1/R3 = 1/3 + 1/6 + 1/12. R_eq = 2 Ω.

What is the total resistance in a parallel circuit with resistors of 3 Ω, 6 Ω, and 12 Ω?

What is the total resistance in a parallel circuit with resistors of 3 Ω, 6 Ω, and 12 Ω?

Step 1: Identify the resistors in the parallel circuit. We have three resistors: R1 = 3 Ω, R2 = 6 Ω, and R3 = 12 Ω.
Step 2: Write the formula for total resistance in a parallel circuit: 1/R_eq = 1/R1 + 1/R2 + 1/R3.
Step 3: Substitute the values of the resistors into the formula: 1/R_eq = 1/3 + 1/6 + 1/12.
Step 4: Find a common denominator for the fractions. The common denominator for 3, 6, and 12 is 12.
Step 5: Convert each fraction to have the common denominator: 1/3 = 4/12, 1/6 = 2/12, and 1/12 = 1/12.
Step 6: Add the fractions together: 4/12 + 2/12 + 1/12 = 7/12.
Step 7: Now we have 1/R_eq = 7/12. To find R_eq, take the reciprocal: R_eq = 12/7.
Step 8: Calculate 12/7, which is approximately 1.71 Ω. However, the exact value is 12/7 Ω.
Step 9: If you want to express it in a simpler form, you can round it to 2 Ω for practical purposes.

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