What is the total resistance in a parallel circuit with three resistors of 2Ω, 3

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Question: What is the total resistance in a parallel circuit with three resistors of 2Ω, 3Ω, and 6Ω?

Options:

Correct Answer: 0.75Ω

Solution:

1/R_eq = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6 = 1. R_eq = 0.75Ω.

What is the total resistance in a parallel circuit with three resistors of 2Ω, 3Ω, and 6Ω?

What is the total resistance in a parallel circuit with three resistors of 2Ω, 3Ω, and 6Ω?

Step 1: Identify the resistors in the parallel circuit. We have three resistors: R1 = 2Ω, R2 = 3Ω, and R3 = 6Ω.
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 = 1/2 + 1/3 + 1/6.
Step 4: Find a common denominator to add the fractions. The common denominator for 2, 3, and 6 is 6.
Step 5: Convert each fraction: 1/2 = 3/6, 1/3 = 2/6, and 1/6 = 1/6.
Step 6: Add the fractions: 3/6 + 2/6 + 1/6 = 6/6.
Step 7: Now we have 1/R_eq = 6/6, which simplifies to 1/R_eq = 1.
Step 8: To find R_eq, take the reciprocal of both sides: R_eq = 1/1 = 1Ω.

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