In a simple series circuit with a 12 V source and three resistors in series (R1 = 4 Ω, R2 = 6 Ω, R3 = 10 Ω), what is the total resistance?

Prepare for the NCCER Introduction to Electrical Circuits exam. Utilize flashcards and multiple choice questions with hints and explanations to excel on your test.

Multiple Choice

In a simple series circuit with a 12 V source and three resistors in series (R1 = 4 Ω, R2 = 6 Ω, R3 = 10 Ω), what is the total resistance?

Explanation:
In a simple series circuit, the total resistance is found by adding all the resistors along the path because the current flows through each one in sequence. So you simply sum R1, R2, and R3: 4 Ω + 6 Ω + 10 Ω = 20 Ω. That 20 Ω is the resistance the voltage source “sees.” The given 12 V source doesn’t change the resistance; it would only determine the current with I = V / R, which would be 12 V / 20 Ω = 0.6 A. The correct total resistance is 20 Ω, matching the sum of the individual resistances in series.

In a simple series circuit, the total resistance is found by adding all the resistors along the path because the current flows through each one in sequence. So you simply sum R1, R2, and R3: 4 Ω + 6 Ω + 10 Ω = 20 Ω. That 20 Ω is the resistance the voltage source “sees.” The given 12 V source doesn’t change the resistance; it would only determine the current with I = V / R, which would be 12 V / 20 Ω = 0.6 A. The correct total resistance is 20 Ω, matching the sum of the individual resistances in series.

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