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?

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Get ready for the NCCER Introduction to Electrical Circuits exam. Study with multiple choice questions, each question comes with hints and explanations. Ace your exam with confidence!

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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