A 12 V source with two resistors in series (2 Ω and 8 Ω). Find the voltage across the 2 Ω resistor.

A 12 V source with two resistors in series (2 Ω and 8 Ω). Find the voltage across the 2 Ω resistor.

• A. 6.0 V
• B. 1.2 V
• C. 4.8 V
• D. 2.4 V

Answer: (D)

In a series circuit the same current flows through all components, and the voltage across each resistor is proportional to its resistance. The total resistance is 2 Ω + 8 Ω = 10 Ω, so the current from a 12 V source is 12 / 10 = 1.2 A. The voltage across the 2 Ω resistor is V = I × R = 1.2 A × 2 Ω = 2.4 V. You can also see this with the voltage-divider idea: V across 2 Ω = 12 V × (2/(2+8)) = 2.4 V. Therefore, the voltage across the 2 Ω resistor is 2.4 V.

In a series circuit the same current flows through all components, and the voltage across each resistor is proportional to its resistance. The total resistance is 2 Ω + 8 Ω = 10 Ω, so the current from a 12 V source is 12 / 10 = 1.2 A. The voltage across the 2 Ω resistor is V = I × R = 1.2 A × 2 Ω = 2.4 V. You can also see this with the voltage-divider idea: V across 2 Ω = 12 V × (2/(2+8)) = 2.4 V. Therefore, the voltage across the 2 Ω resistor is 2.4 V.