Three resistors 3 Ω, 2 Ω, and 5 Ω are in series connected to a 20 V source. What is the current, and what is the voltage drop across the 5 Ω resistor?

Three resistors 3 Ω, 2 Ω, and 5 Ω are in series connected to a 20 V source. What is the current, and what is the voltage drop across the 5 Ω resistor?

• A. I = 3 A; V5Ω = 6 V
• B. I = 1 A; V5Ω = 5 V
• C. I = 2 A; V5Ω = 10 V
• D. I = 4 A; V5Ω = 20 V

Answer: (C)

In a series circuit the same current flows through every resistor, and the total resistance is the sum of the individual resistances. Here the total resistance is 3 + 2 + 5 = 10 Ω. With a 20 V source, the current is I = V/R = 20/10 = 2 A. The voltage drop across the 5 Ω resistor is V5 = I × 5 Ω = 2 A × 5 Ω = 10 V. You can verify by the drops: 3 Ω gets 6 V, 2 Ω gets 4 V, and 5 Ω gets 10 V; these sum to 20 V.

In a series circuit the same current flows through every resistor, and the total resistance is the sum of the individual resistances. Here the total resistance is 3 + 2 + 5 = 10 Ω. With a 20 V source, the current is I = V/R = 20/10 = 2 A. The voltage drop across the 5 Ω resistor is V5 = I × 5 Ω = 2 A × 5 Ω = 10 V. You can verify by the drops: 3 Ω gets 6 V, 2 Ω gets 4 V, and 5 Ω gets 10 V; these sum to 20 V.