In a series circuit with 6 Ω and 3 Ω resistors connected to a 9 V source, what is the current and the voltage drop across each resistor?

In a series circuit with 6 Ω and 3 Ω resistors connected to a 9 V source, what is the current and the voltage drop across each resistor?

• A. I = 0.5 A; V6 = 3 V; V3 = 6 V
• B. I = 1 A; V6 = 6 V; V3 = 3 V
• C. I = 1 A; V6 = 9 V; V3 = 0 V
• D. I = 2 A; V6 = 12 V; V3 = 6 V

Answer: (B)

In a series circuit, the same current flows through every component, and the total supplied voltage is the sum of the voltage drops across all components. The total resistance is 6 Ω plus 3 Ω, which is 9 Ω. With a 9 V source, the current is I = V / R_total = 9 V / 9 Ω = 1 A. The voltage drop across each resistor is V = I × R. For the 6 Ω resistor: V = 1 A × 6 Ω = 6 V. For the 3 Ω resistor: V = 1 A × 3 Ω = 3 V. These drops add up to the source voltage: 6 V + 3 V = 9 V, confirming consistency. So the current is 1 A, and the voltage drops are 6 V across the 6 Ω resistor and 3 V across the 3 Ω resistor.

In a series circuit, the same current flows through every component, and the total supplied voltage is the sum of the voltage drops across all components. The total resistance is 6 Ω plus 3 Ω, which is 9 Ω. With a 9 V source, the current is I = V / R_total = 9 V / 9 Ω = 1 A.

The voltage drop across each resistor is V = I × R. For the 6 Ω resistor: V = 1 A × 6 Ω = 6 V. For the 3 Ω resistor: V = 1 A × 3 Ω = 3 V. These drops add up to the source voltage: 6 V + 3 V = 9 V, confirming consistency.

So the current is 1 A, and the voltage drops are 6 V across the 6 Ω resistor and 3 V across the 3 Ω resistor.