A circuit contains a 6 Ω resistor in series with a 3 Ω resistor across an 18 V supply. What is the total resistance and the current drawn?

A circuit contains a 6 Ω resistor in series with a 3 Ω resistor across an 18 V supply. What is the total resistance and the current drawn?

• A. I = 4 A; Total R = 9 Ω
• B. I = 2 A; Total R = 9 Ω
• C. I = 1 A; Total R = 9 Ω
• D. I = 2 A; Total R = 12 Ω

Answer: (B)

In a series circuit, resistors add to give the total resistance, and the same current flows through every element. So the total resistance is 6 Ω + 3 Ω = 9 Ω. With an 18 V supply, the current drawn is I = V / R = 18 V / 9 Ω = 2 A. A quick check: the 2 A current causes a voltage drop of 2 A × 6 Ω = 12 V across the 6 Ω resistor and 2 A × 3 Ω = 6 V across the 3 Ω resistor, which sums to 18 V. The correct result is total resistance 9 Ω and current 2 A.

In a series circuit, resistors add to give the total resistance, and the same current flows through every element. So the total resistance is 6 Ω + 3 Ω = 9 Ω. With an 18 V supply, the current drawn is I = V / R = 18 V / 9 Ω = 2 A. A quick check: the 2 A current causes a voltage drop of 2 A × 6 Ω = 12 V across the 6 Ω resistor and 2 A × 3 Ω = 6 V across the 3 Ω resistor, which sums to 18 V. The correct result is total resistance 9 Ω and current 2 A.