In a series circuit with a 12 V source and resistors of 4 Ω and 8 Ω, what is the total current?

In a series circuit with a 12 V source and resistors of 4 Ω and 8 Ω, what is the total current?

• A. 0.5 A
• B. 2 A
• C. 1 A
• D. 3 A

Answer: (C)

In a series circuit, the same current flows through every component and the resistances add up to give the total resistance. Here, the 4 Ω and 8 Ω resistors are in series, so their total resistance is 4 + 8 = 12 Ω. With a 12 V source, the current is I = V / R = 12 V / 12 Ω = 1 A. Therefore the total current in the circuit is 1 ampere. The other currents would require different total resistance, which isn’t the case here, since the total resistance is 12 Ω in this setup.

In a series circuit, the same current flows through every component and the resistances add up to give the total resistance. Here, the 4 Ω and 8 Ω resistors are in series, so their total resistance is 4 + 8 = 12 Ω. With a 12 V source, the current is I = V / R = 12 V / 12 Ω = 1 A. Therefore the total current in the circuit is 1 ampere. The other currents would require different total resistance, which isn’t the case here, since the total resistance is 12 Ω in this setup.