Two resistors 4 Ω and 6 Ω are in series across a 12 V source. What is the current, and the voltage across each resistor?

Two resistors 4 Ω and 6 Ω are in series across a 12 V source. What is the current, and the voltage across each resistor?

• A. I = 1.2 A; V4Ω = 4.8 V; V6Ω = 7.2 V
• B. I = 0.8 A; V4Ω = 3.2 V; V6Ω = 6.8 V
• C. I = 1.5 A; V4Ω = 6.0 V; V6Ω = 6.0 V
• D. I = 2.0 A; V4Ω = 8.0 V; V6Ω = 4.0 V

Answer: (A)

In a series circuit, the same current flows through every component and the voltages add up to the supply. The resistors in this arrangement add to a total resistance: 4 Ω + 6 Ω = 10 Ω. With a 12 V source, the current is I = V/R = 12/10 = 1.2 A, and this current runs through both resistors. Now find the voltage across each resistor using V = I R. For the 4 Ω resistor: V = 1.2 A × 4 Ω = 4.8 V. For the 6 Ω resistor: V = 1.2 A × 6 Ω = 7.2 V. The sum is 12 V, as expected. The larger resistor drops more voltage because the same current passes through both; voltage drop is proportional to resistance. Other options would require either a different total resistance or a different current that doesn’t satisfy V = IR with the given resistances, or would imply equal voltage drops for unequal resistances, which isn’t consistent with a single current in series.

In a series circuit, the same current flows through every component and the voltages add up to the supply. The resistors in this arrangement add to a total resistance: 4 Ω + 6 Ω = 10 Ω. With a 12 V source, the current is I = V/R = 12/10 = 1.2 A, and this current runs through both resistors.

Now find the voltage across each resistor using V = I R. For the 4 Ω resistor: V = 1.2 A × 4 Ω = 4.8 V. For the 6 Ω resistor: V = 1.2 A × 6 Ω = 7.2 V. The sum is 12 V, as expected. The larger resistor drops more voltage because the same current passes through both; voltage drop is proportional to resistance.

Other options would require either a different total resistance or a different current that doesn’t satisfy V = IR with the given resistances, or would imply equal voltage drops for unequal resistances, which isn’t consistent with a single current in series.