In a parallel circuit with a 6 Ω and a 3 Ω resistor across a 12 V source, what is the total current?

In a parallel circuit with a 6 Ω and a 3 Ω resistor across a 12 V source, what is the total current?

• A. 4 A
• B. 8 A
• C. 12 A
• D. 6 A

Answer: (D)

In a parallel circuit, the full source voltage appears across every branch, and the total current drawn from the source is the sum of the currents through each branch. So each resistor has 12 V across it, giving currents of 12 V divided by each resistance: 12/6 = 2 A through the 6 Ω branch and 12/3 = 4 A through the 3 Ω branch. Adding these branch currents gives 2 A + 4 A = 6 A total from the source. You can also see this by finding the equivalent resistance of the parallel combination: 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R_eq = 2 Ω. The total current is then I = V / R_eq = 12 / 2 = 6 A. The total current in a parallel circuit is the sum of the branch currents, not the current through any single resistor.

In a parallel circuit, the full source voltage appears across every branch, and the total current drawn from the source is the sum of the currents through each branch. So each resistor has 12 V across it, giving currents of 12 V divided by each resistance: 12/6 = 2 A through the 6 Ω branch and 12/3 = 4 A through the 3 Ω branch. Adding these branch currents gives 2 A + 4 A = 6 A total from the source.

You can also see this by finding the equivalent resistance of the parallel combination: 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R_eq = 2 Ω. The total current is then I = V / R_eq = 12 / 2 = 6 A. The total current in a parallel circuit is the sum of the branch currents, not the current through any single resistor.