In a parallel circuit with 6 Ω and 3 Ω across a 9 V source, find total current and branch currents.

In a parallel circuit with 6 Ω and 3 Ω across a 9 V source, find total current and branch currents.

• A. I_total = 4 A; I6 = 1 A; I3 = 3 A
• B. I_total = 6 A; I6 = 3 A; I3 = 3 A
• C. I_total = 4.5 A; I6 = 1.5 A; I3 = 3 A
• D. I_total = 5 A; I6 = 0.5 A; I3 = 4.5 A

Answer: (C)

In a parallel circuit, the voltage across each branch is the same as the source, and the total current is the sum of the branch currents. With a 9 V source across both resistors, use Ohm’s law for each branch: the current through 6 Ω is 9 V / 6 Ω = 1.5 A, and the current through 3 Ω is 9 V / 3 Ω = 3 A. Adding these gives the total current: 1.5 A + 3 A = 4.5 A. You can also see this from the equivalent resistance of parallel resistors: R_eq = (6 × 3) / (6 + 3) = 18/9 = 2 Ω, so I_total = 9 V / 2 Ω = 4.5 A, which matches the sum of the branch currents. Therefore, I_total = 4.5 A, I6 = 1.5 A, I3 = 3 A.

In a parallel circuit, the voltage across each branch is the same as the source, and the total current is the sum of the branch currents. With a 9 V source across both resistors, use Ohm’s law for each branch: the current through 6 Ω is 9 V / 6 Ω = 1.5 A, and the current through 3 Ω is 9 V / 3 Ω = 3 A. Adding these gives the total current: 1.5 A + 3 A = 4.5 A. You can also see this from the equivalent resistance of parallel resistors: R_eq = (6 × 3) / (6 + 3) = 18/9 = 2 Ω, so I_total = 9 V / 2 Ω = 4.5 A, which matches the sum of the branch currents. Therefore, I_total = 4.5 A, I6 = 1.5 A, I3 = 3 A.