Three resistors—2 Ω, 6 Ω, and 3 Ω—are connected in parallel across a 12 V source. Which set of branch currents and total current is correct?

• A. I_2Ω = 6 A; I_6Ω = 1 A; I_3Ω = 5 A; Itotal = 12 A
• B. I_2Ω = 6 A; I_6Ω = 2 A; I_3Ω = 4 A; Itotal = 12 A
• C. I_2Ω = 4 A; I_6Ω = 2 A; I_3Ω = 6 A; Itotal = 12 A
• D. I_2Ω = 6 A; I_6Ω = 2 A; I_3Ω = 3 A; Itotal = 11 A

Answer: (B)

In a parallel circuit, each resistor has the same voltage across it, so current through each branch is found using I = V/R with V = 12 V.

• For 2 Ω: I = 12/2 = 6 A

• For 6 Ω: I = 12/6 = 2 A

• For 3 Ω: I = 12/3 = 4 A

The total current is the sum of the branch currents: Itotal = 6 + 2 + 4 = 12 A. This also matches the total resistance in parallel (1/Rtotal = 1/2 + 1/6 + 1/3 = 1, so Rtotal = 1 Ω, and Itotal = 12/1 = 12 A).

So the correct set has currents 6 A, 2 A, and 4 A with a total of 12 A. The other options involve currents that don’t satisfy I = V/R for the given resistors or give a total that doesn’t equal the sum of the branch currents.