In a circuit with a parallel combination of 8 Ω and 12 Ω across a 30 V source, what is the total current from the source?

In a circuit with a parallel combination of 8 Ω and 12 Ω across a 30 V source, what is the total current from the source?

• A. 3.75 A
• B. 6.25 A
• C. 2.50 A
• D. 4.00 A

Answer: (B)

When resistors are in parallel, the same voltage from the source appears across each branch, and the total current drawn from the source is the sum of the currents in each branch (which is equivalent to V divided by the total parallel resistance). Compute the parallel resistance: 1/Rt = 1/8 + 1/12 = 5/24, so Rt = 24/5 = 4.8 Ω. The total current from the 30 V source is It = 30 / 4.8 = 6.25 A. You can verify by adding branch currents: I through 8 Ω is 30/8 = 3.75 A, through 12 Ω is 30/12 = 2.5 A, sum = 6.25 A. Therefore, the total current is 6.25 A.

When resistors are in parallel, the same voltage from the source appears across each branch, and the total current drawn from the source is the sum of the currents in each branch (which is equivalent to V divided by the total parallel resistance).

Compute the parallel resistance: 1/Rt = 1/8 + 1/12 = 5/24, so Rt = 24/5 = 4.8 Ω. The total current from the 30 V source is It = 30 / 4.8 = 6.25 A.

You can verify by adding branch currents: I through 8 Ω is 30/8 = 3.75 A, through 12 Ω is 30/12 = 2.5 A, sum = 6.25 A.

Therefore, the total current is 6.25 A.