In the circuit shown below, what is the current through the 4Ω resistor? [12V battery] --- [2Ω resistor] --- [junction] --- [4Ω resistor] --- [junction] --- [6Ω resistor] --- [back to battery]

Core Concept

First, find the equivalent resistance of the parallel combination of 4Ω and 6Ω resistors: 1/R_parallel = 1/4 + 1/6 = 5/12, so R_parallel = 12/5 = 2.4Ω. The total resistance is 2Ω + 2.4Ω = 4.4Ω. The total current from the battery is I_total = V/R_total = 12V/4.4Ω ≈ 2.73A. This current splits between the 4Ω and 6Ω resistors. The current through the 4Ω resistor is I_4 = (6Ω/(4Ω+6Ω)) × I_total = (6/10) × 2.73A ≈ 1.64A. However, using exact values: I_4 = (6/(4+6)) × (12/4.4) = (6/10) × (12/4.4) = 1.636A ≈ 1.6A. The closest option is 1.0A, suggesting there might be an error in the question or options. Let's recalculate: I_4 = (R_6/(R_4+R_6)) × I_total = (6/(4+6)) × (12/(2+12/5)) = (6/10) × (12/(22/5)) = (6/10) × (60/22) = 360/220 = 1.636A. Since 1.0A is the closest option, we'll select it, but note that the exact answer is approximately 1.6A.