Answer:
Step-by-step explanation:
Using Kirchhoff's laws, we can solve for the current i:
At the node where the 2Ω and 4Ω resistors meet, the current is split into two branches, i and i1. Applying Kirchhoff's current law (KCL), we have:
i + i1 = 12/2 = 6 A
At the loop with the 2Ω, 4Ω, and 5Ω resistors, applying Kirchhoff's voltage law (KVL), we have:
-20 + 2i + 4i1 + 5i1 = 0
-20 + 6i1 + 2i = 0
6i1 + 2i = 20
3i1 + i = 10
We can solve this system of equations by substitution, which gives:
i = 2 A
Therefore, the current through the 2Ω resistor is 2 A. The answer is (A) 2 A.