Kirchhoff's Voltage Law Application

Steps:

• Assign currents: I1 through left branch, I2 right branch, I3 bottom branch; I1 = I2 + I3 by KCL at junction.
• For left loop: 10V - 2Ω(I1) - 4Ω(I3) = 0 → 2I1 + 4I3 = 10.
• For right loop: 20V - 4Ω(I2) - 4Ω(I3) = 0 → 4I2 + 4I3 = 20.
• Solve: I3 = -5A, I2 = 1A, I1 = 3.5A (negative I3 indicates opposite direction).

Why C is correct:

• Values satisfy KVL equations, ensuring voltage drops balance emfs per Kirchhoff's second law.

Why the others are wrong:

• A: Negative I1 violates assumed direction and doesn't balance left loop equation.
• B: Positive low values fail to account for 20V battery driving higher currents.
• D: Overestimates I1, exceeding total emf contribution in combined loops.

Final answer: C