Question

Question with circuits

Answer 1

My question is, when I apply KVL to loop 1, shouldn't it be 6V - 4(2) - 6(i) = 0. But if that was so, my current across 6 ohm resistor would be 1/3. But it's actually 2, because solving loop 2 gives the current across 6 ohm resistor as 2 amps. Why doesn't the first loop add up to 0?

Answer 2

Have you tried KVL after converting the current source to an equivalent voltage source?