Step 1: Label currents using point rule | I_{2} = 3I_{1} - 5 -> I_{2} = -2A | |
Step 2: Write loop rule equations | 0 = -12 + I_{1} - 5.5(3I_{1}-5) | |
0 = - 1V - (I_{1}-I_{2})1 + 6V - I_{1}2 | 0 = -12 + 27.5 + I_{1} - 16.5I_{1} | |
0 = - I_{2}3 - 12V - I_{2}1.5 + (I_{1}-I_{2})1 | 0 = 15.5 - 15.5I_{1} -> I_{1} = 1A | |
Step 3: Solve | Step 4: Check point and loop rules | |
0 = 5 - 3 I_{1} + I_{2} | 0 = -1 -3 +6 -2 (left loop) | |
0 = -12 + I_{1} - 5.5I_{2} |