3Ω 17 V A3 V3
9Ω A2 V4 6Ω A1 10Ω 12Ω V2 8Ω 37 V A3 V1 V3 7Ω 11Ω A4 13Ω
9Ω A2 V4 A1 6Ω 10Ω 12Ω V2 8Ω 37 V A3 V1 V3 7Ω 11Ω A4 13Ω
9Ω A2 V4 A1 6Ω 5.4545Ω 8Ω 37 V V1 V3 7Ω 11Ω A4 13Ω
9Ω A2 V4 A1 6Ω 5.4545Ω 8Ω 37 V V1 V3 7Ω 11Ω A4 13Ω
9Ω A2 V4 A1 6Ω 5.4545Ω 8Ω 37 V V1 V3 7Ω 5.9583Ω
9Ω A2 V4 A1 6Ω 5.4545Ω 8Ω 37 V V1 V3 7Ω 5.9583Ω
A2 A1 6Ω 20.4129Ω 8Ω 37 V V3 7Ω
A2 A1 6Ω 20.4129Ω 8Ω 37 V V3 7Ω
Finally we have a simple series circuit: 6Ω 5.7475Ω 37 V V3 7Ω
We will use this voltage to solve the rest of the subcircuit… Solving: Rtot = 6 + 7 + 5.7475 = 18.7475Ω I = V/R = 37/18.7475 = 1.9736 A So A2 reads 1.9736 A and V3, the voltage across the subcircuit = IR = 1.9736*5.7475 = 11.3432 V We will use this voltage to solve the rest of the subcircuit… A2 6Ω 5.7475Ω 37 V V3 7Ω
Solving the subcircuit: The current through the 8Ω is just V/R = 11.3432/8 = 1.4179A which is the reading on A1 A1 20.4129Ω 8Ω 11.3432 V
Now we have: 20.4129Ω 11.3432 V
Which is really 9Ω V4 10Ω 12Ω V2 11.3432 V A3 V1 11Ω A4 13Ω
But let’s go back to: 9Ω V4 V2 5.4545Ω 11.3432 V V1 5.9583Ω
Using IR to pick off the voltages: Solving: Rtot = 9 + 5.4545 + 5.9583 = 20.4129Ω And I = V/R = 11.3432/20.4129 = .5557 A Using IR to pick off the voltages: V4 = .5557*9 = 5.0012 V V2 = .5557*5.4545 = 3.0310 V V1 = .5557*5.9583 = 3.3110 V These are also the voltages across the remaining subcircuits. 9Ω V4 V2 5.4545Ω 11.3432 V V1 5.9583Ω
The first sub circuit: 3.0310 V 5.4545Ω
Which is really: 3.0310 V 10Ω 12Ω A3
The current through the 10Ω is just V/R = 3.0310/10 = .3031 A 12Ω A3
The last sub circuit: 3.3110 V 5.9583Ω
Which is really: 3.3110 V 11Ω 13Ω A4
The current through the 13Ω is just V/R = 3.3110/13 = .2547 A 11Ω 13Ω A4
Ta Daaa! The current through the 13Ω is just V/R = 3.3110/13 = .2547 A 11Ω 13Ω A4