3Ω 17 V A3 V3

17 V 5Ω 8Ω 9Ω 7Ω A2 V3 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω

17 V 5Ω 8Ω 9Ω 7Ω A2 V3 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω

17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω

17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω

17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 A3 23Ω 4Ω

17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 A3 23Ω 4Ω

17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 V1 A1 V2 A3 6.9696Ω 4Ω

17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 V1 A1 V2 A3 6.9696Ω 4Ω

17 V 5Ω 7Ω A2 3Ω 6Ω V1 A1 V2 A3 23.9696Ω 4Ω

17 V 5Ω 7Ω A2 3Ω 6Ω V1 A1 V2 A3 23.9696Ω 4Ω

17 V 5Ω 5.4178Ω A2 3Ω 6Ω V1 A1 V2 A3 4Ω

17 V 5Ω 5.4178Ω A2 3Ω 6Ω V1 A1 V2 A3 4Ω

17 V 5.4178Ω A2 12Ω 6Ω V2 A3

17 V 5.4178Ω A2 12Ω 6Ω V2 A3

Finally we have this last series circuit: 5.4178Ω 4Ω V2 A3

17 V Solving: Rtot = 5.4178 + 4 = 9.4178 5.4178Ω 4Ω V2 A3

(This is the reading on A3) 17 V Solving: Rtot = 5.4178 + 4 = 9.4178 I = V/R = 17/9.4178 = 1.8051A (This is the reading on A3) 5.4178Ω 4Ω V2 A3

(This is the reading on A3) Picking off voltages: Solving: Rtot = 5.4178 + 4 = 9.4178 I = V/R = 17/9.4178 = 1.8051A (This is the reading on A3) Picking off voltages: V2 = 1.8051*5.4178 = 9.7796 V V4Ω = 1.8051*4 = 7.2204 V 5.4178Ω 4Ω V2 A3

Solving the subcircuit on the left: 4Ω

7.2204 V Which is really A2 12Ω 6Ω

The current through the 6Ω is just V/R = 7.2204/6 = 1.2034A Which is the reading on A2. A2 12Ω 6Ω

7.2204 V This leaves: 12Ω

7.2204 V 5Ω Which is really: 3Ω V1 A1 4Ω

7.2204 V 5Ω Solving (series) Rtot = 5 + 3 + 4 = 12Ω 3Ω V1 A1 4Ω

Which is the reading on A1 7.2204 V 5Ω Solving (series) Rtot = 5 + 3 + 4 = 12Ω I = 7.2204/12 = .6017A Which is the reading on A1 3Ω V1 A1 4Ω

Which is the reading on A1 7.2204 V 5Ω Solving (series) Rtot = 5 + 3 + 4 = 12Ω I = 7.2204/12 = .6017A Which is the reading on A1 V1 = IR = .6017*3 = 1.8051V 3Ω V1 A1 4Ω

Now let’s look at the right subcircuit: 17 V Now let’s look at the right subcircuit: 5.4178Ω 4Ω V2 A3

Now let’s look at the right subcircuit: (the 5.4178Ω resistor) 17 V Now let’s look at the right subcircuit: (the 5.4178Ω resistor) V2 = 1.8051*5.4178 = 9.7796 V 5.4178Ω 4Ω V2 A3

9.7796 V So it looks like this: 5.4178Ω

9.7796 V Which is really 8Ω 9Ω 7Ω V3 A4 10Ω 12Ω 11Ω

9.7796 V But let’s go back to 7Ω 17Ω A4 6.9696Ω

9.7796 V Which is really 8Ω 7Ω 9Ω V3 A4 6.9696Ω

Solving the right side which is a series circuit (ignore the 7Ω) 8Ω 7Ω 9Ω V3 A4 6.9696Ω

Solving the right side which is a series circuit (ignore the 7Ω) Rtot= 8+9+6.9696 = 23.9696Ω 8Ω 7Ω 9Ω V3 A4 6.9696Ω

Solving the right side which is a series circuit (ignore the 7Ω) Rtot= 8+9+6.9696 = 23.9696Ω I = V/R = 9.7796/23.9696 = .408A which is the reading on A4 8Ω 7Ω 9Ω V3 A4 6.9696Ω

Solving the right side which is a series circuit (ignore the 7Ω) Rtot= 8+9+6.9696 = 23.9696Ω I = V/R = 9.7796/23.9696 = .408A which is the reading on A4 And finally, V3 = IR = .408*9 = 3.6720V 8Ω 7Ω 9Ω V3 A4 6.9696Ω

9.7796 V Solving the right side which is a series circuit (ignore the 7Ω) Rtot= 8+9+6.9696 = 23.9696Ω I = V/R = 9.7796/23.9696 = .408A which is the reading on A4 And finally, V3 = IR = .408*9 = 3.6720V Ta Daa! 8Ω 7Ω 9Ω V3 A4 6.9696Ω