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3Ω 17 V A3 V3
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17 V 5Ω 8Ω 9Ω 7Ω A2 V3 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω
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17 V 5Ω 8Ω 9Ω 7Ω A2 V3 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω
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17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω
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17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω
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17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 A3 23Ω 4Ω
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17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 A3 23Ω 4Ω
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17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 V1 A1 V2 A3 6.9696Ω 4Ω
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17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 V1 A1 V2 A3 6.9696Ω 4Ω
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17 V 5Ω 7Ω A2 3Ω 6Ω V1 A1 V2 A3 Ω 4Ω
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17 V 5Ω 7Ω A2 3Ω 6Ω V1 A1 V2 A3 Ω 4Ω
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17 V 5Ω 5.4178Ω A2 3Ω 6Ω V1 A1 V2 A3 4Ω
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17 V 5Ω 5.4178Ω A2 3Ω 6Ω V1 A1 V2 A3 4Ω
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17 V 5.4178Ω A2 12Ω 6Ω V2 A3
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17 V 5.4178Ω A2 12Ω 6Ω V2 A3
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Finally we have this last series circuit:
5.4178Ω 4Ω V2 A3
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17 V Solving: Rtot = = 5.4178Ω 4Ω V2 A3
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(This is the reading on A3)
17 V Solving: Rtot = = I = V/R = 17/ = A (This is the reading on A3) 5.4178Ω 4Ω V2 A3
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(This is the reading on A3) Picking off voltages:
Solving: Rtot = = I = V/R = 17/ = A (This is the reading on A3) Picking off voltages: V2 = * = V V4Ω = *4 = V 5.4178Ω 4Ω V2 A3
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Solving the subcircuit on the left:
4Ω
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V Which is really A2 12Ω 6Ω
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The current through the 6Ω is just V/R = 7.2204/6 = 1.2034A
Which is the reading on A2. A2 12Ω 6Ω
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V This leaves: 12Ω
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V 5Ω Which is really: 3Ω V1 A1 4Ω
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V 5Ω Solving (series) Rtot = = 12Ω 3Ω V1 A1 4Ω
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Which is the reading on A1
V 5Ω Solving (series) Rtot = = 12Ω I = /12 = .6017A Which is the reading on A1 3Ω V1 A1 4Ω
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Which is the reading on A1
V 5Ω Solving (series) Rtot = = 12Ω I = /12 = .6017A Which is the reading on A1 V1 = IR = .6017*3 = V 3Ω V1 A1 4Ω
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Now let’s look at the right subcircuit:
17 V Now let’s look at the right subcircuit: 5.4178Ω 4Ω V2 A3
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Now let’s look at the right subcircuit: (the 5.4178Ω resistor)
17 V Now let’s look at the right subcircuit: (the Ω resistor) V2 = * = V 5.4178Ω 4Ω V2 A3
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V So it looks like this: 5.4178Ω
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V Which is really 8Ω 9Ω 7Ω V3 A4 10Ω 12Ω 11Ω
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V But let’s go back to 7Ω 17Ω A4 6.9696Ω
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V Which is really 8Ω 7Ω 9Ω V3 A4 6.9696Ω
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Solving the right side which is a series circuit (ignore the 7Ω)
8Ω 7Ω 9Ω V3 A4 6.9696Ω
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Solving the right side which is a series circuit (ignore the 7Ω)
Rtot= = Ω 8Ω 7Ω 9Ω V3 A4 6.9696Ω
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Solving the right side which is a series circuit (ignore the 7Ω)
Rtot= = Ω I = V/R = / = .408A which is the reading on A4 8Ω 7Ω 9Ω V3 A4 6.9696Ω
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Solving the right side which is a series circuit (ignore the 7Ω)
Rtot= = Ω I = V/R = / = .408A which is the reading on A4 And finally, V3 = IR = .408*9 = V 8Ω 7Ω 9Ω V3 A4 6.9696Ω
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V Solving the right side which is a series circuit (ignore the 7Ω) Rtot= = Ω I = V/R = / = .408A which is the reading on A4 And finally, V3 = IR = .408*9 = V Ta Daa! 8Ω 7Ω 9Ω V3 A4 6.9696Ω
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