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3 Ω. 17 V. A 3. V 3. A 1. 17 V. 5 Ω. 8 Ω. 9 Ω. 7 Ω. A 2. V 3. 3 Ω. 6 Ω. A 4. 10 Ω. V 1. V 2. 12 Ω. A 3. 11 Ω. 4 Ω. 17 V. 5 Ω. 8 Ω. 9 Ω. 7 Ω. A 2. V 3. 3 Ω. 6 Ω. A 4. 10 Ω. V 1. A 1. V 2. 12 Ω. A 3. 11 Ω. 4 Ω. 17 V. 5 Ω. 7 Ω. 17 Ω. A 2. 3 Ω. 6 Ω. A 4. - PowerPoint PPT Presentation
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17 V
5.4178Ω
4Ω
A3
V2
Solving:Rtot = 5.4178 + 4 = 9.4178
I = V/R = 17/9.4178 = 1.8051A(This is the reading on A3)
17 V
5.4178Ω
4Ω
A3
V2
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 VV4Ω = 1.8051*4 = 7.2204 V
7.2204 V
12Ω
6Ω
A2
The current through the 6Ω is just V/R = 7.2204/6 = 1.2034AWhich is the reading on A2.
7.2204 V5Ω
3Ω
4Ω
A1
V1
Solving (series)
Rtot = 5 + 3 + 4 = 12Ω
I = 7.2204/12 = .6017AWhich is the reading on A1
7.2204 V5Ω
3Ω
4Ω
A1
V1
Solving (series)
Rtot = 5 + 3 + 4 = 12Ω
I = 7.2204/12 = .6017AWhich is the reading on A1
V1 = IR = .6017*3 = 1.8051V
17 V
5.4178Ω
4Ω
A3
V2
Now let’s look at the right subcircuit: (the 5.4178Ω resistor)
V2 = 1.8051*5.4178 = 9.7796 V
9.7796 V Solving the right side which is a series circuit (ignore the 7Ω)
Rtot= 8+9+6.9696 = 23.9696Ω7Ω
6.9696Ω
9Ω
A4
8Ω
V3
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
7Ω
6.9696Ω
9Ω
A4
8Ω
V3
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
7Ω
6.9696Ω
9Ω
A4
8Ω
V3