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Published byGordon Waters Modified over 9 years ago
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Series/Parallel Circuits
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I1I1 + - VR2R2 R3R3 I2I2 I3I3 ITIT R1R1
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Series/Parallel Circuit 1.Reduce the Parallel Branches into One Series Resistor. 2.This resistor is called an Equivalent Resistance. 3. Do this for all the Parallel Paths. 4. Do your calculation on the Series Circuit.
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} 1 R 23 1R21R2 1R31R3 =+ 1 R 23 1414 1414 =+ R 23 = 2Ω + - V=10v R 3 =4Ω R 1 =8Ω R 2 =4Ω Block
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+ - V=10v R 1 =8Ω R 23 =2Ω R 23 is an equivalent resistor to R 2 and R 3. R T = R 1 + R 23 R T = 8Ω + 2Ω = 10Ω
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CompRIV R1R1 R2R2 R3R3 R 23 Total 8Ω8Ω 4Ω4Ω 4Ω4Ω 10v
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I T = V/ R T I T = 10v / 10Ω I T = 1A I T = I 1 = I 23 = 1A V 1 = I 1 X R 1 V 1 = 1A X 8Ω V 1 = 8v V 23 = I 23 X R 23 V 23 = 1A X 2Ω V 23 = 2v
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} + - V=10v R 1 =8Ω R 23 =2Ω V 23 = 2v V 2 = V 3 = 2v
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I 2 = V 2 / R 2 I 2 = 2v / 4Ω I 2 = 0.5A I 3 = V 3 / R 3 I 3 = 2v / 4Ω I 3 = 0.5A + - V=10v R 3 =4Ω R 1 =8Ω R 2 =4Ω
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+ - V=60v R 1 = 10kΩ R 2 = 15kΩ R 3 = 30kΩ R 5 = 24kΩ R 4 = 24kΩ R 6 = 3kΩ
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R 1 = 10kΩ R 2 = 15kΩ R 3 = 30kΩ 1 R 123 1R11R1 1R21R2 =+ 1R31R3 + 1 R 123 1 10 1 15 =+ 1 30 +
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1 R 123 1 10 1 15 =+ 1 30 + 1 R 123 3 30 2 30 =+ 1 30 + 1 R 123 6 30 = R 123 30 6 = 5 kΩ =
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R 123 5kΩ=
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+ - V=60v R 5 = 24kΩ R 4 = 24kΩ R 6 = 3kΩ R 123 = 5kΩ
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+ - V=60v R 5 = 24kΩ R 4 = 24kΩ R 6 = 3kΩ R 123 = 5kΩ
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R 5 = 24kΩ R 4 = 24kΩ 1 R 45 1R41R4 1R51R5 =+ 1 R 45 1 24 1 24 =+ 1 R 45 2 24 = R 45 24 2 = 12 kΩ =
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R 45 12kΩ=
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V=60v R 6 = 3kΩ R 123 = 5kΩ + - R 45 =12kΩ R T = R 123 + R 45 + R 6
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R T = 5kΩ + 12kΩ + 3kΩ R T = 20kΩ I T = 60v 20kΩ I T = ETET RTRT 3mA
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V=10v R 6 = 3kΩ R 123 = 5kΩ + - R 45 =12kΩ I T =3mA V 123 = I 123 X R 123 = 3mA X 5kΩ V 123 = 15v V 45 = I 45 X R 45 =3mA X 12kΩ V 45 = 36v
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V 6 = I 6 X R 6 = 3mA X 3kΩ V 6 = 9v
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I 1 = E1E1 R1R1 = 10v 10kΩ I 1 = 1mA
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I 2 = E2E2 R2R2 = 15v 10kΩ I 2 = 1.5mA
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I 3 = E3E3 R3R3 = 15v 30kΩ I 3 = 0.5mA
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I 4,5 = E4E4 R4R4 = 36v 24kΩ I 4,5 = 1.5mA
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I 6 = E6E6 R6R6 = 9v 3kΩ I 3 = 3mA
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