Circuit Analysis Final Project KEITH PARKER EECT 111-51C
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Circuit Analysis Final Project
KEITH PARKER
EECT 111-51C
1. Resistors – Series & Parallel
For Resistors In Series
Resistors In Series
Impedence Is CumulativeR1 + R2 + R3 + … R(n) = RT
R1 1.0E+3 ΩR2 2.2E+3 ΩR3 3.3E+3 ΩRT 6.5E+3 Ω
R13.3kΩ
R21kΩ
R32.2kΩ
XMM1
For Resistors In Series, Resistance Is Cumulative.
That is to say R1 + R2 + … R(n) = Rt
For Resistors In Parallel
Resistors In Parallel
Equal Value
R1 2.2E+3R2 2.2E+3R3 2.2E+3Rt 733.3E+0
R12.2kΩ
R22.2kΩ
R32.2kΩ
XMM1For Any Number Of Equal Value Resistors
The total resistance will be equal to the Value of any individual resistor divided byThe total number of resistors
2200Ω / 3 resistors = 733.3 Ω
For Resistors In Parallel
Resistors In Parallel
Two Unequal
R1 2.2E+3R2 3.3E+3
Rt 1.3E+3
R12.2kΩ
R33.3kΩ
XMM1
For Any Two Unequal Resistors
Resistance is given by
(R1 * R2) / (R1 + R2)
For Resistors In Parallel
Resistors In Parallel
3< Resistors
R1 2.2E+3R2 3.3E+3R3 3.3E+3R4 4.7E+3Rt 785.3E+0
R12.2kΩ
R33.3kΩ
XMM1
R23.3kΩ
R44.7kΩ
For Any Number Of Unequal Resistors
Resistance can be found by the reciprocal method
1/((1/R1)+(1/R2)+(1/R3)+(1/R4))
For Parallel and Series Combined
Combination Parallel and Series
R1 2.2E+3R2 3.3E+3R3 3.3E+3R4 4.7E+3R(1234) 785.3E+0R5 4.7E+3Rt 5.5E+3
R12.2kΩ
R33.3kΩ
XMM1
R23.3kΩ
R44.7kΩ
R5
4.7kΩ
For A Circuit With Parallel and Series
Combine your resistors. Take the Resistance of the Parallel set by1/((1/R1)+(1/R2)+(1/R3)+(1/R4)) With that you may think of it as oneResistor. Now its in series with yourFirst resistor.
Add for your total resistance.
R1234 + R5 = Rt
2. Resistor Network
Resistance
R6 4.7E+3 ΩR7 4.7E+3 ΩR8 4.7E+3 ΩR9 4.7E+3 ΩR10 4.7E+3 ΩR(789) 1.6E+3 ΩRt 11.0E+3 Ω
R6
4700Ω R74700Ω
R84700Ω
R94700Ω
R10
4700Ω
XMM4
Resistance is found by the reciprocalMethod for the parallel part, then Adding them to the series part of theCircuit
Rt = R(789) = R6 + R10
Branch Currents and Voltages
R1 4.70E+03 ΩR2 4.70E+03 ΩR3 4.70E+03 ΩR4 4.70E+03 ΩR5 4.70E+03 ΩR6 4.70E+03 ΩR box 1.18E+03 ΩRt 1.06E+04 Ω
Voltages CurrentsV1 9 VProbe 1 5 V 6.38E-04 mAProbe 2 5 V 4.26E-04 mAProbe 3 5 V 2.13E-04 mAProbe 4 4 V 6.38E-04 mA P3 = P6, P2 = P5, and P1 = P4 because current in has to equal current outProbe 5 4 V 4.26E-04 mAProbe 6 4 V 2.13E-04 mAProbe 7 5 V 8.51E-04 mAProbe 8 4 V 8.51E-04 mA
It 8.51E-04 mA * same as probe 7 because amperage in series is constantI Drop 2.13E-04
R1
4.7kΩ
V19V
R3
4.7kΩ
R2
4.7kΩ
R44.7kΩ
R5
4.7kΩ
R6
4.7kΩ
Probe1
V: 5.00 V I: 638 uA
Probe2
V: 5.00 V I: 426 uA
Probe3
V: 5.00 V I: 213 uA
Probe4
V: 4.00 V I: 638 uA
Probe5
V: 4.00 V I: 426 uA
Probe6
V: 4.00 V I: 213 uA
Probe7
V: 5.00 V I: 851 uA
Probe8
V: 4.00 V I: 851 uA
Currents were found using Ohm’s law. I = VR
For current total it was I = 9v / 10600 Ohms
For the branches there were 4 branches so eachDrop should be Itotal / 4 to get 213uA. So each Branch total had 213uA taken from it.
For the opposite branches the totals were the same.In parallel current in = current out.
For voltages, voltage drop across the series resistor wasTaken as V = (R5 / Rt ) * Vt
On the parallel part voltage is constant.
3. Thevenin Resistance / Voltage of a Resistor Network
V(T) 9.00E+00 VR1 4.00E+01 ΩR2 4.00E+01 ΩR3 1.00E+01 ΩR4 2.00E+01 ΩR5 2.00E+01 ΩRL 1.00E+02 ΩR12 2.00E+01 ΩR45 1.00E+01 ΩR345 2.00E+01 ΩR(T) 4.00E+01 ΩI(T) 2.25E-01 AV(A) 4.50E+00 VV(B) 0.00E+00 VVth 4.50E+00 V
RTH = 10Ω
VTH = 4.5 VRL = 100Ω
R1
40Ω
V19V
R2
40Ω
R310Ω
R420Ω
R520Ω
R6100Ω
XMM1XMM2
R7
40Ω
R8
40Ω
R910Ω
R1020Ω
R1120Ω
XMM3
XMM4
RL is calculated as the Resistance of the LoadIn this caseRL = 100 Ω
Vth is calculated as Va – VbVa = R12 * ItotalVb = ((R2/(R1+R2))*Vtotal) - ((R45/(R3+R45))*Vtotal)Vth = 4.5V
Rth is the resistance looking back into theCircuit with the power source shorted outRth = 10Ω
4. Multiple Capacitors in Parallel and Series
Series ParallelC1 10 10 uFC2 27 27 uFC3 47 47 uFCT 6.32 84 uF
The numbers in the worksheet were determined using the same formulas.
Multisim shows the same capacitance numbers.
C110µF
C222µF
C347µF
V11Vpk 1kHz 0°
C410µF
C522µF C6
47µFV21Vpk 1kHz 0°
R1
1kΩ
R2
1kΩ
Vout2Vout1
Capacitance in Parallel isC1 + C2 + C3 = Ctotal
Capacitance in Series is1/(1/C1)+(1/C2)+(1/C3) = Ctotal
Multisim had to have an expression added to the Single Frequency ACAnalysis to allow it to measure capacitance. The expression is as follows.Vout1 : 1/((abs(imag(V(vout1)/I(R1))))*1000*2*pi)Vout2 : 1/((abs(imag(V(vout2)/I(R1))))*1000*2*pi)
5. RC Circuits
Time Constant is R * C
1000 Ω* 10uF = .001
For RC as a function of timeFor Vc
Vs*(1-EXP(-Time/TC))
To Graph RC as a function of time for Vc
Vs = 1 2t = 2.0E-3R = 1.0E+3 3t = 3.0E-3C = 1.0E-6 4t = 4.0E-3t = 1.0E-3 5t = 5.0E-3
Deta T = 100.0E-6
Time Vc000.0E+0 000.0E+0100.0E-6 95.2E-3
Time constant is calculated as R * C 200.0E-6 181.3E-3300.0E-6 259.2E-3
R 1.00E+03 Ω 400.0E-6 329.7E-3C 1.00E-05 uF 500.0E-6 393.5E-3RC 1.00E-02 600.0E-6 451.2E-3
700.0E-6 503.4E-3It takes 5 time constants to fully charge 800.0E-6 550.7E-3
900.0E-6 593.4E-3RC2 2.00E-02 1.0E-3 632.1E-3RC3 3.00E-02RC4 4.00E-02RC5 5.00E-02
C110µF
V11Vpk 1kHz 0°
R1
1kΩ
Vin
Vout
0.0E
+0
100.
0E-6
200.
0E-6
300.
0E-6
400.
0E-6
500.
0E-6
600.
0E-6
700.
0E-6
800.
0E-6
900.
0E-6
1.0E
-30.0E+0
100.0E-3
200.0E-3
300.0E-3
400.0E-3
500.0E-3
600.0E-3
700.0E-3
RC as a Function of Time
RC As A Function Of Time - Multisim
V10V 1V 10ms 20ms
R1
100Ω
XSC1
A B
Ext Trig+
+
_
_ + _
Iin
C110µF
Xc is calculated asXc = (1/(2PI*Freqency*Capacitance)
Time constant is calculated as R * C
R 1.00E+03 ΩC 1.00E-05 uFRC 1.00E-02
For Xc
Xc = (1/(2∏fC))
F Xc
10 1.59E+0350 3.18E+02
100 1.59E+02500 3.18E+011000 1.59E+013000 5.31E+005000 3.18E+007000 2.27E+00
10000 1.59E+00
C110µF
V11Vpk 1kHz 0°
R1
1kΩ
Vin
Vout
1.00E+00
1.00E+01
1.00E+02
1.00E+03
1.00E+04
0 2000 4000 6000 8000 10000 12000
Xc
Xc As A Function Of Frequency
Xc As A Function Of Frequency - Multisim
6. Inductors In Series And Parallel
Inductors in series are additive in nature, that is L1 + L2 + L3 = LtotalL1 1 mHL2 2.2 mHL3 4.7 mH
Ltotal 7.9 mH
Inductors in parallel are give by the reciprocal value. 1/((1/L1)+(1/L2)+(1/L3))L1 1 mHL2 2.2 mHL3 4.7 mH
Ltotal 0.599768 mH
L11mH
L22.2mH
L34.7mH
L41mH
L52.2mH
L64.7mH
R1
1mΩ
R2
1mΩ
Vout1
V11Vpk 1kHz 0°
V21Vpk 1kHz 0°
R3
1mΩ
R4
1mΩ
Vout2
Inductors in series are additive.L1 + L2 + L3 = Ltotal
Inductors in parallel are reciprocal.1 / ((1/L1) + (1/L2) + (1/L3)) = Ltotal
7. Simple RL Circuit
The Time Constant is calculated as L/RFor this circuit, 1mH / 1kΩ
R1
1kΩ
V11Vpk 1kHz 0°
L11mH
Vout1
RL Time
The current out is calculated as (R/L)*(1-EXP(-R*Time/(L/R)
To Graph RL as a function of time for Xl
Vs = 1 V 2t = 200.0E+0R = 1.0E+3 Ω 3t = 300.0E+0L = 100.0E-3 H 4t = 400.0E+0t = 100.0E+0 5t = 500.0E+0
Deta T = 10.0E+0I = 1.0E-3 A
Time I Out000.0E+0 000.0E+0
10.0E+0 951.6E+0Time constant is calculated as R * C 20.0E+0 1.8E+3
30.0E+0 2.6E+3R 1.00E+03 Ω 40.0E+0 3.3E+3L 1.00E-01 H 50.0E+0 3.9E+3RL 1.00E-04 60.0E+0 4.5E+3
70.0E+0 5.0E+3It takes 5 time constants to fully charge 80.0E+0 5.5E+3
90.0E+0 5.9E+3RC2 2.00E-04 100.0E+0 6.3E+3RC3 3.00E-04RC4 4.00E-04RC5 5.00E-04
V10V 1V 10ms 20ms
R1
1mΩ
XSC1
A B
Ext Trig+
+
_
_ + _
L11mH
XCP1
100 mV/mAIout
Iin
000.0E+0
1.0E+3
2.0E+3
3.0E+3
4.0E+3
5.0E+3
6.0E+3
7.0E+3
I Out
RL Time - Multisim
Xl As A Function Of Frequency
Time constant is calculated as R * C
R 1.00E+02 ΩL 1.00E-01 HRL 1.00E-03
For Xc
Xl = 2*PI*Freqency*L
F Xl
10 6.28E+0050 3.18E-02
100 1.59E-02500 3.18E-031000 1.59E-033000 5.31E-045000 3.18E-047000 2.27E-04
10000 1.59E-04 1.00E-04
1.00E-03
1.00E-02
1.00E-01
1.00E+00
1.00E+01
0 2000 4000 6000 8000 10000 12000
Xl
R1
100Ω
V11Vpk 1kHz 0°
Vin
L1100mH
Xl is calculated as
Xl = 2*PI*Frequency*L
Xl Multisim
R1
100Ω
V11Vpk 1kHz 0°
Vin
L1100mH
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