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| 1-Obtain Z equivalent and Norton Equivalent for the whole circuit.
note: Z is the same as the impedance of the circuit. 2-obtain the average power and the power factor of the capacitor. Values: L(1,2)=20 L(4,5)=20 L(7,6)=10 L(8,g)=10 L(9,10)=15 R(2,3)=20 R(3,7)=10 R(3,4)=20 R(6,11)=5 R(6,8)=10 R(g,9)=5 C(10,11)=5 voltage oscillator(1,6)=10*sin(t) V voltage oscillator(5,6)=10*sin(t-45) V current oscillator(6,g)=sin(t) A note: the numbers in parenthesis i.e. (1,2) are the nodes where the device is located. For example, L(1,2) is an inductor between nodes 1 and 2 with value= 20 L=inductor R=resistor C=capacitor |
This is the main circuit
Zoomed top section of above circuit. 
Zoomed bottom section of above circuit. 
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| First lets reduce the circuit to its Norton equivalent.
The easiest way to do this is by breaking the circuit into a sub-circuit and then finding Norton equivalent across of this sub-circuit. Step1- Choose the two terminals of a device where you would like to find the Norton equivalent looking "into" these two terminals. It doesn't really matter which two terminals are chosen. For our example lets pick the two terminals of the R(3,7) device. Step2- Remove the R(3,7) device from the circuit. What is left is the sub-circuit shown to the right. |
sub-circuit  |
| Step3-Now solve for the Norton equivalent of this sub-circuit.
Circuit4 can do this easily. Place the Rin device between terminals 3 and 7 as shown and press enter. |
Solving for the Norton equivalent of the sub-circuit
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| After "thinking" for some time Circuit4 will display the answers shown.
Therefore the sub-circuit has
Rin=(22.3611\_63.4327) Thevenin Voltage=9.2388*cos(t-67.5) Norton Current=.413163*cos(t-130.933) Remember that this are only the Norton-Thevenin equivalent values for the sub-circuit. |
Values of Rin, Norton Current, and Thevenin voltage of sub-circuit |
| With the answers of above we can now draw the Norton equivalent of the sub-circuit.
Our big sub-circuit has now been reduced to only a current oscillator and resistor. |
Norton equivalent of sub-circuit
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| It is time now to re-attach the resistor that we removed earlier. | Resistor has been re-attached to Norton equivalent of sub-circuit.
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| The last step is to combine the two parallel resistors.
Their combined resistance is = [(22.3611\_63.4327) * 10 ]/[ (22.3611\_63.4327) +10 ] =(7.90565\_18.4341) The Norton equivalent of the circuit is shown to the right with values: Rin=Z=Impedance of circuit=(7.90565\_18.4341) ohms Norton Current=.413163*cos(t-130.933) Amps |
Norton equivalent of main circuit
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| Now lets solve for the average power of the capacitor.
The formula for average power is:
Where Vm= magnitude of voltage. Im= magnitude of current. angleVoltage= angle of voltage angleCurrent= angle of current Use Circuit4 to solve for the above variables then plug them into the formula of above. Finally solve for the power factor. The formula for the power factor is: Pf=angleVoltage-angleCurrent; End of Problem |
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