Linearity and superposition theorem

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Ivan Antunovic
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Homework Statement


For the network of constant current shown in Figure 4 it is known that R1 = 50 Ω and , R = 10 Ω. When the switch P is
in the 1-position , current I = 50 mA and Ip = 70 mA known i . When the switch P is in
the 2-position , current I' = 40 mA and Ip' = 90 mA are known . Determine the current Ip''
when the switch switches to the third position.
Handwritten derivation with three linear equations Ip = x U1 + y, Ip' = x U1 + y, Ip'' = x U1 + y, with U1 = I3R, U1 = I'4R, U1 = I''.R, and solved constants x = 0.2, y = -0.23; beside it, an active network block with resistors R and 3R, 4R, and voltage U1 and current Ip labelled.

Homework Equations

The Attempt at a Solution


By linearity and superposition theorem we have a linear network and if we have our active load network we can write I = xU + y , where current I is the response and voltage U represents excitation , constants x and y are determined by the resistance of the active network ,where x has dimension of A/V and y has dimension of A.

By solving equation 1 and equation 2 I get that,
x = 0.2 and y = -0.23,
for the 3rd equation I get:
Ip" = xU1 + y = 0.2 U1 - 0.23
I have two unknowns Ip" ,U1 and only 1 equation.
What am I doing wrong?
 
Attachments
  • Circuit schematic with a current source I feeding a switch P that selects among three resistors R, 3R and 4R, each connected to a common node on the right.
    upload_2016-4-24_11-23-26.png
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What if you put the parallel network of R, 3R and 4R inside the active network box? You know the value of R and currents I and I'. Is it possible to find I'' using the same method you used above? I am not sure, it's just a suggestion. I'm not familiar with your method.
 
cnh1995 said:
What if you put the parallel network of R, 3R and 4R inside the active network box? You know the value of R and currents I and I'. Is it possible to find I'' using the same method you used above? I am not sure, it's just a suggestion. I'm not familiar with your method.
It would only change constants x and y since the resistance of the active network would be changed , I would have form Ip = xI + y , where x would be non dimensional constant and y would be constant with dimension of amperes , still I would end up with 1 equation with 2 unknowns.
 
Hm if my logic is right
Handwritten graph of Ip [mA] against I [mA], a straight line sloping downward from the upper left to the lower right. Labelled points: 4R at I = 50 with Ip = 90; 3R at I = 50 with Ip = 70; 2R = ? at I = 60 with Ip = 50; R at I = 70 = I" with Ip = 30, marked Ip" = 30 in blue.


Answer should be for R I"p = 30 mA ?
 
@cnh1995
For example look at those simulations, imagine this network in the black box is the active network and you don't know what is in it ( active since it has active sources, if there were only resistors equation would have y = 0)
Proteus schematic of a network with a black box labelled active network, containing unseen contents. Outside the box: source V1 on the left, V2 source with resistors R2, R3, R4, R5 and R1 on the right, and a voltmeter.

Proteus schematic capture screenshot of a circuit with voltage source V1 feeding a black-boxed network containing resistors R1 to R5 and voltage sources V2, with a measured output voltage at the right.

capture

Let the V1 be excertation and the voltage measured by the voltmeter be response V2 .
So for the first two pictures we have
V1 = 200 V , V2 = 14.3 V
V1 = 300 V , V2 = 28.6 V
V2 = xV1 + y , and solving this equation you get
x = 0.142 and y = -14

Now suppose you wanted to find the voltage for V1 = 400 V
you already know constants so it's
V2 = V1*0.142 -14 = 400 * 0.142 -14 = 42. 8 V .

Look at the picture below , we get 42.8 V the same result.
Proteus schematic capture of a circuit: source V1 (400 V) feeds a black-boxed network, with R2, R3, R4, R5 and R1 resistors, source V2 (10 V), and voltmeters measuring the output.

screen shot on windows
That is the whole idea about this theorem as I said if there were only resistors in the black box we would have constant y = 0 , so equation would look like V2 = x*V1 ,linear homonogeous equation.
 
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Try writing Ip in terms of that switched resistance (I'll denote it R* for convenience, i.e.,
Ip = kR* + y

If you need to find I'' then determine I as a similar linear relationship to R*.
 
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NascentOxygen said:
Try writing Ip in terms of that switched resistance (I'll denote it R* for convenience, i.e.,
Ip = kR* + y

If you need to find I'' then determine I as a similar linear relationship to R*.
70 m = x30 +y
90 m = x40 +y

x = 0.002 , y = 0.01

Ip" = xR + y = 0.002 * 10 + 0.001 = 30mA , may I ask you how did you come up with this?
I was thinking it could be since R becomes excitation and by compensation theorem
http://www.electrical4u.com/compensation-theorem/

if a current or voltage is known on the resistor it can be replaced by ideal voltage/current source and vice versa
 
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The only independent variable (called "input") in this system is the switch position, i.e., the resistance R❇. So it seemed that you should be looking for the linear relationship between output Ip and input R*.
 
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NascentOxygen said:
The only independent variable (called "input") in this system is the switch position, i.e., the resistance R❇. So it seemed that you should be looking for the linear relationship between output Ip and input R*.
By the way I think it's called proportionality theorem in english literature.