Linearity and superposition theorem

AI Thread Summary
The discussion revolves around applying the linearity and superposition theorem to determine the current Ip'' when a switch is in a third position in a circuit with known resistances and currents. Participants explore how to express the relationship between current and voltage using linear equations, leading to constants x and y that define the system's behavior. They suggest that introducing a parallel network of resistors would alter these constants but still result in an equation with two unknowns. The final calculations indicate that Ip'' could be determined as 30 mA based on the established relationships. The conversation highlights the importance of understanding the underlying principles of linear networks in circuit analysis.
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.
Linearity_theorem.png

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?
 

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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
fad.png


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)
active_network_1.png

active_network_2.png

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.
active_network_3.png

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.
 

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