Engineering Help understanding the solution to a circuit problem

[doktorwho]

Homework Statement

When the switch is open, the voltage is $U_{ab}$. When the switch is closed the calculated max current is $I_{max}$. Calculate the resistance of $R_p$ when its power is max and the current of $I_g$
$R1=100,R3=200,R4=400$,$0<Rp<100$ and $U_{ab}=2V$,$I_{max}=10mA$
This is the problem we had today in school and i have solved the second part of it but am not clear about the first part. I will post the problem and the solution and hope you can help answer what's bugging me.

Homework Equations

3. The Attempt at a Solution [/B]
The solution: The equivalent simplified system:
$E_t=U_{ab}=2V$ {This part i don't get. Why is it U_{ab} when we also have the thevenin resistor so the total voltage doesn't equal $U_{ab}$ right?}
$I_{max}=\frac{E_t}{R_t+R_p}=\frac{E_t}{R_t}$
$R_t=200$
$R_t=\frac{R1R3}{R1+R3} + \frac{R2R4}{R2+R4}=200$
$R_2=200$
$U_{ab}=I_gR_t$
$I_g=I_{max}=10mA$ {This should be because if $I_g$ was higher than the resistor could not handle it? Could have we stated this from the start then?}
since $R_{pmax}<R_t$ then $R_p=R_{pmax}=100$ {How did they deduce that $R_{pmax}$ is smaller than $R_t$. Cant see they derived it anywhere? What am i missing?

 

[gneill]

The solution: The equivalent simplified system:
$E_t=U_{ab}=2V$ {This part i don't get. Why is it U_{ab} when we also have the thevenin resistor so the total voltage doesn't equal $U_{ab}$ right?}

The Thevenin voltage is the open-circuit voltage for the network that's being replaced. With the switch open you're told that the potential across $U_{ab}$ is 2 V. So that's the open circuit potential for the network, and thus by definition it is $V_{th}$.

$I_{max}=\frac{E_t}{R_t+R_p}=\frac{E_t}{R_t}$
$R_t=200$
$R_t=\frac{R1R3}{R1+R3} + \frac{R2R4}{R2+R4}=200$
$R_2=200$

Okay, your derivations of $R_{th}$ and $R_2$ look good.

$U_{ab}=I_gR_t$
$I_g=I_{max}=10mA$ {This should be because if $I_g$ was higher than the resistor could not handle it? Could have we stated this from the start then?}
since $R_{pmax}<R_t$ then $R_p=R_{pmax}=100$ {How did they deduce that $R_{pmax}$ is smaller than $R_t$. Cant see they derived it anywhere? What am i missing?

Can you explain what $I_g$ is? Is it the Norton current?

 

[doktorwho]

The Thevenin voltage is the open-circuit voltage for the network that's being replaced. With the switch open you're told that the potential across $U_{ab}$ is 2 V. So that's the open circuit potential for the network, and thus by definition it is $V_{th}$.

Okay, your derivations of $R_{th}$ and $R_2$ look good.

Can you explain what $I_g$ is? Is it the Norton current?

No, its the current from the generator in the picture (the leftest part of the circuit). :)

 

[gneill]

No, its the current from the generator in the picture (the leftest part of the circuit). :)

Ah. Okay. It's not obvious (at least to me) that the voltage source E won't contribute to the short circuit current and that it'll be supplied only by $I_g$, so I don't think you can simply conclude that $I_g = I_{max}$ without showing some justification. Fortunately the cuurent can be found by analyzing the circuit and making use of the given value of 2 V for the top node when the switch is open. I suggest using nodal analysis where the voltage source E makes a supernode of the two middle nodes.

$R_p$ has a defined range, right? $0<Rp<100$. So $R_{pmax}$ must be essentially 100 Ohms, which is half of the $R_{th}$ that you found.

 

[doktorwho]

Ah. Okay. It's not obvious (at least to me) that the voltage source E won't contribute to the short circuit current and that it'll be supplied only by $I_g$, so I don't think you can simply conclude that $I_g = I_{max}$ without showing some justification. Fortunately the cuurent can be found by analyzing the circuit and making use of the given value of 2 V for the top node when the switch is open. I suggest using nodal analysis where the voltage source E makes a supernode of the two middle nodes.

$R_p$ has a defined range, right? $0<Rp<100$. So $R_{pmax}$ must be essentially 100 Ohms, which is half of the $R_{th}$ that you found.

Yeah, its not clear to me to, you see this is a solution which was finished in school so i just posted it so you can help me understand how it was got. They just stated that tha last part without any calculation.

 

[gneill]

Yeah, its not clear to me to, you see this is a solution which was finished in school so i just posted it so you can help me understand how it was got. They just stated that tha last part without any calculation.

Well, you'e right to be skeptical of any solution that's presented without justification ("and then some magic happens"). Try the nodal analysis approach and I think you'll find what you're looking for.

Edit: It just occurred to me that another approach to show that the source E doesn't contribute to the short circuit (maximum) current would be to use superposition. Short the output terminals, suppress the current source, and use loop analysis (mesh analysis) to find the current through the short due to voltage source E.