When faced with a circuit like this, when calculating thevenin equivalents or anything. We can always say that the current ix is zero. Correct ?

 

View attachment 20062

When faced with a circuit like this, when calculating thevenin equivalents or anything. We can always say that the current ix is zero. Correct ?

When faced with a circuit where the current is shown going into a terminal which isn't connected to anything, then, yes, the current will be zero as long as the circuit topology remains as shown.

But, to calculate the Thevenin equivalent, you may need to short the a-b terminals. Then the current Ix probably won't be zero.

 

View attachment 20062

When faced with a circuit like this, when calculating thevenin equivalents or anything. We can always say that the current ix is zero. Correct ?

I think when you speak about thevenin almost find load between a and b terminals Why? because we want calculate max. power for load . Of couse, if nothing between a and b terminals Ix will be zero

 

So... in summary no, it is not zero for Thevenin / Norton calculations.

Short circuit current (ix) is not zero.
Open circuit voltage has ix = 0 (since it's open circuit) and depends on the 10 mA source and V2 .

Using those two you can get the Thevenin resistance as R = V / I

 

This circuit lends itself to a useful trick in finding a solution.

Rather than shorting the terminals a & b you can apply a current drain of 10 mA.

In this situation the left hand 10 mA source can be ignored in finding Vab.

Which would yield

Vab_10mA=100*(0.5-V2)/(R1+100)=(50-100V2)/(R1+100)

We know Vth=Vab open cct which can be found from

Vth=1-100(1+V2)/(R1+100)=(R1-100V2)/(R1+100)

Hence

Rth=(Vth-Vab_10mA)/10mA
Rth={(R1-50)/(R1+100)}/10E-3
Rth=100(R1-50)/(R1+100)

If say R1=100 then Rth=100(100-50)/200=25Ω

Interestingly, it would seem this circuit would have a negative Rth if R1 is less than 50Ω!

 

When faced with a circuit where the current is shown going into a terminal which isn't connected to anything, then, yes, the current will be zero as long as the circuit topology remains as shown.

That's what I mean't, I misphrased my question.

Thank you for your replies.


@t_n_K
I'm a civil major, so all of that was wasted on me thank you though

 

@t_n_K
I'm a civil major, so all of that was wasted on me thank you though

No problem - it was probably of little interest most forum members, but sometimes there are ideas that may be worth posting just in case .....

At least you have a general formula for Rth as a check on you work.

 

This can also be solved by traditional mesh and nodal methods.

For the mesh method, replace the 10 mA current source and 100 Ω resistor with a 1 volt voltage source in series with a 100 Ω resistor. Now you have two meshes, the left one (I1 current) and the right one (I2 current) involving the a-b terminals.

If the a-b terminals are shorted, we have:

\(\left[ \begin{array}{2}100+R_1 & 50-R_1\\-R_1 & R_1-50\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\\end{array}\right]=\left[ \begin{array}{1}1+V2\\-V2\\\end{array}\right]\)

Solving gives \(I_t_h=I_2=\frac{R_1-100*V2}{100*R1-5000}\)

To find Rth, we can set all the internal sources to zero and apply a 1 volt source across the a-b terminals, giving:

\(\left[ \begin{array}{2}100+R_1 & 50-R_1\\-R_1 & R_1-50\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\\end{array}\right]=\left[ \begin{array}{1}0\\1\\\end{array}\right]\)

Solving, we get:

\(I_2 = \frac{R1+100}{100*R1-5000}\)

Since this is the current resulting from the application of 1 volt to a-b, Rth is the reciprocal of this value:

\(R_t_h = \frac{100*R1-5000}{R1+100}\)

Since Vth = Rth*Ith, we have:

\(V_t_h =\frac{100*R1-5000}{R1+100}*\frac{R_1-100*V2}{100*R1-5000}=\frac{R_1-100*V2}{R1+100}\)


Or, we could use the nodal method. There is only one node, so to calculate the open circuit voltage, Vth, we would have:

\(\left[ \begin{array}{1}\frac{1}{100}+\frac{1}{R_1}\\\end{array}\right]\left[ \begin{array}{1}V_a_b\\\end{array}\right]=\left[ \begin{array}{1}\frac{1}{100}-\frac{V2}{R_1}\\\end{array}\right]\)

Solving, we get:

\(V_t_h = V_a_b = \frac{R_1-100*V2}{R_1+100}\)

To get Rth, we zero all internal sources and apply a 1 amp current source to the a-b terminals:

\(\left[ \begin{array}{1}\frac{1}{100}+\frac{1}{R_1}\\\end{array}\right]\left[ \begin{array}{1}V_a_b\\\end{array}\right]=\left[ \begin{array}{1}1-\frac{50}{R_1}\\\end{array}\right]\)

Solving, we get:

\(R_t_h = V_a_b = \frac{100*R1-5000}{R1+100}\)