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Why can't I just add 1A current source through terminals a to b in this circuit to calculate the Thevenin's resistance.

I tried doing this using Norton's approach of short circuiting and it gave the desired answer as well.

But since I saw dependent source 30i, my instinct first kicked me to go for adding 1A current source and then solve for Voltage at node A and divide it by 1 to get the resistance (Rth). But it doesn't work, any reasons why?

Edit:

I had messed a couple of resistor values in previous picture. I have updated it now. Because I know the answer to the circuit with these values, I have updated to this, Apologies for the error.

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  • \$\begingroup\$ But you can add a current source to those terminals to work out the Thevenin resistance. Why don't you think so? \$\endgroup\$ Commented Oct 26, 2024 at 6:23
  • \$\begingroup\$ @periblepsis, but you need to do it with two different values and find the slope between the two I-V points. You can't just calculate the voltage with 1 A applied and take V/I. \$\endgroup\$ Commented Oct 26, 2024 at 6:26
  • \$\begingroup\$ Yesk, What's the desired answer that you have for Rth? \$\endgroup\$ Commented Oct 26, 2024 at 6:43
  • \$\begingroup\$ @periblepsis the desired/true answer is 20kohms, while I got 300kohms with 1A source. \$\endgroup\$ Commented Oct 26, 2024 at 6:45
  • \$\begingroup\$ @Yesk Thanks for the correction. \$\endgroup\$ Commented Oct 26, 2024 at 7:07

3 Answers 3

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Let me draw the circuit and show the test current:

schematic

simulate this circuit – Schematic created using CircuitLab

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } ans = solve( [ kclx, kcly, kcla ], [ vx, vy, va ] ) va = ans[ va ].subs( vars ) va.subs( itest, 0 ) 280.000000000000 va.subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

enter image description here

As you can see, there's no difficulty using an external current source.

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The problem appears straightforward. Connect a 1A current source between nodes a and b. Connect node b to GND. Assume voltages at the top nodes be a, c, d from right to left. Short circuit the voltage source on left.

Using nodal analysis,

$$\frac{V_a}{40k} +\frac{V_a - V_c}{10k} = 1 + 30.\frac{V_d}{20k} \tag{1}$$

$$\frac{V_c}{50k} +\frac{V_c - V_a}{10k} = -30.\frac{V_d}{20k}\tag{2} $$

$$\frac{V_d}{20k || 2k} +\frac{V_d - V_c}{5k} = 0 \tag{3}$$

Solve these equations using Kramers rule to get \$V_a\$ , \$V_c\$ and \$V_d\$. \$V_a\$ would give value of Thevenin resistance in ohms.

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Calculation of the resistance and voltage of the Thevenin equivalent circuit. Results obtained on the basis of the topology of the electrical networks and with the method of voltages at the nodes:

enter image description here

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