-1
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I have to find V Thevenin in that circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

I did :

\$(\frac{V_1}{R_1}-I_1)*R_2 \$

Appeared its not right.

Also Im not sure I can divide the volt source by the resistor to find the current there.

Thanks for your help

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  • \$\begingroup\$ Why don't you use superposition in this simple case? Determine \$V_{th1}\$ with \$I_1\$ turned off (open circuited) while \$V_1\$ is alive and then \$V_{th2}\$ with \$V_1\$ turned off (a short circuit) but \$I_1\$ alive. Then \$V_{th}\$ is the sum of both intermediate results. For \$R_{th}\$, short the voltage source and open-circuit the current source: the resistance you "see" from the left-side terminals is your small-signal resistance. \$\endgroup\$ – Verbal Kint Dec 28 '19 at 13:04
  • \$\begingroup\$ In this circuit one simple way to find Vth is first find the Norton equivalent of R1 and V1, with the Norton current simply V1/R1 (note the current direction) and the Norton resistance R1. This reduces the circuit to two parallel current sources and two parallel 5k ohm resistors. The Vth should now be easy to calculate. \$\endgroup\$ – dirac16 Dec 28 '19 at 13:30
  • \$\begingroup\$ Nodal analysis is easiest. Vth is the only unknown node voltage. \$\endgroup\$ – Chu Dec 28 '19 at 15:35
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Using KCL and KVL you can find:

$$\frac{\text{V}_1-\text{V}_\text{th}}{\text{R}_1}+\text{I}_1=\frac{\text{V}_\text{th}}{\text{R}_2}\tag1$$

Using the given values, we get:

$$\frac{-30-\text{V}_\text{th}}{5000}+10^{-3}=\frac{\text{V}_\text{th}}{5000}\space\Longleftrightarrow\space\text{V}_\text{th}=-\frac{25}{2}=-12.5\space\text{V}\tag2$$

I also checked the answer using LTspice and -12.5 is the correct answer.

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  • \$\begingroup\$ We don't give solutions to homework. \$\endgroup\$ – Chu Dec 28 '19 at 17:19
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Apply Superposition theorem and find the voltage across R2.

Analyze the circuit for voltage source first. Here, you have to replace the current source with open circuit. This gives you a simple voltage divider. Find the voltage across the R2. Consider in V1.

Next, replace the voltage source with short circuit to detemine the effect of current source. You have a current divider now. Find the current going through R2. Use Ohm's to find the voltage across it. Consider it V2.

Now, the final answer is V1+V2.

I am not gonna do the whole math for you but I can say the answer is -12.5V.

For finding Rth, replace the voltage source with short circuit and current source with open circuit. Find the equivalent resistance.

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  • \$\begingroup\$ Using your method (which I agree with), I get -14.6 V, not 17.5 V. \$\endgroup\$ – The Photon Dec 28 '19 at 16:11
  • \$\begingroup\$ The answer is not 17.5, the correct answer is -12.5 \$\endgroup\$ – Jan Dec 28 '19 at 16:15
  • \$\begingroup\$ I am really sorry for not considering the polarity of battery. I also got 12.5V @The Photon. \$\endgroup\$ – Giga-Byte Dec 28 '19 at 16:25
  • \$\begingroup\$ @JuneStar_2918, did you consider that when finding the contribution from the current source, current can flow through both R1 and R2? \$\endgroup\$ – The Photon Dec 28 '19 at 16:33
  • \$\begingroup\$ @ThePhoton I did. I divided the current in exactly half and 0.5m* 5k=2.5V \$\endgroup\$ – Giga-Byte Dec 28 '19 at 16:51

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