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In order to find the Thevenin voltage here, why is it that the 3k Οhm resistor is ignored and no current goes through it? (hence \$V_{TH}=8mA \cdot 7kΩ = 56V\$)

  • 2
    \$\begingroup\$ The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem \$\endgroup\$
    – thece
    Commented Jan 4, 2019 at 16:36
  • 1
    \$\begingroup\$ The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V \$\endgroup\$ Commented Jan 4, 2019 at 16:41

3 Answers 3


Computing a Thevenin Equivalent requires two steps:

  1. Obtain the Thevenin Impedance \$Z_{TH}\$
  2. Obtain the Thevenin Voltage \$V_{TH}\$

In order to compute \$Z_{TH}\$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you \$Z_{TH} = 7k\Omega + 3k\Omega = 10k\Omega\$, as the resistors are in series.

In order to obtain \$V_{TH}\$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3k resistor, as it would violate Kirchhoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!

Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V \$V_{AB} = \alpha\$ you would get: $$I = \lim_{R \to \infty}\frac{V_{AB}}{R} = \lim_{R \to \infty}\frac{\alpha}{R} = 0$$ No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3k resistor means we will have no voltage drop there. Remember that we want to find \$V_{TH} = V_A - V_B\$. Following the circuit we see that the only drop involved is the one in the 7k resistor. By Ohm's Law, that drop will be \$V_{DROP} = I·R\$, so: $$V_A = V_B - I·R \to V_A - V_B = V_{TH} = - I·R$$ Being I = 8mA and R = 7k\$\Omega\$.

This is the Thevenin Equivalent and why the 3k resistor doesn't play a role in the Thevenin Voltage.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-) \$\endgroup\$ Commented Jan 5, 2019 at 17:33
  • \$\begingroup\$ Thanks a lot for the corrections. Is it ok now? ;) \$\endgroup\$
    – Collado
    Commented Jan 5, 2019 at 23:21

The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.


The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is \$8mA \cdot 7K\$ = 56V (A is negative wrt B).

When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.


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