I have the circuit shown below. I need to find the Thevenin voltage for it. If the dependent voltage source was a resistor instead, I think I could have disregarded it when using KVL (since the current equals zero). According to the solutions manual, however, I need to consider the dependent voltage source when using KVL, even though there is no current through it. Can someone explain why this is correct?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ This is not a dependent current source, but a dependent voltage source where the voltage depends on the value of a given current. \$\endgroup\$
    – Roger C.
    Commented Mar 11, 2015 at 19:11
  • \$\begingroup\$ But is it possible to have voltage when the wire is open and there is no current in it? \$\endgroup\$ Commented Mar 11, 2015 at 19:13
  • \$\begingroup\$ Yes, it's possible. \$\endgroup\$
    – Null
    Commented Mar 11, 2015 at 19:13

2 Answers 2


If the dependent voltage source was a resistor there would be no current through it since one end of it is connected to an open circuit (at node a). A resistor with no current through it has no voltage through it (since \$V = IR\$), so a resistor would not affect \$V_{\text{TH}}\$.

But the dependent voltage source has a non-zero voltage because its value is \$30\times 10^3 i_0\$ (where \$i_0\$ is the current through the upper resistor), and \$i_0\$ is non-zero.

There is still no current through the dependent source because it is connected to an open circuit at node a. That means the current through both resistors (\$i_0\$) is the same: $$i_0 = -\frac{100\text{V}}{20\text{k}\Omega + 80\text{k}\Omega} = -1\text{mA}$$

\$i_0\$ is negative based on the direction indicated in the circuit. The dependent voltage source's voltage is therefore $$V_D = 30\times 10^3 i_0 = -30\text{V}$$

This needs to be added to the voltage at the node between the resistors (which is a simple voltage divider) to calculate \$V_{\text{TH}}\$.


You can have voltage without current. That's what an open circuit is, actually. \$i_o\$ is non-zero due to the independent voltage source, so \$30000*i_o\$ is also non-zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.