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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
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.
