Electrical – Assigning Current Direction in a Circuit + Resistors

You can solve this circuit by more or less the same method you've given in the question; however you need to plug in one more equation (\$V_1=V_2+12\$) into the system and introduce an unknown current variable. So I'm not sure if we can call it a 'pure' nodal analysis.

This is what you've got to do:

• Write KCL on the left node (\$I_s\$ is the current through the voltage source): $$6A=\frac { { V }_{ 1 }-{ V }_{ 2 } }{ { R }_{ 3 } } +\frac { { V }_{ 1 } }{ { R }_{ 3 } } +{ I }_{ s }$$

• Do it again on the node on the right side: $$4A={ I }_{ s }-\frac { { V }_{ 2 } }{ { R }_{ 2 } } +\frac { { V }_{ 1 }-{ V }_{ 2 } }{ { R }_{ 3 } } $$

• So far, we're in line with the method described in the question. As a last step, write down this one: $$V_1=V_2+12$$

Now we're left with 3 equations in three unknowns, which you can easily solve to get \$V_1, V_2\$ and \$I_s\$

There are two steps to finding the Thevenin equivalent circuit: finding the Thevenin voltage and the Thevenin resistance.

Thevenin voltage is the voltage across the two points you interested in (Vin). In this case it is easy to calculate as there is no current flowing in the 43 and 60 \$\Omega\$ resistors thus no voltage drop. Thus the voltage at Vin is the same as the voltage form the source, 72 V.

Thevenin resistance is calculated by 'turning off' all independent current and independent voltage sources and calculating the resistance between the two points. Turning off a voltage source sets the voltage across it to 0, which results in a short (0 \$\Omega\$) in parallel with the 275 \$\Omega\$ resistor. Any resistor combined in parallel with a short results in a short, leaving you with the 43 and 60 \$\Omega\$ resistors now in series, giving a Thevenin resistance of 103 ohms.

Putting the two together gives you a voltage source of 72 V in series with a 103 \$\Omega\$ resistor for you Thevenin equivalent circuit.