Circuit Description: 6V source series 3 Ohm, parallel 6 Ohm. Output node taken after the parallel combo, then series 2 Ohm to terminals.
Show Hint
Thevenin Voltage (Vth). Find the open-circuit voltage across the terminals of interest. Use circuit analysis techniques like voltage divider, node analysis, or mesh analysis on the source circuit (with the load removed). No current flows through any component connected in series directly to the open-circuit terminals.
The Thevenin voltage (Vth) is the open-circuit voltage across the output terminals To find Vth, we remove any load and calculate the voltage at the terminals
First, consider the circuit without the 2 \(\Omega\) resistor connected to the output terminals We need the voltage across the 6 \(\Omega\) resistor, as this is the voltage at the node before the 2 \(\Omega\) resistor The 6 V source is connected to the 3 \(\Omega\) and 6 \(\Omega\) resistors which form a voltage divider (as they appear in series from the source's perspective when calculating the voltage across the 6 \(\Omega\) part)
Voltage across the 6 \(\Omega\) resistor (let's call this \(V_{node}\)) is:
$$ V_{node} = V_{source} \times \frac{R_{parallel}}{R_{series} + R_{parallel}} = 6 \, \text{V} \times \frac{6 \, \Omega}{3 \, \Omega + 6 \, \Omega} $$
$$ V_{node} = 6 \, \text{V} \times \frac{6}{9} = 6 \, \text{V} \times \frac{2}{3} = 4 \, \text{V} $$
Now, consider the output terminals where Vth is measured Since the terminals are open-circuited, no current flows through the 2 \(\Omega\) resistor Therefore, there is no voltage drop across the 2 \(\Omega\) resistor The voltage at the output terminal is the same as the voltage at the node before the 2 \(\Omega\) resistor
$$ V_{th} = V_{node} = 4 \, \text{V} $$
The Thevenin voltage is 4 V
