# How to apply Thevenin's Theorem to this circuit?

I have been trying to solve this question with either Thevenin's Theorem or Norton's Theorem but I got stuck at finding the Vth .

Load Resistor is 5Ohm resistor in the middle .

I have removed the load and circuit became :

I calculated Rth as =(3+1+3+1)//4//4 = 1.6Ω

Simplifying the circuit :

Now I need to find the Voltage between points Req . But I don't know how to .

Thanks

• Pretty easy! Consider your 3rd figure. Find the source current $I_s = \frac{V_s}{R_{total}}$ and then apply a voltage divider. The voltage you are trying to find is the same with the voltage across the 2Ω resistor. Feb 7, 2019 at 19:15
• How do you get 0V for Vth?
– Chu
Feb 7, 2019 at 21:25

You have correctly calculated the thevenin resistance. Next you need to calculate the thevenin voltage source. This can be done by replacing the source in the circuit while leaving the load out, and solving for the open-circuit voltage across the load nodes.

Adding the source back in, you can find the current from the source by dividing the voltage by the simplified resistance seen by the source.

From there, simplifying the four 2ohm resistors to a single 2 ohm resistor will leave you with a single loop carrying the current you calculated. Calculating the voltage drops over each resistor in the loop will give you the voltage across the load nodes, which is your thevenin open-circuit voltage.

Finally, you can re-draw everything as a thevenin equivalent circuit with a single source, and a single resistance in series with your load. Solving the load voltage and current should then be simple.

Simplify each resistor in the network by adding serial resistors and by using the appropriate formula for parallel resistors until you have a circuit that looks like what's in your textbook, not a mess of resistors :)

How did you get from your 3rd schematic to the 4th? You were almost there.

You have a 10 V source and two equivalent resistors in series, 8 ohms and 2||5 ohms. Use Ohm's Law to find the voltage across the parallel resistors, and Ohm's Law again to find the current in the 5 ohm part.

• The 5 ohm resistor is the load, which is removed for Thevenin analysis.
– Chu
Feb 7, 2019 at 21:23