# Finding the Thevenin equivalent circuit for the following one From my poor drawing , it's obvious that the resistance between A and B terminal is just

$$((((((100 || 100)+30) ||80)+20)||60)+20) = 50\Omega$$

But I am getting confused about determining the Thevenin voltage across the AB terminal.

If I remove the $50\Omega$ resistance from AB then I can ignore the rightmost $20\Omega$ resistor as no current is passing through it.

Then the equivalent resistance for the circuit would be (across the 80V source) $$((((60+20)||80)+30||100)+100) = 141.176\Omega$$

So, the current through the leftmost $100\Omega$ resistor is $80/141.176 = .567$A.

Then the current through the $30\Omega$ resistor is $.567 \times 100/130 = .436$ A and the current through the $60\Omega$ one is $.436/2 = .218$A. But then the $V_{\text{TH}} = V60 = 60 \times .218 \text{V} = 13 \text{V}$.

But the actual answer in my book is 10 V. What am I doing wrong here?

Source: It's an example problem from "V.K Mehta - Principles of Electronics".

Then the current through the $30\Omega$ resistor is $.567 \times 100/130 = .436$ A
It looks like you are attempting to use a current divider (which is a correct approach) but you are using the wrong resistance. You need to use the equivalent resistance looking into the $30\Omega$ resistor, not just the $30\Omega$ resistor itself. That equivalent resistance is $30\Omega + 40\Omega = 70\Omega$ so the correct current divider is $$0.567\text{A} \times \frac{100\Omega}{100\Omega + 70\Omega} = 0.333\text{A}$$
You should be able to repeat this process to find the current through the leftmost $20\Omega$ resistor and then arrive at $V_{\text{TH}}$. Doing so I arrived at $V_{\text{TH}} = 10$V.