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Circuit Diagram

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".

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This statement and what follows is false:

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.

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  • \$\begingroup\$ Can't believe I did this!Thanks a lot for pointing it out. \$\endgroup\$ Aug 14, 2015 at 17:31

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