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Actually I am not able to understand how the author arrived at this equation using voltage divider formula ?

enter image description here

what I tried is , I combined the two 2R resistor connected to ground in series and wrote 4R then, it is in parallel with the single R resistor .

enter image description here

I am confused where I made the mistake, and also I request you to please provide any general rule for this method .

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    \$\begingroup\$ The 2R resistors are not in series with each other. \$\endgroup\$
    – The Photon
    Commented Mar 24, 2022 at 14:09

3 Answers 3

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Here's a hint. Using Thevenin's theorem, you can simplify the circuit like this: -

enter image description here

It should be a walk in the park for the next step.

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Here is my method, which I think is valid.

First, let's label the resistors: enter image description here

Then redraw the circuit so that it's easier to see what's going on:

schematic

simulate this circuit – Schematic created using CircuitLab

Now, find the voltage at node A. From there it's easy to find Vin.

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If you just want to use the voltage divider formula you can first calculate the voltage at the junction point which is:

Vx = \$ \frac{+V\cdot (3 || 2)}{3||2 + 2} \$ = \$+V\cdot\frac{3}{8}\$

Then apply the divider formula again to get Vin

2/3 * 3/8 = 1/4

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