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Circuit in question

This is the solution I saw for the equivalent resistance: "12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus R = 5 + 12.5 + 15 = 32.5 ohm"

I would really appreciate a circuit diagram solution, one that shows the circuit after each step so that I can properly understand the solution.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. CircuitLab version for OP to work on [by Transistor].

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    \$\begingroup\$ Why don’t you redraw the circuit after each step. Get a crayon and mark the nodes and redraw / combine so that it makes more sense. Don’t fall at the first hurdle. \$\endgroup\$ – Andy aka Jun 2 '18 at 15:40
  • \$\begingroup\$ There's a schematic button on the editor toolbar. Redraw you're schematic in that and then copy it showing the changes for each parallel or series combination you do. We'll help you along. \$\endgroup\$ – Transistor Jun 2 '18 at 15:45
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    \$\begingroup\$ I've added the schematic for you. Double-click to edit component values. Drag ends to create wires. \$\endgroup\$ – Transistor Jun 2 '18 at 15:50
  • \$\begingroup\$ shorten the wire that runs from the top of 12Ω resistor to the bottom of the 60Ω resistor until the two resistors almost touch \$\endgroup\$ – jsotola Jun 2 '18 at 16:16
  • \$\begingroup\$ I can't edit properly, the buttons are too small on my mobile. \$\endgroup\$ – lekarane Jun 2 '18 at 16:16
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I know that Michael took a numerical solution approach, using tools. But sometimes an ape doesn't have their tools with them and have only their fingers with which to draw in the sand and a brain to apply.

In such a situation you might consider the following initial steps:

schematic

simulate this circuit – Schematic created using CircuitLab

Now, from your own post above:

12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus R = 5 + 12.5 + 15 = 32.5 ohm

You should be now able to see why the highlighted line above might be a first step towards a solution.

Note that the use of a convenient, temporary ground doesn't violate anything. All it does is help a little in "seeing" how you might combine things. Sometimes, lots of wires get in the way of being able to see better. If so, just cut them away and name the end-point, instead. Getting rid of wires can really help, at times. (You can always re-connect them back up and un-name them, later.)

At this point, it should be easy to see how the next step you quoted might occur. Remember?

12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus R = 5 + 12.5 + 15 = 32.5 ohm

Can you now see how that happens, looking at the schematic in the lower right corner?

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  • \$\begingroup\$ @lekarane If it is an answer, you are permitted to select it as such (I think there are points for that, though whether you'd care or not is a different question.) However, also, if you want to wait for a better answer then please do. Selecting an answer usually stops the addition of more. \$\endgroup\$ – jonk Jun 2 '18 at 19:22
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Here is how I picture it. Bias the input nodes between A and B with 1V and then let the computer derive the operating point.

enter image description here

enter image description here

Now the final calculation is easy. Knowing that the voltage of the V1 source is 1V and the current sourced from V1 is 0.0307692A it can be shown, using Ohms law, that the equivalent resistance is:

R = 1V / 0.0307692A = 32.5000325 ohm

(note that the computer uses floating point binary so there may be a net error term in the value calculation)

You can use the same technique to solve manually using circuit mesh equations.

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