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I am solving objective questions from my book but the main problem is that it doesn't have solutions and in following question I don't know what type of question is this. First I am thinking its Wheatstone bridge related but its not. And resistance are in series but its not. So please tell me what type of question is this and how to solve it.

enter image description here

Here's one I draw before but above one is confusing.

enter image description here

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  • \$\begingroup\$ Did you try redrawing it? \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 9 '18 at 17:11
  • \$\begingroup\$ Yes I try but totally puzzled. I can draw where connection also has resistance but here not. \$\endgroup\$ – Kanwaljit Singh Feb 9 '18 at 17:14
  • \$\begingroup\$ Do you feel like sharing any of your tries so that we can tell you where you went wrong? \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 9 '18 at 17:14
  • \$\begingroup\$ First of all I give names to all points from A to F. Then I try to redraw like any other question but I am unable. \$\endgroup\$ – Kanwaljit Singh Feb 9 '18 at 17:18
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    \$\begingroup\$ Here's a clue - count the independent nodes in the circuit and what do you get. Then ask yourself how many ways you can arrange 3 resistors with these nodes. \$\endgroup\$ – Andy aka Feb 9 '18 at 17:25
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schematic

simulate this circuit – Schematic created using CircuitLab

Now why was that a problem? ( lack of numbering Nodes? )

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  • \$\begingroup\$ @MichelKeijzers look more closely at your original circuit. Every resistor has V1+ at one end and V1- at the other. They are all in parallel, just drawn in a way to confuse you. \$\endgroup\$ – Trevor_G Feb 9 '18 at 17:49
  • \$\begingroup\$ @Trevor_G Sorry I removed my comment because I saw that they are indeed equal :-) \$\endgroup\$ – Michel Keijzers Feb 9 '18 at 17:50
  • \$\begingroup\$ How you changed them into parallel? \$\endgroup\$ – Kanwaljit Singh Feb 9 '18 at 17:54
  • \$\begingroup\$ I identified Node 1 and 2 at each end of each resistor and redrew. You can do this with more complex ccts by drawing nodes in more logical positions then connecting between same nodes. This cct has only 2 nodes but 5 dots, so 3 are redundant. remember this. Nodes must be unique. \$\endgroup\$ – Sunnyskyguy EE75 Feb 9 '18 at 17:59
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Re-drawing these circuits helps a great deal to find a simple solution - Andy makes a good suggestion to reduce the apparent number of nodes.
At first glance, it appears that there are four nodes...

schematic

simulate this circuit – Schematic created using CircuitLab
Andy suggests that Node 1 is not independent from Node 3 - they are the same. And similarly, Node 2 is the same as Node 4.
So there are really only two nodes in this circuit. One of them connects to the "+" end of the battery, the other connects to the "-" end of the battery.

No one would publish a circuit diagram like that shown above. It is a book exercise to help you reduce circuits to their simplest form. Becoming adept at this skill is an important one for anyone dealing with circuits.

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  • \$\begingroup\$ Node 1 and 3, node 2 and 4 same but how to proceed further? \$\endgroup\$ – Kanwaljit Singh Feb 9 '18 at 17:59
  • \$\begingroup\$ What is the benefit of finding independent nodes? \$\endgroup\$ – Kanwaljit Singh Feb 9 '18 at 17:59
  • \$\begingroup\$ @KanwaljitSingh Once you realize that Node 1 and Node 3 collapse to a single node, then you should see that R1 & R2 are in parallel. And the junction point of R1 & R2 is the same as the "-" end of the battery. And the collapsed Node 1, Node 3 is the "+" end of the battery. \$\endgroup\$ – glen_geek Feb 9 '18 at 18:17

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