Redrawing complicated schematics.

Thread Starter


Joined Sep 18, 2011
Hey guys. Im really having problems translating the more complicated schematics we get and reducing them. This all stems from learning the reduce and return approach to which am competent in the return and somewhat of the reducing but sometimes my mind goes haywire when the circuit is drawn in certain ways.

The hardest part for me basically is redrawing it correctly and accurately. Is there a more systematic approach to this? For instance in the pictures below my total Resistances are off...the first one I get is off by .100 or so and the second one I dont know what to do with the right side of the schematic....any help would be appreciated...

by the way I have been trying to do it on my own Im only here at this point because I for the life of me can't figure it out...



Joined Aug 15, 2007
"my total Resistances are off...the first one I get is off by .100 or so" Don't worry about that. Honest! Unless you're doing really sensitive stuff, that 0.1 ohm(?) is not going to make a bit of difference. It's probably a rounding error in your calculator.

"and the second one I dont know what to do with the right side of the schematic"

What effect will the diode have on the current flow?


Thread Starter


Joined Sep 18, 2011
perhaps its better to rephrase and condense my problem....I have a hard time seeing the less obvious parallel connections, and in order to reduce and return this schematics I need to be able to redraw them. Unfortunately this is where my mind doesnt work so well....because I'm never sure if my redrawn schematic is accurate or not...

Usually what I try to do is to "stretch" out the circuit so one end will have the source voltage and the other will have 0V. Inbetween are all the relevant resistors and series/parallel connections....but for some of the schematics I find it hard to do this an idiots approach and is there a more effective way to take apart schematics and simplifiy them?


Joined Dec 26, 2010
There is no substitute for practice. People vary in their individual aptitudes for this sort of thing, but most will improve with repeated exercises. The French have a saying, in loose translation something like "It's by beating the iron that one becomes a blacksmith". Whether we are interested in circuit theory or learning a foreign language, this is a good maxim.

To spot parallel paths, simply look for any branches which have the same starting an ending points. Are they really in parallel, or do connections at other points along the branches make a more complex circuit. For instance, the resistors in a Wheatstone bridge circuit cannot always be reduced to a single resistance by simple series-parallel conversion, except in certain cases.

  1. If there is no resistance in the central "detector" position, i.e. only four resistors are present. One of your examples shows this.
  2. If the "detector" or central branch position is short-circuited (wired across).
  3. If the bridge is in perfect balance (the central branch passes no current, so may be neglected).
  4. If one or more resistors is zero or infinite (a bit artificial), or so large or small that it may be regarded as such.
Last edited:

panic mode

Joined Oct 10, 2011
what method are you trying to use?
if you are not experienced, stick with nodal or loop analisys (even if it means one or two more steps).

i suggest being systematic and starting by marking all components, nodes etc.
for example making center and upper right corner node 'c' in question 5.
positive terminal of the 35V source can be marked 'd'.

it is cleaner (and easier to communicate) by refering to some resistor as "R6" instead of "the 4.7k in upper right corner".

then write equations....

and then solve the system of equations.

I agree with previous post about rounding errors. if you get result within 5% of answers in the book, you did it right.

as for combining circuit elements, use systematic aproach, mark what you did (on the schematic enclose resistors you combined and mark new sysmbol, for example R1+R2=R12 or R3||R6=R36 etc.)

in the second question (Q6) i would eliminate resistors starting from upper left corner:
R1=(1200+2700)=3900 Ohm
R2=(1800+3900)=5700 Ohm

R12=R1||R2=3900*5700/(3900+5700)=2316 Ohm

R123= R12+R3=2316+1500=3816 Ohm (R3=1k5)

then we have 4.7k and 6.8k in parallel
R45=4700*6800/(4700+6800)=2779 Ohm

R12345= 3816 + 2779 = 6595 Ohm

This is something you can redraw and notice that is in parallel with 5.6k

R123456=5600*6595/(5600+6595)=3028 Ohm

this is now in series with 2.2k so
R1234567= 2200+3028=5228 Ohm

current through 2k2 resistor is

this splits into I1 (actually -I1) and current through 5.6k



Joined Dec 26, 2010
What you say is good advice, generally... However the OP may be in the business of having to answer questions rapidly, perhaps in an on-line situation in which exact numerical answers are all that counts. In that case 5% accurate may be no good at all.

We don't know where the OP is, but for instance in India there seems to be a strong emphasis on this kind of exercise, maybe not terribly relevant to real life practice, but the course requirements are outside a student's control.

Of course, I may be entirely wrong about this, but if the student is expected to know short-cut methods to solve these puzzles quickly, he had better learn these tricks.