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I'm working out textbook examples to get better at beginner analysis. I'm having trouble with this circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

Steps I've tried:

1) 10k*20k / 30k = 6.6666...k ohms
2) 5k*10k / 15k = 3.3333...k ohms
3) those are now in series, for 10k ohms equiv on the left side
4) from the right, 8k+4k = 12k ohms
5) 144k / 24k = 6k ohms
6) 6k + 4k = 10k ohms equiv. on the right side
7) now the 3k ohm resistor is in parallel with both the 10k ohm equiv resistors
   on both side, so I know the voltage across both 10k ohm resistors is 16V.  
   Calculating current, I find that current is 1.6mA across each combined 10k 
   ohm resistor
8) now, splitting the 10k ohm resistor into the 6k and 4k in series, i know they
   both share 1.6mA current; thus V1 = 1.6mA * 4000 ohms = 6.4 V.

I stopped here because I got the wrong answer. Will someone please help me figure this out?

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  • \$\begingroup\$ step 7: both 10K are in parallel, so you get 5k which is in series with 3K. I= 16/8 = 2mA. step 8: splitting 2mA in 10K resistors, you get 1mA. so, V1 = -1*4 = -4V \$\endgroup\$
    – hassansin
    Jun 12, 2014 at 5:48
  • \$\begingroup\$ Thanks! I get it... 2mA at 8k ohm resistor, split that into the 5k + 3k in series, they share the current, so the voltage across the 5k is 10V. Split into both 10k ohm equiv resistances and they are in parallel, sharing common voltage at 10V, so the current across both 10k ohm resistors is 1mA. Split the right branch into 4k + 6k and V1 = 4k*1mA = 4V, but because of the polarity marked, it's -4V. ------> Now, I'm sure I can do the rest. However, is there a faster way or is this the only way? Also, can you explain why the 3k ohm resistor is not in parallel with both 10k ohm equivs? \$\endgroup\$
    – asdf
    Jun 12, 2014 at 5:57

1 Answer 1

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After step 6, your circuit will be as shown below.

schematic

simulate this circuit – Schematic created using CircuitLab

Which can be further reduced as a 16V battery in series with 3k and 5k resistors.

Now, you will have 10V across this 5k resistance, ie, 10V across each parallel 10k. Splitting 10k into 6k+4k, you will have 4V across 4k resistance. From the polarity of v1 marked, V1= -4V.

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  • \$\begingroup\$ Thanks! Can you please explain why the 3kohm is not in parallel with the two 10k ohms? Because of the voltage source? (Brand new to circuits..!) \$\endgroup\$
    – asdf
    Jun 12, 2014 at 6:01
  • \$\begingroup\$ Starting from one terminal of a source, if current has more than one independent path to reach the other terminal, then those two paths are parallel. And these paths will have same voltage across them. \$\endgroup\$
    – nidhin
    Jun 12, 2014 at 6:08
  • \$\begingroup\$ Aha, thanks! I'm going to YouTube a video on it, but your explanation makes it more clear. \$\endgroup\$
    – asdf
    Jun 12, 2014 at 6:10
  • \$\begingroup\$ One more Q: Is there a better way to do this or does this circuit require all this collapsing to utilize the 16V source? \$\endgroup\$
    – asdf
    Jun 12, 2014 at 6:16
  • \$\begingroup\$ Sorry about that. I don't have a better way to suggest. \$\endgroup\$
    – nidhin
    Jun 12, 2014 at 6:20

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