Engineering How Can You Find V1 and V2 in a Circuit Without Mesh Analysis?

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To find V1 and V2 in a circuit without mesh analysis, KCL can be effectively applied. By analyzing the node above the 20k and 5k resistors, it is determined that 10 mA flows in and out, leading to a voltage drop across the equivalent 4k resistor of 40 volts. However, upon further calculation, V1 is found to be 20 volts, not 40 volts, as the two resistors in series have an equivalent resistance of 4k in parallel with another set of 4k resistors. The current splits equally due to the equal resistances, confirming the voltage across V1. Using the known V1, V2 can be calculated through voltage division.
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1. I made a quick sketch
http://img352.imageshack.us/img352/7629/testwv7.jpg
Find V1 and V2






3. Is there an easier way to do this question without applying mesh analysis, because this question was in my chapter 2 (intro to KVL/KCL), but i cannot think of a way to find this unless i use mesh analysis. Any advice
 
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Well assuming I'm doing it right of course, you wouldn't need any simultaneous equations or anything, that'd be over complicating it, just need KCL really

so at the node above the 20 and 5 k resistors(or the equivalent 4k resistor!)you have 10 milliamps flowing in, and you need 10 mA flowing out, obtained with a voltage drop across that 4 k of just V=IR=10*4=40 Volts

You can check to see that puts 8 mA across the 5 k resistor and 2 across the 20 k, for a total of 10 mA
 
the answer for v1 = 20 V

not 40 V

so hmmmmmmmm
 
ok, so what you basically have are two resistors in series, and those two resistors(with eq. resistance of 4k) are in parallel with the other two resistors which I mentioned have an equiv. resistance of 4k as well

Since they're in parallel the current splits down both paths and equal amounts both ways since the resistances are ultimately equal, sooo it's actually V=I*R=5*4=20Volts
 
If you know V1, then all you need to do is "voltage-divide" to find V2.
 
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