# I can't understand the result of this circuit

Discussion in 'General Electronics Chat' started by Oussama Zaidi, Apr 24, 2016.

1. ### Oussama Zaidi Thread Starter Member

Mar 1, 2016
53
0

Can anyone explain to me how can I get 5V from this circuit?? I simulated it in different programmes and I get the same results but I can't figure why, can anyone explain to me ??

2. ### hrs Member

Jun 13, 2014
86
7
R4 and R5 are parallel so you get a 5k/1k voltage divider.

3. ### Lestraveled Well-Known Member

May 19, 2014
1,957
1,218
Because one end of R4 and R5 are at the same voltage and the other ends are tied together, they are effectively in parallel.

Do you understand this?

4. ### crutschow Expert

Mar 14, 2008
13,498
3,374
5V across 1kΩ is 5mA
(30V-5V) = 25V across 10kΩ is 2.5mA.
2.5mA +2.5mA = 5mA.
Thus the current through the two 10kΩ resistances balances the current through the 1kΩ resistor as required for the observed voltages.
What did you think the voltage should be?

5. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Apply KCL at that node. What does the voltage at that node have to be in order for KCL to be satisfied?

6. ### DGElder Member

Apr 3, 2016
347
87
If you draw the schematic with V3 and R4 on the right side you can more easily see that the circuit is symmetric about R6. So it becomes obvious that the current through V1 has to be the same as the current through V3. You can also see that both currents meet and flow through R6. So if we call the magnitude of the currents through each voltage source X, then that relationship can be expressed as......

30 - (10K*X) = 1k*2*X

then solve for x

30 - 10KX = 2KX

30= 12KX

X= 30/12K = 2.5 ma

So...

V = 2 * 2.5ma * 1K = 5V