# I can't understand the result of this circuit

#### hrs

Joined Jun 13, 2014
369
R4 and R5 are parallel so you get a 5k/1k voltage divider.

#### Lestraveled

Joined May 19, 2014
1,946
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?

#### crutschow

Joined Mar 14, 2008
32,041
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?

#### WBahn

Joined Mar 31, 2012
28,472
Apply KCL at that node. What does the voltage at that node have to be in order for KCL to be satisfied?

#### DGElder

Joined Apr 3, 2016
351
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