# Which technique do I use to solve this problem?

I'm currently working on open questions for my Physics 2 exam about eltric fields, magnetic field, circuits and so on.

I'm currently stuck with this problem:

I'm asked to solve for the voltage $$U_X$$ In our previous circuits I was always able to use Kirchhoffs Law to solve for the voltage but now I'm stuck because I don't really how to apply it to this one.

It would be great if someone could help me out with this one.

• Write a KCL equation at the center node. Solve. Done. May 4, 2020 at 15:53
• Another approach is to use superposition. Since this is an exam, I won't say much more than that. May 4, 2020 at 16:30
• I think the answer is U/6 volts - do you know the required answer? May 4, 2020 at 16:35
• Source transformation and thevenins. (1/6 was a mistake - it's U/6 by my scribbled reckoning). May 4, 2020 at 16:38
• I get U/6 too by superposition. If you want to try that method (which may be what they want, since it is a physics class), set one terminal at a time to its stated value, the other three to zero. Then compute the value at the center node. Add up the four solutions. There are only two distinct circuit configurations, so it is very quick. May 4, 2020 at 17:43

Hi, I just directly realised that there is no "COM" on the circuit, Volts are mesured in relation to another point or reference. So i chosed my own reference, the lower voltage level so i have only positive VOLTS, the redraw the circuit and VOILA i have a simple Passif audio summing circuit, i think with millman analysis or node analysis you can get the voltage easily. Sometimes is just a matter of perspective to see a solution to a problem.

1. Define current flow direction, I simply define current go from 4 node to Ux
2. Set KVL Equations

(1) $$\U_x = 2U-i_1 R \$$

(2) $$\U_x = U-i_2 2R \$$

(3) $$\Ux = -2U-i_32R \$$

(4) $$\Ux = -U-i_4R \$$

1. set KCL

(5) $$\ i1+i2+i3+i4 = 0\$$

1. calculate for 5 variables with 5 equation

reform (1) to (4)

(6) $$\i_1 = \frac{(2U-U_x)}{R}\$$

(7) $$\i_2 = \frac{(U-U_x)}{2R}\$$

(8) $$\i_3 = \frac{(-2U-U_x)}{2R}\$$

(9) $$\i_4 = \frac{(-U-U_x)}{R}\$$

then put to (5)

$$\\frac{(4U-2U_x)}{2R} + \frac{(U-U_x)}{2R} + \frac{(-2U-U_x)}{2R}+\frac{(-2U-2U_x)}{2R} = 0\$$

$$\ (4U+U-2U-2U)+(-2U_x-U_x-U_x-2U_x)=0\$$

$$\U = -6U_x\$$

there for

$$\U_x = \frac{U}{6}\$$

I run into variations of this problem in real life frequently enough that I've come up with a method that is very straightforward- it's basically an extension of the voltage divider formula for n resistors with voltages (could be 0V if the resistor is grounded) and a single common point. .

$$\V_{NODE} = (V_1/R_1 + V_2/R_2 +...+ V_n)\cdot (R_1||R_2||..||R_n)\$$

In this case we can remove the common factor of U/R from the left (..) and R from the right (..) giving us:

$$\U_x= U\cdot (1/2 + 2 -1 -1) \cdot (1/2 || 1) \$$ = $$\ U \cdot (1/2) \cdot (1/3)\$$ = $$\ U/6\$$.