Voltage divider (circuit analysis)

[Miike012]

How am I suppose to use voltage divider to solve for V1?

The first pic is the original pic
The second pic is me solving for V1,Vx, and V2

The third pic is and equivalent circuit of the first pic... I did this to reason that V1 is the voltage across the 10k ohm resistor.

Therefore using voltage divider...
V1 = (Vx - V1)(10/15) = -30(10/15) = -20V.

Is that the correct approach or is there an easier way?

 

[lightgrav]

Presuming that the polarity shown on the 10k device is the polarity that appears across the diamond,
loop rule is 25V - V_10k - ½V_10k + ¼V_10k = 0 , so that V_10k = 20V = -Vx. (otherwise Vx=14.28V)
(but why isn't their V1 = 25V - V_10k = 25V - 20V = 5V?)

 

[Miike012]

Presuming that the polarity shown on the 10k device is the polarity that appears across the diamond,
loop rule is 25V - V_10k - ½V_10k + ¼V_10k = 0 , so that V_10k = 20V = -Vx. (otherwise Vx=14.28V)
(but why isn't their V1 = 25V - V_10k = 25V - 20V = 5V?)

im confused as to how you came up with the equation 25V - V_10k - ½V_10k + ¼V_10k = 0 .

shouldnt it be...
Eq 1.
25 - V_x/4 + V₅ + V₁₀ = 0 ?
and
V₁₀ = V_x.


Next...

V₅ = I(5kohm) and I = V_x/(10k ohm).

Now from Eq 1.

25 - V_x/4 + V_x/2 + V_x = 25 + 5/4V_x = 0

and V_x = -25*4/5 = -20



(but why isn't their V1 = 25V - V_10k = 25V - 20V = 5V?)...

I added the third picture in my paint document that shows that V1 is the voltage across the 10 ohm resis.