Trying to find the voltage of a circut

  • Thread starter Thread starter LocationX
  • Start date Start date
  • Tags Tags
    Voltage
AI Thread Summary
To find the voltage in the circuit, it's crucial to track the signs correctly when analyzing loops, as this affects the results significantly. The voltage across V3 is influenced by the 10V source being shorted, indicating a specific relationship. The effective voltage from the 10V and 12V sources acting in series opposes the 24V source, which informs the voltage across V1. Algebraic errors were noted in the original equations, particularly regarding V3 and the changing meanings of variables. Understanding the signs and directions in the equations is essential for accurate voltage calculations.
LocationX
Messages
141
Reaction score
0
I'm trying to find the voltage of a circut.

Could someone verify if I did this correct?

Questions and answers http://photobucket.com/albums/b242/physicslatex/?action=view&current=3bd3873d.jpg
 
Last edited by a moderator:
Physics news on Phys.org
You're not keeping proper track of the signs. It makes a difference when you make a loop whether you're going through from + to - or from - to +. (You also have a couple of small algebra errors. By your equations, V3 should be a -10V, and I'm not sure about the others because your variables seem to change meaning partway through. This is a common problem, especially when you're using a lot of things that look very much alike.)

Let's look at this logically. Considering your diagram, the 10V source is shorted across v3. What does that tell you about the voltage across v3?

Now - if we removed the v2 and v3 tracks, you have the 24V source opposite to the 10 and 12v sources, which are acting in series. This means that the 10 and 12v sources are acting like an effective 22V source, opposing a 24V source. What would this tell you about the remaining voltage to be dissipated across v1?

The equations I used:

10 - v3 = 0 (loop inside the right-hand circuit, going counter-clockwise)

12 + 10 + v1 - 24 = 0 (loop around the outside, going counter-clockwise)

24 - v1 + v2 = 0 (loop inside the upper left-hand circuit, going counter-clockwise)

It's more a coincidence than anything else that I went counter-clockwise on all three. All that would change would be the signs on the numbers. Notice also that v1 changes sign between equation 2 and equation 3, reflecting the fact that I'm going a different direction through it in each case.

One other point: if you end up with a negative voltage somewhere, it simply means you chose the wrong signs for either end of the circuit element. This is clearly the case for V2 - can you see why V2 is obviously mislabelled?

Does this help?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top