Potential -- Find the potential at node 3

  • Thread starter Thread starter Rectifier
  • Start date Start date
  • Tags Tags
    Potential
AI Thread Summary
The discussion centers on determining the potential at node 3, given that the potential at node 1 is 0. Two approaches yield different results: moving from node 1 to node 3 suggests a potential of -8V, while moving from node 3 to node 1 indicates a potential of 8V. The inconsistency arises from the labeling of voltages, which confuses the direction of potential changes. Ultimately, using Kirchhoff's Voltage Law (KVL) confirms that the potential at node 3 is 8V. The confusion highlights the importance of standard voltage labeling in circuit analysis.
Rectifier
Gold Member
Messages
313
Reaction score
4
The problem
What is the potential at node 3, if the potential at node 1 is 0?

XEOtcWF.jpg


The attempt
If I start at node 1 and move towards node 3
1->2
-5V
2->3
+(-3V)
1->3
-5+(-3)=-8V

If I start at node 3 and move towards node 1
3->2
-(-3)
2->1
5
3->1
3+5=8V

I am not really sure what is right here.
 
Physics news on Phys.org
Rectifier said:
The problem
What is the potential at node 3, if the potential at node 1 is 0?

XEOtcWF.jpg


The attempt
If I start at node 1 and move towards node 3
1->2
-5V
2->3
+(-3V)
1->3
-5+(-3)=-8V

If I start at node 3 and move towards node 1
3->2
-(-3)
2->1
5
3->1
3+5=8V

I am not really sure what is right here.

The problem looks overconstrained and inconsistent. Where is it from?

And when you go in the direction of - to +, that is an increase in potential. I get 2 different answers for going the 2 different ways from node 1 to node 3...
 
KVL in the loop, starting at 3..
3->2
-3v(rise)
2->1
-5v(fall)
1->4
-9v(fall)
4->3
17V(rise)
-3-5-9+17=0, so the voltage in the loop adds up.

Now from V3->V2->V1
V3-3-5=0
V3=8V
Now from V3->V4->V1
V3-17+9=0
V3=8V

Seems okay to me?
 
bennyq said:
KVL in the loop, starting at 3..
3->2
-3v(rise)
2->1
-5v(fall)
1->4
-9v(fall)
4->3
17V(rise)
-3-5-9+17=0, so the voltage in the loop adds up.

Now from V3->V2->V1
V3-3-5=0
V3=8V
Now from V3->V4->V1
V3-17+9=0
V3=8V

Seems okay to me?

Ah, I see what fooled me. They labeled the right-hand voltage as negative, instead of relying on the +/- signs to determine the direction of the voltage. That's very non-standard in electronics, but I guess I should have seen that trick part of the question. Thanks for clearing that up! :smile:
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top