Is voltage drop different in parallel circuits?

AI Thread Summary
Voltage drop in parallel circuits is equal to the voltage of the power supply, not the same as in series circuits. Each resistor in a parallel circuit experiences the same voltage drop as the power supply. There is confusion regarding how to calculate voltage drop and current flow when components like ammeters are involved. Removing an ammeter from the circuit raises questions about whether current will continue to flow. Additional resources, such as educational websites, can provide further clarification on these concepts.
TonyG247
Messages
2
Reaction score
0
★★ members must not remove essential diagrams after receiving help on these forums ★★
Screen+Shot+2016-08-13+at+1.45.47+PM.png

https://puu.sh/qzv6o/5abc45494b.png

I was under the impression that voltage drop was the same in parallel which would make it the power supply, but apparently not? Any insight would help. I know this isn't a hard a question, I'm just horrible at physics.
 
Last edited by a moderator:
Physics news on Phys.org
TonyG247 said:
I was under the impression that voltage drop was the same in parallel which would make it the power supply, but apparently not?

Can you reword this? I think I know what you're saying, but I don't think you're saying it very well.

The same in parallel compared to what? To series? No, but I don't think this is what you meant. I think you meant it's the same as the power supply voltage. Is this correct?
 
Student100 said:
Can you reword this? I think I know what you're saying, but I don't think you're saying it very well.

The same in parallel compared to what? To series? No, but I don't think this is what you meant. I think you meant it's the same as the power supply voltage. Is this correct?

My apologies. Yes, the same as the power supply. And if that isn't the case, at least the same in each of the resistors. But the answer doesn't seem to be correct? I'm lost on how I should go about calculating the voltage drop with the provided information.
 
Say I pulled out what I can only assume is an ammeter, breaking the circuit there? Would current still flow?
 
Last edited by a moderator:
Thread reopened.
 
Last edited:
  • Like
Likes informatory
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top