Solving Circuits: Understanding Parallel and Series Connections

  • Thread starter Thread starter smarthbehl
  • Start date Start date
AI Thread Summary
The discussion centers around a request for help with a physics problem involving a circuit with resistors in series and parallel configurations. The user seeks to find the current through a 6-ohm resistor and the potential difference across a 12-ohm resistor, as well as the ammeter reading after a switch is closed. Respondents emphasize the importance of showing effort in solving the problem before seeking assistance, highlighting that simply copying answers does not facilitate learning. The community encourages users to draw the circuits and attempt solutions independently. Overall, the focus is on understanding circuit principles rather than just obtaining answers.
smarthbehl
Messages
3
Reaction score
0
hi guys
Need some help
If u can provide
Plzzzzzzzzzzz

I think its a very tough ques
anyone here has guts to solve this
its Physics
electricity
The circuit

////////-represents wire
restances of resistors have been written
Total pd is 4v

///////////////_4v__///////////////
/ /
///////////6ohms///////4ohms///
/ /
/ /
////////12ohms////////3ohms///

U have to find the
most important: Explain this entire circuit that where its parallel and series
1. current through:6ohm resistor
2.pd across 12 ohm resistor




Please also try this
in a diagram given below the cell and ammetre have negligible resistance. The resistors are identical.With the switch k open ammetre reads 0.6 ampere
What will be ammetre reading after the switch is closed
DIagram:
/////=== represents wire



///Ammetre////////////////battery///////////////////
/ /
////////////////Resistance////////////////////////////
/ /
////////////////Resistance////////////////////////////
/ /
///////////Resistance/////K open switch///////////








I ADVICE U TO DRAW THESE CIRCUITS IN UT COPY AN THEN SOLVE
and explain details of both circuits

Advance thanks
For details pm me
Or contact me immediately if u can solve this at::
<personal email deleted>
Advance thanx
 
Last edited by a moderator:
Physics news on Phys.org
Please have the "guts" to do your own work and show it before asking for help. We are not here to do your work for you.
 
no tat not the point
i just need solution of that
its not like that
 
If you read the https://www.physicsforums.com/showthread.php?t=5374", which you agreed to when you signed up, you'll find
Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

Besides, don't you realize that you're not going to learn anything by just copying the answers? If you show your thoughts/ideas on this problem, we'll be glad to help :smile:
 
Last edited by a moderator:
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