Give magnitude and direction of current flow

Click For Summary

Homework Help Overview

The discussion revolves around a circuit analysis problem involving current flow, voltage across resistors, and the application of Kirchhoff's rules. Participants are attempting to determine the current through various components of the circuit and clarify the configuration of the batteries.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating current and voltage using Ohm's law and Kirchhoff's rules. There are attempts to clarify the configuration of batteries and their effects on total voltage. Questions arise about the direction of current flow and how to analyze the circuit correctly.

Discussion Status

Some participants have provided checks on calculations and clarified misunderstandings regarding battery configurations. There is ongoing exploration of how to apply Kirchhoff's rules to the circuit, with various interpretations being discussed. Guidance has been offered regarding the relationships between currents in different branches of the circuit.

Contextual Notes

Participants note confusion regarding the setup of the circuit and the implications of opposing battery voltages. There is mention of the problem being outside the scope of what has been taught, which may affect understanding and analysis.

illidari
Messages
46
Reaction score
0

Homework Statement



Problem 3
diagram.png


Homework Equations


The Attempt at a Solution



1) Find I and voltage across resistor 5

V=IR
12.6 = I * 5.08
I=2.48

.08*2.48= .1984
12.6-.1984= 12.4 V across battery.

So would I and V across the resistor be I=2.48 and V=12.4?

2) Find current I
(Forgot to mark, I is going clockwise)
Total resistance is 20. Would I simply add the two batteries making V=20?

V=IR 20=I20 I=1?

3) Find current through each resistor. Give magnitude and direction of current flow.

Kind of lost on this one. I have a feeling I need to do kirchhoffs rule. How would I handle the battery in the center? I only have examples with the center having a resistor to go off from. The flow was also specified for me. How would I determine the flow if I am not given that information? :(

So far I took the loop counter clockwise from positive end for the top part and current heading towards the right in the center.

12V-3I(1) -6I(2) + 3V= 15V -3I(1)-6I(2) = 0

Bottom counter clockwise from negative end.
-6V-2I(3) +6I(2) -3V = 0
-9V - 2I(3)+6I(2)=0

Look anywhere right so far?
 
Last edited:
Physics news on Phys.org


Number one looks OK. Checking the rest now.

For number 2, the total of the batteries voltage is 18V not 20. The total resistance is correct at 20 ohms.

(Note, your batteries would resist each other in this configuration you have drawn giving a total voltage of 6V. I take it that they aren't wired like you have drawn it? Remember, positive is the long leg on the symbol.)

V=IR and you now know V = 18V and R = 20ohms. You can take it from there.
 
Last edited:


jarednjames said:
Number one looks OK. Checking the rest now.

For number 2, the total of the batteries voltage is 18V not 20. The total resistance is correct at 20 ohms.

(Note, your batteries would resist each other in this configuration you have drawn giving a total voltage of 6V. I take it that they aren't wired like you have drawn it? Remember, positive is the long leg on the symbol.)

V=IR and you now know V = 18V and R = 20ohms. You can take it from there.

Teehee got the 20 from missadding that.
Yeah I wasn't sure which configuration would result in them opposing or not. I drew the diagram correctly. 6/20 = I

Thanks so much
 


illidari said:
Teehee got the 20 from missadding that.
Yeah I wasn't sure which configuration would result in them opposing or not. I drew the diagram correctly. 6/20 = I

Thanks so much

So they should be opposing? Then it is 6 not 18 as in my example.

Also, are number 3 supposed to be opposing too? I'll look at 3 once you can confirm this.
 


yeah it is opposing as well
 


For circuit 3, the voltage will be equal across all items in the circuit. It is the current through each item which changes.

You will have i1, i2 and i3. Where i1 = i2 + i3.

i1 = Vtotal / R1

and so on for the whole circuit.
 


You need to analyse this circuit using Kirchoff’s Second Rule (loop rule).
 


Would the V total be 3?

12 opposed by the 3 and 6 , 12-3-6= 3

V=IR

3=3I
I=1 across top resistor

I(2) + I(3) = 1

Would the I(1)=I(2)+I(3) look something like this?
hw3.png


Got pulled away from computer, working on this now. If you happen to see is this, was this what you were trying to say?
 
Last edited:


In a parallel circuit the voltage is the same across it. So no matter which resistor you measure the voltage across you will read the same value.

It is the current which will vary around the circuit.

In your diagram above, instead of V you should have I, aside from that it is correct.

V1 in your diagram = V2 + V3 But remember, this is current not voltage (again, should be I not V).
 
Last edited:
  • #10


So far as voltage goes, I'm not sure how it works in parallel if you have different voltages.

https://www.physicsforums.com/showthread.php?t=67509 shows it isn't a good idea and so far I've found nothing that explains it.

I think you need an electrical engineer to help with this one. Seems a bit complicated for a homework question.

Sorry I can't help you further.

Jared
 
  • #11


Yeah I got in a hurry copying from paper to paint and meant the V to be I.

Its the only homework problem from my lab like this and this wasn't even taught to us. *sad face*
 
  • #12


This circuit wouldn't work (even without the 12V battery) in real life, it is a pointless exercise.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
14
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K