Current in circuit with resistor and two batteries

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  • #1
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Homework Statement



what is the current in each of the branches in this circuit:

[PLAIN]http://img593.imageshack.us/img593/893/resisotrcurrentquestion.th.jpg [Broken]

http://img593.imageshack.us/i/resisotrcurrentquestion.jpg/

Homework Equations



v=ir
r1+r2 = r3
1/r1 + 1/r2 = 1/r3
v1 + v2 = v3 ?

The Attempt at a Solution



im guessing this has something to do with simultaneous equations but im not sure about the extra battery. And which circuits am i supposed to pick, because i can see 4 different paths (2 for each battery) and am not sure if i need to have the battery's in different paths or not.

basically i'm very confused any help would be appreciated
thank you
 
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Answers and Replies

  • #2
gneill
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Yes, there will be simultaneous equations to solve.

What circuit analysis methods have you studied so far? Is this question from a section of your coursework where a particular method is being discussed?
 
  • #3
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he explains by saying imagine theres a guy walking around the circuit and you record the voltage rises and drops so +20 for a battery -10v and -10v for two resistors, but he has never done this with a parallel circuit let alone one with two batteries, its an assignment to make us learn by ourselves, something about learning the same thing in different ways
 
  • #4
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This is a pretty basic "super position" problem...

basically solve it as if were a series parallel circuit.... twice
remove V2 and replace it as a "short" then solve for current through each resistor...
then replace V2 and remove V1 replacing it as a short...

Current flows from the negative side of the battery.... Give me like 10min to run the numbers and I can show a better example...
 
  • #5
gneill
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he explains by saying imagine theres a guy walking around the circuit and you record the voltage rises and drops so +20 for a battery -10v and -10v for two resistors, but he has never done this with a parallel circuit let alone one with two batteries, its an assignment to make us learn by ourselves, something about learning the same thing in different ways
Okay, that would be a KVL method (Kirchhoff Voltage Loop).

Your circuit comprises two obvious loops. Sketch in a separate current loop for each, then do the "walk" for each. Be sure to take into account that components shared by two loops will have both currents going through them, so both currents will appear in the terms for the voltage drops (or rises) for such components. This will result in two equations in two unknowns (the two independent loop currents).
 
  • #6
gneill
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Here's your circuit with two current loops sketched in:
 

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  • #7
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gneill's sketch kinda shows the current loop for V1 the left circle is should go clockwise...as current flows from the negative side.. basically the direction of current flow...and with V1 in the circuit and no V2...
*edit start
for V1 R4 R3 would have total current...your current would follow clockwise on the left.. counter clockwise on the right path... for V2 your current would flow clockwise on both sides of the sketch as it will flow out the negative and split at the parallel branches.. with R5 and R6 having total current of V2
*end edit
example... solved current for V1...
I2 153.85mA up
I3 461.54mA down
I4 461.54mA down
I5 307.69mA left
I6 307.69mA up

then Solve for current through each resistor for V2 with no V1... and where current flow is the same... the current add's (in this circuit would only be the currents through R2 adding) if current flow is opposite of each other.. they subtract... when your done... you have your current totals for each resistor...
 
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  • #8
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so the equations for the two loops you indicated would be

+ 12V – 4V – (I*3) – I*5 – I*1 = 0
+4V – I*1 – I*5 – I*8 = 0

i've done something wrong as these equations do not work
 
  • #10
gneill
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First, use two separate currents. I've indicated i1 and i2 on the diagram.

In your first equation (which appears to be for loop 2), you should have an entry for every component that the loop current passes though. So, walking around the loop,

+12V - i2*R6 - i2*R5 - (i2 - i1)*R4 - (i2 - i1)*R3 - 4V = 0

Once you've done that you can do the algebra to collect the i1 and i2 terms.
 
  • #11
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V1 and V2 are only used to find "their" current through each resistor...
more like...
V1|V2 Total
I2+I2=I2total (will be the ONLY one that adds)
I3-I3=I3total
I4-I4=I4total
etc..
etc...
with real numbers
(V1I2) + (V2I2)
153.85mA+692.3mA = I2 total
(V2I3) - (V1I3)
923.07mA-461.54mA = I3 total
Since there can not be negative current flow... Just subtract from the larger number of each current value...
 
  • #12
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ok i think i understand it now

so i2 would be the reading on xmm3
and i1 would be xmm1
and xmm2 is i2 - i1
 
  • #13
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It maybe helpful to google "superposition circuits" as that is what this problem is...
 
  • #14
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yeah you got it!

most useless circuit ever... Haven't been able to find any real world examples where that is used...
just made to prove that current is able to oppose each other and there are other ways to control it than the normal ones...
 
  • #15
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ok so these are the two equations

+12V - i2*R6 - i2*R5 - (i2 - i1)*R4 - (i2 - i1)*R3 - 4V = 0
+4V – (i2-i1)R3 - (i2-i1)R4 – i1*R2 = 0

which gives me i1 = -0.1429A and i2 = 0.7143A

which is not correct
 
  • #16
gneill
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For your second equation you should have

+4V - (i1 - i2)R3 - (i1 - i2)R4 - i1*R2 = 0.

Note that you always count a voltage drop when the current passes through the component in the direction that you "walk" around the loop. So you must have -i1*R3 and -i1*R4 and since the current in the other loop flows in these components in the opposite direction (causing voltage *rises* rather than drops), +i2*R3 and +i2*R4.
 
  • #17
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well multisim understands it....
i don't know what type of course you are learning this in.... but assume you can break down a series parallel circuit...
try it like this... 2 charts fill in the missing information...
Code:
        V              I         R
R2 1.23v       153.85mA   8
R3 461.54mV 461.54mA       1
R4 2.31V       461.54mA   5
R5 923.07mV 307.69mA       3
R6 307.69mV 307.69mA       1
T  4V            461.54mA   8.67
*edit
with V1 the voltage drop for R2 will be equal to R5+R6 since they are parallel to R2
with V2 the voltage drop for R2 will be equal to R4+R5 since they are parallel to R2
*edit end
     V    I     R
R2             8
R3             1
R4             5
R5             3
R6             1
T  12V
otherwise known as a VIRP chart... it should be treated as two independent circuits with one voltage source each.... THEN just simple math to find the values... nothing near so complex... simple ohms law..
 
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  • #18
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thank you both for your help

i managed to get the correct answers and more importantly gained more understanding on the topic.
 
  • #19
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For your second equation you should have

+4V - (i1 - i2)R3 - (i1 - i2)R4 - i1*R2 = 0.

Note that you always count a voltage drop when the current passes through the component in the direction that you "walk" around the loop. So you must have -i1*R3 and -i1*R4 and since the current in the other loop flows in these components in the opposite direction (causing voltage *rises* rather than drops), +i2*R3 and +i2*R4.
can you say whats the answer and how to solve the equation?
 
  • #20
gneill
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can you say whats the answer and how to solve the equation?
Nope. Refer to the Forum Rules that you agreed to abide by when you signed up :wink:

We can, however, help by guiding you to solve the problem yourself if you first show some effort.
 

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