Can someone explain how to find stuff with 2 Voltmeters?

[doublea500]

Need help finding Current on one resistor when there are 2 batteries

Homework Statement

Find Current in the 3 ohm resistor

Sorry it didnt come out right. ill explain what it is. 12V meter on the left, + on top - on bottom. then if you go counter-clockwise you will hit a 6ohm resistor, keep going and you will hit a 1 ohm resistor. on the right there's a 6v +top - bottom. in the middle between the 6ohm and 1 ohm there's a 3 ohm separating it comming out.

Imma take a picture

Homework Equations

V=IR... and others

The Attempt at a Solution

I Tried using kirchhoffs law and i kinda got lost. i was trying to do the 2 boxes seperately like 3I+6I=12V so I1=12/9 (1.3333) and 3I+1I=6V so 41=6v so I2=6/4. (1.5). i got stuck there because i didnt know what to do next


The answer for this is 1.8

can someone answer how to do this and tell me the basic rules when there are 2 voltmeters? because the 2 voltmeters really confuses me.

 

[jhae2.718]

Those aren't voltmeters, but rather batteries--sources of electric potential.

You need to use Kirchoff's loop and junction rules, [tex]\textstyle \sum \Delta V = 0[/tex] and [tex]\textstyle \sum i = 0[/tex] (at a junction).

You know the potentials of the batteries and the resistances of the resistors. How can you use Ohm's law and Kirchoff's laws to create a set of equations in terms of voltage? Note that the current changes at the junctions.

 

[doublea500]

Those aren't voltmeters, but rather batteries--sources of electric potential.

You need to use Kirchoff's loop and junction rules, [tex]\textstyle \sum \Delta V = 0[/tex] and [tex]\textstyle \sum i = 0[/tex] (at a junction).

You know the potentials of the batteries and the resistances of the resistors. How can you use Ohm's law and Kirchoff's laws to create a set of equations in terms of voltage? Note that the current changes at the junctions.

eh, not sure but is it 3I1+6I2=12V and 3I1+I3=6v?

Thats for [tex]\textstyle \sum \Delta V = 0[/tex] how can i get the [tex]\textstyle \sum i = 0[/tex] part though? and I am not sure what you mean by the the sum of I is 0 at a junction

 

[jhae2.718]

Look at the "T" shaped part of wire above the 3 Ohm resistor. If current i is flowing in, and two different currents (say, i₁ and i₂) flow out, if the sum has to be zero, what does this say about the relationship between the three currents?

(This requirement derives from conservation of charge.)

 

[doublea500]

so 3I1=6I2+1I3?

i keep using the loop rule but i am getting nowhere

Also, should i be using the same I or different Is like I've been doing?

 

[jhae2.718]

Different currents.

 

[doublea500]

ok, how bout I1=I2+I3 whereas I1 is amp for 3 ohm resistor, I2 is amp for 6 ohm resistor, i3 is amp for 1 ohm resistorSo here's the formulas i have so far.

12=3I1+6I2
6=3I1+I3
I1=I2+I3
Ahh i got it... here's what i did, tell me if there's a simpler way. i combined these formulas. and used subsitution first to get rid of I3, made I2=2-.5I1 in the first equation then subsitute that into the second equation then i got the I1 value on one side and everything else on the side. (4.5I1=8V) then solved for I1 and got 1.7777777 which i guess got rounded to 1.8. thanks for the help. is there a simpler way of doing this though?

 

[jhae2.718]

You usually have to solve the system. In some cases, you can use equivalent resistance to simplify the circuit.