Circuit with multiple batteries finding the current through each one

[tmint]

Diagram: http://gyazo.com/0b719e6d399ad894d2c48ce06d994672

Homework Statement

In the figure the resistances are R1 = 0.77 Ω and R2 = 1.8 Ω, and the ideal batteries have emfs ε1 = 2.1 V, and ε2 = ε3 = 3.5 V. What are the (a) current (in A) in battery 1, (b) the current (in A) in battery 2 and (c) the current (in A) in battery 3? (d) What is the potential difference Va - Vb (in V)?

Homework Equations

V=IR
Kirchoff's Loop Rules

The Attempt at a Solution

Okay, for some reason I solved a part of this or two before and hoped to get back to it.
For part (a), I set up the equation of I being the current and R being the resistance; B being battery:

I= B2-B1/(4R1+R2)

And I got the answer .287A and then multiplied that by 2 and got the first part which is .587A. I understand the logic of its the potential difference divided by the equivalent resistance but why is that multiplied by 2, any reason for that? I didn't understand.

For parts (b)+ (c), I tried to set it up in a similar way but they are both wrong:

For part (b): B2-B1/(4R1+R2) and I got .287A

For part (c): I did the same equation as above putting in B3 instead of B2 and got .287APart (d) I was able to do by finding the potential difference between points (a) and (b) which I took the .287A value that I calculated in (b) and (c) and set it up like this:

ΔV=-I2R2^2+(3.5V) which comes out to 2.98V which was correct. I could use some help with this problem, I understand how the equation to solve B is found out, but I don't understand why you would have the EQ resistance to find the first battery, rather than the resistance the battery goes through. Could use a nice explanation on that. Thanks!

If someone could help with doing it the Kirchhoff rule way, I tried to assign currents with directions but for some reason I get something completely different from the correct answer I got in part (a).
I have a test on this material Tuesday and I REALLY need to do well, I blanked out on the first exam and forgot everything and did really, really bad :uhh:.

I really appreciate it! Thank you!

 

[SammyS]

Diagram: http://gyazo.com/0b719e6d399ad894d2c48ce06d994672

Homework Statement

In the figure the resistances are R1 = 0.77 Ω and R2 = 1.8 Ω, and the ideal batteries have emfs ε1 = 2.1 V, and ε2 = ε3 = 3.5 V. What are the (a) current (in A) in battery 1, (b) the current (in A) in battery 2 and (c) the current (in A) in battery 3? (d) What is the potential difference Va - Vb (in V)?

Homework Equations

V=IR
Kirchoff's Loop Rules

The Attempt at a Solution

Okay, for some reason I solved a part of this or two before and hoped to get back to it.
For part (a), I set up the equation of I being the current and R being the resistance; B being battery:

I= B2-B1/(4R1+R2)

Hello tmint. Welcome to PF !


I get something different for I₁, the current through battery 1.

Show what equations you get from Kirchhoff's Laws.

(Also, your notation is a bit confusing.)

And I got the answer .287A and then multiplied that by 2 and got the first part which is .587A. I understand the logic of its the potential difference divided by the equivalent resistance but why is that multiplied by 2, any reason for that? I didn't understand.

For parts (b)+ (c), I tried to set it up in a similar way but they are both wrong:

For part (b): B2-B1/(4R1+R2) and I got .287A

For part (c): I did the same equation as above putting in B3 instead of B2 and got .287A


Part (d) I was able to do by finding the potential difference between points (a) and (b) which I took the .287A value that I calculated in (b) and (c) and set it up like this:

ΔV=-I2R2^2+(3.5V) which comes out to 2.98V which was correct.


I could use some help with this problem, I understand how the equation to solve B is found out, but I don't understand why you would have the EQ resistance to find the first battery, rather than the resistance the battery goes through. Could use a nice explanation on that. Thanks!

If someone could help with doing it the Kirchhoff rule way, I tried to assign currents with directions but for some reason I get something completely different from the correct answer I got in part (a).
I have a test on this material Tuesday and I REALLY need to do well, I blanked out on the first exam and forgot everything and did really, really bad :uhh:.

I really appreciate it! Thank you!

 

[tmint]

Thank you.

Sorry for the notation I'll write it out to make it easier to follow. When I use Kirchhoff's rules I get the following equations:
Let c be left of a, d left of b, f right of b, and e right of a

Oh, and Current 1 is current from Battery 1, Current 2 from Battery 2 in the middle branch and so on.

Current 3= Current 1+ Current 2

cabdc: -.77(Current 1)+1.8(Current 2)- 3.5-.77(Current 1)+2.1=0

aefba: -.77(Current 3)-3.5-.77(Current 3)+3.5-1.8(Current 2)

I use the first equation and solve for Current 2 since Current 2 is apart of both equations.

Current 2= Current 3- Current 1

I then plug this into the above equations after I combine common factors and I get:

-1.54(Current 1)+1.8(Current 3- Current 1)=1.4
-1.54(Current 3)-1.8(Current 3- Current 1)=0

I get the Current 3-1 to cancel out, but I'm left with two different currents and I'm stuck. Any idea where I went wrong?

 

[SammyS]

Thank you.
...

Current 3= Current 1+ Current 2

cabdc: -.77(Current 1)+1.8(Current 2)- 3.5-.77(Current 1)+2.1=0

aefba: -.77(Current 3)-3.5-.77(Current 3)+3.5-1.8(Current 2)

I use the first equation and solve for Current 2 since Current 2 is apart of both equations.

Current 2= Current 3- Current 1

I then plug this into the above equations after I combine common factors and I get:

-1.54(Current 1)+1.8(Current 3- Current 1)=1.4
-1.54(Current 3)-1.8(Current 3- Current 1)=0

I get the Current 3-1 to cancel out, but I'm left with two different currents and I'm stuck. Any idea where I went wrong?

That all looks fine.

(By the way, you have chosen Current 3 to be opposite the direction shown in the figure. No problem however.)

You still have three equation in three unknowns but the last two are now in a convenient form.

Subtract the last two equations. (Bottom from top works nicely.)

What does that give for (Current3 - Current1) ?

 

[tmint]

Okay,

1.54[Current 3- Current 1]= 1.4

So [Current 3- Current 1]= 1.1 A

So would Current 2 by that equal 1.1 A?

Right?

Would I then sub this back in for Current 3- Current 1 and solve for Current 1 in the first equation or is that wrong?

Like so,

-1.54[Current 1]+ 1.8[1.1 A] = 1.4 and solve for Current 1?

Current 1 is .376A, that doesn't seem right? I solved this earlier, and really want to understand how its solved. Current 1 came out to .59A not sure exactly how though. My answer before fell within the 5 % tolerance

Did I go about it wrong?


Really appreciate your help.

 

[SammyS]

Okay,

1.54[Current 3- Current 1]= 1.4
...

Really appreciate your help.

Check your Algebra/Arithmetic .

1.8(Current 3- Current 1) - (-1.8(Current 3- Current 1) ) ≠ 0

 

[tmint]

Oh my mistake. So 3.6[Current 3-Current 1]=1.4

Oh you're right, this does work somewhat nicely.

So 3.6[Current 3- Current 1]=1.4

Current 3- Current 1= .389

So Current 2= .389 , I'm kind of suspicious of this value. Its differs from what I got previously. Did I mess up again in my arithmetic again?

and with that value you plug in that for the other ones right into the previous two equations to find Current 1 and Current 3, right?

 

[SammyS]

Oh my mistake. So 3.6[Current 3-Current 1]=1.4

Oh you're right, this does work somewhat nicely.

So 3.6[Current 3- Current 1]=1.4

Current 3- Current 1= .389

So Current 2= .389 , I'm kind of suspicious of this value. Its differs from what I got previously. Did I mess up again in my arithmetic again?

and with that value you plug in that for the other ones right into the previous two equations to find Current 1 and Current 3, right?

3.6[Current 3- Current 1] is only part of what you get by subtracting those equations.

It's what you left off of the left hand side when you got

1.54[Current 3- Current 1]= 1.4 .​

 

[tmint]

Oh okay,

So 1.54(Current 3-Current 1)+3.6(Current 3-Current 1)=1.4

5.14(Current 3-Current 1)=1.4

5.14(Current 2)=1.4

Current 2= .27A!

And I take this value and plug it in back,

I get Current 1= -.593 A change the direction

and I get Current 3= -.323 A change the direction!

And solve for part D which is the EMF of battery 2- (Current 2 times Resistance 2) and I got 3.01 V

Thank you so much! Makes so much more sense now! A big problem of mine was the substituting and your steps helped clear that up. Thanks for not just giving me the solution and helping me figure it out.

I'm actually having another circuit I'm having trouble with, and if you could me that'd be great. I didn't feel creating a new thread for the specific problem was the best way to go about it.

Here is the circuit I'm having trouble with: http://gyazo.com/1e11d9208da0aa29d10b72128844ad40

I set up the following Kirchhoff's equations following the same naming as the last circuit:

cabdc: -217(Current 1) +100(Current 2)+ 6.76=0

aefba: -312(Current 3)-11.8-100(Current 2)=0

Letting Current 2= Current 3-1

Which simplifies to the two equations:

-217(Current 1)+ 100(Current 3-1)=-6.76

-312(Current 3)-100(Current 3-1)=11.8

When I try to simplify this I get

(312-273)(Current 3- Current 1) + 200(Current 3- Current 1)= -18.56

I'm stuck. Could use some help. Don't know if I went about it wrong.

 

[SammyS]

...
I'm actually having another circuit I'm having trouble with, and if you could me that'd be great. I didn't feel creating a new thread for the specific problem was the best way to go about it.

Here is the circuit I'm having trouble with: http://gyazo.com/1e11d9208da0aa29d10b72128844ad40

I set up the following Kirchhoff's equations following the same naming as the last circuit:

cabdc: -217(Current 1) +100(Current 2)+ 6.76=0

aefba: -312(Current 3)-11.8-100(Current 2)=0

Letting Current 2= Current 3-1

Which simplifies to the two equations:

-217(Current 1)+ 100(Current 3-1)=-6.76

-312(Current 3)-100(Current 3-1)=11.8

When I try to simplify this I get

(312-273)(Current 3- Current 1) + 200(Current 3- Current 1)= -18.56

I'm stuck. Could use some help. Don't know if I went about it wrong.

You really should start a new thread for a new problem.

Now would be a good time to familiarize yourself with the rules & guidelines for Physics Forums.

https://www.physicsforums.com/showthread.php?t=414380

In particular look at Homework Help Guidelines .