Calculate the three currents through the battery

[james123]

Homework Statement

Calculate the value of the current through the 12 V battery shown in FIGURE 3.

attachment-php-attachmentid-41345-stc-1-d-1322543216-jpg.147932

Homework Equations

KCL
KVL

The Attempt at a Solution

So, I1+I2=I3
[/B]
Loop A (Left hand loop):
+13-3*I3-12-I1=0
3*I3+I1=1
Sub: 4I1+3I2=1

Loop B (Right hand loop):
+14-3I3-12-2I2=0
3I3+2I2=2
Sub: 3I1+5I2=2

4I1+3I2=1 (x3) gives: 12I1+9I2=3 (equation 1)

3I1+5I2=2 (x4) gives: 12I1+20I2=8 (equation 2)

Subtract equation 2 from equation 1 giving:
-11I2=-5

So, I2=-5/-11
I2=0.45A

For I1:
4I1+3I2=1
4I1+(3*0.45)=1
4I1+1.35=1
I1=1-1.35/4=-0.0875A

I3=I1+I2
I3=-0.0875+0.45=0.3625A

Overall:
I1=-0.0875A
I2=0.45A
I3=0.3625ACan anybody confirm if these answers are correct?
Also, Why is I1 negative? I remember reading that if I get a negative answer that it doesn't matter but do I need to do anything to it at all?

Any advice is much appreciated!

 

[Merlin3189]

I1=-0.0875A
I2=0.45A
I3=0.3625A

If voltage at bottom is 0, then voltage at top is
13-1xI₁ = 13.0875
or 12 + 3x I₃ = 12 +3x0.3625 = 13.0875
or 14 - 2x I₂ = 14 - 2x 0.45 = 13.1
So near enough, if you round off your answers.
I got $-\frac {1} {11} , \frac {5} {11} and \frac {4} {11}$ giving:
$13 \frac {1} {11}, 12 + 3 \times \frac {4} {11} = 13 \frac {1} {11} \ \ and 14 - 2\times \frac {5} {11} = 13 \frac {1} {11}$

 

[phinds]

I have not followed you math or checked your answers but in general it is advisable to write a single equation for each loop. I find it odd that you have written multiple equations per loop. I think you are making this harder than it is.

A negative answer just means the direction shown in the arrow is the reverse of what is actually happening

 

[Merlin3189]

Just checking the maths - I used slightly different route, but essentially like yours. ( My substitution is because capital I on my default font is just a vertical bar)

I substituted A=I₁, B=I₂ and (A+B)= I₃
So, I1+I2=I3 I use this from the start and eliminate I₃ from my calculations.

Loop A (Left hand loop):
+13-3*I3-12-I1=0 +13 -3(A+B) -12 -A=0 so 1 = 4A +3B
3*I3+I1=1
Sub: 4I1+3I2=1

Loop B (Right hand loop):
+14-3I3-12-2I2=0 +14 -3(A+B) -12 -2B =0 so 2 = 3A +5B
3I3+2I2=2
Sub: 3I1+5I2=2

4I1+3I2=1 (x3) gives: 12I1+9I2=3 (equation 1) 1 = 4A +3B (x3 bs) gives 3 = 12A +9B

3I1+5I2=2 (x4) gives: 12I1+20I2=8 (equation 2) 2 = 3A +5B (x4 bs) gives 8 = 12A +20B

Subtract equation 2 from equation 1 giving:
-11I2=-5 -5 = -11B

So, I2=-5/-11 So B = 5/11 = 0.4545
I2=0.45A (Rounded to 2sf, leading to eventual slight inaccuracy in another result)

For I1:
4I1+3I2=1 4A +3B =1 so 4A + 15/11 =1 so 4A = - 4/11 so A = -1/11 = 0.09090
4I1+(3*0.45)=1 3*0.4545 = 1.3635
4I1+1.35=1
I1=1-1.35/4=-0.0875A 1 -1.3635/4 = 1 - 0.34087 = 0.65913 but (1-1.3635)/4 = -0.3635/4 = -0.09087

I3=I1+I2 I₃ = A+B = -1/11 +5/11 = 4/11 = 0.3636 or -0.09.87 + 0.4545 = 0.36363 (=0.364 to 3sf)
I3=-0.0875+0.45=0.3625A

Overall:
I1=-0.0875A I₁=0.9090 A = 0.909 A to 3sf or 0.91 A to 2sf
I2=0.45A I₂=0.4545 A = 0.455 A to 3sf or 0.45 A to 2sf
I3=0.3625A I₃=0.3636 A = 0.364 A to 3sf or 0.36 A to 2sf

So I agree with all your maths. I just think you need to be careful about early rounding: always use unrounded answers for continuing calculation.
Just as a matter of style, I'd be inclined to give similar answers to the same accuracy, like all to 3sf say.

phinds has covered the negative value. Just to add, you often can't know in advance which way the current flows in each loop. So you have to assign an arbitrary direction. If it comes out positive you were right and if negative, you guessed wrong, but no matter, you know now.

 

[magoo]

I couldn't follow your equations either. Keep in mind the three currents (I1, I2 and I3) are branch currents not loop currents.

Your loop equations should involve only two different currents. Yours involve three.

 

[james123]

Hi there, thanks for all the replies.

The reason I've written multiple equations for each loop is me simplifying really, and also substituting (I1+I2) for I3. It's also the way my learning materials showed me to do it.

Is there a setting on my calculator that I could change to give me more significant figures as 0.45A was the full figure it gave me.

Where did I use 3 currents in my loop equations?

 

[phinds]

Where did I use 3 currents in my loop equations?

I'm puzzled as to why you would ask this, when you clearly used i1, i2, and i3. Our point is that I3 is redundant for the loop equations and should not be there.

 

[james123]

It's just the way my learning materials have shown me, I've also seen YouTube videos which do it a similar way to this.

Has it affected my answers in a negative way? Apart from the rounding errors which I will go back and change

 

[phinds]

It's just the way my learning materials have shown me, I've also seen YouTube videos which do it a similar way to this.

Has it affected my answers in a negative way? Apart from the rounding errors which I will go back and change

If you like doing it the hard way, then you should continue doing it the hard way.

And by the way, I don't know about these days, but in my day you would likely have been dinged for this method because it implies that you do not understand loop currents.

 

[james123]

Was my first go at a question like this so I may have made a meal out of it. I will go back and revise the method I used.
Many thanks for your advice

 

[Merlin3189]

Is there a setting on my calculator that I could change to give me more significant figures as 0.45A was the full figure it gave me.

AFAICS your first result of 0.45 came from -5/-11
I can't believe any calculator gave just 0.45 for this, unless there is some setting telling it to round off to 2dp. Even a Poundshop 4 function calculator gives more figures than that. And yours gave more figures for your other calculations, so I'm puzzled.

 

[james123]

That's what I can't understand, will have a play around with it! I'm sure it's something I've done wrong so I'll redo all the calculations!

 

[Babadag]

< Mentor Note -- Even though this post below is a full solution, after a Mentor discussion we have decided to allow it since the OP is pretty much done except for figuring out their rounding errors >

The mesh voltage circuit will be :
Ia*(R1+R3)-Ib*R3=E1-E3 for loop a
-Ia*R3+Ib*(R3+R2)=E3-E2 for loop b
4*Ia-3*Ib=1
-3*Ia+5*Ib=-2
The general determinant is:
4 -3
-3 5
and the result is 20-9=11
The Ia determinant is:
1 -3
-2 5
result: 5-6=-1
The Ib determinant is:
4 1
-3 -2
-8+3=-5
Ia=-1/11=-0.0909
Ib=5/11=04545
I1=Ia=-0.0909
I2=Ib=0.4545
I3=-(I1+I2)=-0.3636
Check:
I1+I2+I3=-0.0909+0.4545-0.3636=0
The result voltages in each branch will be:
V1=E1-I1*R1=13.0909
V2=E2-I2*R2=13.0909
V3=E3-I3*R3=13.0909

 

[cnh1995]

I believe nodal analysis will be an easier approach. Assuming the bottom wire to be the ground node and using KCL at the top node, you need only one equation to find the top node voltage. Then using Ohm's law, you can find the three branch currents.

 

[Babadag]

At first, I have to apologize for my incorrect behavior. I forgot that the purpose of this forum is to establish how computing and compute not instead of solicitant.
I agree with you, cnh1995. This way, it seems to me, is better than mine. At least it is a more elegant solution to this problem.