Kirchhoff's law: Circuit problem with multiple batteries and resistors

[Bolter]

Homework Statement

See image attached

Relevant Equations

Kirchhoff's 1st and 2nd law

Struggling to see if I am doing this right or not from a section of a question I have been given

I have drawn out a sketch of the circuit again, and labelled as to what my 1st and 2nd closed loop to be to then apply Kirchhoff's 2nd law

I get a negative I_2 current which just indicates that my original assumption of the currents direction was wrong and it is in fact going the opposite way

So I believe that the current going through:

6 ohm resistor = 0.703 A
1 ohm resistor = 1.162 A
4 ohm resistor = 0.459 A

Is this right or have I miserably gone wrong somewhere?

Any help would be really appreciated! Thanks

 

[scottdave]

I followed through your workings and arrived at the same conclusion as you.
I found your error. It is in the step where you expand the parenthesis:
8(I₁ + I₂) + I₁ = 10

Check how you expand the parenthesis and take care of the 8 coefficient.

 

[scottdave]

Hi @Bolter I originally made the same error, and didn't catch your mistake. It was when I took those currents, plugged them back into the original circuit - I realized that the loop voltage drops did not sum to zero, so then started looking for an error.

Lesson: always go back and plug your answers into your original problem, as a check.

 

[Leo Liu]

Hi @Bolter I originally made the same error, and didn't catch your mistake. It was when I took those currents, plugged them back into the original circuit - I realized that the loop voltage drops did not sum to zero, so then started looking for an error.

Lesson: always go back and plug your answers into your original problem, as a check.

Hi, can I know what value you got for I? I got $47\over44$.

 

[Bolter]

I followed through your workings and arrived at the same conclusion as you.
I found your error. It is in the step where you expand the parenthesis:
8(I₁ + I₂) + I₁ = 10

Check how you expand the parenthesis and take care of the 8 coefficient.

Thank you for catching in on my mistake :)

I have redone solving the simultaneous equations again using the corrections and I got these as my currents

So I can say that the current through:

6 ohm resistor = 47/44 A
1 ohm resistor = 16/11 A
4 ohm resistor = –17/44 A

Is it also possible if you could help me out with part c) of this question?

I first worked out total charge that passes in 10 seconds, then divided that by the charge of a single electron to get total number of electrons in 10s?

 

[scottdave]

All those calculations (current in resistors & number of electrons) look correct.