Using Kirchoff's Voltage Rule to find currents in a system

[Rexx]

Homework Statement

Then solve these equations for i1-4

Homework Equations

V=IR

The Attempt at a Solution

80i_1-50i_2-30i_3 = -120
-50i_1+100i_2-10i_3-25i_4 = 0.
-30i_1-10i_2+65i_3-20i_4 = 0.
-25i_2-20i_3+100i_4 = 0.

i_1=-4.18239492
i_2=-2.66455194
i_3=-2.71213323
i_4=-1.20856463

I used python to solve the matrix (using gauss elimination) and I am pretty sure I have done that bit correct (it was setting up the linear equations that I was unsure of if they were correct and how to explain my workings)

 

[Rexx]

Homework Statement

View attachment 214358
Then solve these equations for i1-4

Homework Equations

V=IR

The Attempt at a Solution

80i_1-50i_2-30i_3 = -120
-50i_1+100i_2-10i_3-25i_4 = 0.
-30i_1-10i_2+65i_3-20i_4 = 0.
-25i_2-20i_3+100i_4 = 0.

I feel I have done this completely wrong but:

i_1=-4.18239492
i_2=-2.66455194
i_3=-2.71213323
i_4=-1.20856463

I used python to solve the matrix (using gauss elimination) and I am pretty sure I have done that bit correct (it was setting up the linear equations that I was unsure of)

 

[ehild]

Homework Statement

View attachment 214358
Then solve these equations for i1-4

Homework Equations

V=IR

The Attempt at a Solution

80i_1-50i_2-30i_3 = -120
-50i_1+100i_2-10i_3-25i_4 = 0.
-30i_1-10i_2+65i_3-20i_4 = 0.
-25i_2-20i_3+100i_4 = 0.

i_1=-4.18239492
i_2=-2.66455194
i_3=-2.71213323
i_4=-1.20856463

I used python to solve the matrix (using gauss elimination) and I am pretty sure I have done that bit correct (it was setting up the linear equations that I was unsure of if they were correct and how to explain my workings)

Check the first equation; the RHS is wrong. The potential changes between 120 V and -120 V.

 

[Rexx]

Check the first equation; the RHS is wrong. The potential changes between 120 V and -120 V.

Thanks, I am getting a bit confused with it all now. Just to check, am I correct in saying that across a component, if it is going from high to low potential, we can indicate this as a positive voltage.

hence, for loop 1:
considering the 'battery' the current would be flowing from low to high potential and hence would be +120,
then considering the 50 ohm resistor, the current i1 is flowing from low to high potential and hence would be negative so -(50i1)
again, for the 50 ohm resistor, the current i2 is flowing from high to low potential, so would be +(50i2)
for the 30 ohm resistor, the current i1 is flowing from low to high potential, so would be -(30i1)
again for the 30 ohm resistor, the current i3 is flowing from high to low potential, so would be +(30i3)

assuming the above is correct, by KVR:
0=120-50i1+50i2-30i1+30i3
-120=-80i1+50i2+30i3

if i were to flip the minus signs I would get
120=80i1-50i2-30i3

Is this the correct logic for the problem? if so how would i even know the whether the other currents in the lower loops were going from low to high or high to low. When I did it, i was half guessing.

e.g. if i try to compute loop 2, i know
for the 50 ohm resistor i2 goes from high to low potential, so must be -(50i2)
50 ohm- i1 goes from low to high, so must be +(50i1)
for the 15 ohm resistor- how will i determine whether i2 goes from high to low potential ( i assume it goes from high to low) and hence is -15i2
for 25 ohm- i2 again I am assuming it goes from high to low and is -(25i2)
25 ohm- i4 I am assuming it goes from low to high potential, so would be (25i4)
for the 10 ohm resistor- i2 I am assuming goes from high to low potential hence is -10i2
for i3 I am assuming it goes from low to high and would be (10i3)

hence by kvr loop 2 would give
0=50i1-100i2+10i3+25i4
inverting minus signs would give
0=50i1-100i2+10i3+25i4

which is what I had. Like i said, this is me half guessing whether the currents are going from high to low potential, which is why my method is flawed

p.s. sorry for the essay, but I'd really appreciate any help to explain the logic behind this

 

[ehild]

Thanks, I am getting a bit confused with it all now. Just to check, am I correct in saying that across a component, if it is going from high to low potential, we can indicate this as a positive voltage.

Yes, voltage usually means potential drop, and can be confusing.
According to your picture, i1 flows from -120 V potential to + 120 V potential. What is the potential drop? Is not it -120-(+120) = -240 V?

 

[Rexx]

Yes, voltage usually means potential drop, and can be confusing.
According to your picture, i1 flows from -120 V potential to + 120 V potential. What is the potential drop? Is not it -120-(+120) = -240 V?

Thank you, that makes sense. How am i supposed to check the potential drop across the other resistors

 

[ehild]

Thank you, that makes sense. How am i supposed to check the potential drop across the other resistors

Your other equations were correct.
The original formulation of Kirchhoff's Voltage Law said that the sum of voltages is equal to the sum of emf-s in a closed loop. But it is easier to follow the potential changes going round a loop. If you go in the same direction as the current flows through a resistor R, the potential will decrease by IR. If the current flows in the opposite direction, the potential will increase by IR. Going from the negative pole of a battery to the positive one, the potential increases by the emf, and decreases in the opposite direction.
So in the top loop of the circuit, we start at the point with -120 V potential and follow the potential in the direction of current i1, reaching the point with 120 V potential:
-120-50(i₁-i₂)-30(i₁-i₃)=120 which is equivalent to
80 i₁-50 i2-30 i₃ = -240
The net change of potential in the other loops is zero, your equations were correct.