Kirchhoff's First Law to solve a circuit with 2 loops

[Daniel2244]

Homework Statement

I've recently watched a video which explains Kirchhoff's first law however, I'm not sure about his calculations. I got the same answer just wanting to make sure I am doing it right. the guy wrote L1: -75I2 +3V=0 ⇒ I2= -3/-75=0.04A
He calculated L2 like -12V -125I3 +75I2=0 ⇒ -12 -125I3 + 75(0.04)=0 ⇒ -12 - 125I3 +3=0⇒ I3=12-3.0/-125=0.072A
Another question, does this -12V -125I3 +75I2=0 equal zero becasue of Kirchhoff's second law?

Homework Equations

I1=I2+I3

The Attempt at a Solution

starting in the top left corner - L1: -75I2 +3V=0 ⇒ 3V=75I2⇒ 3/75=75I2/75⇒ I2=3/75=0.04A

starting at J1 - L2: -12V -125I3 +75I2=0 ⇒ -12V+75I2=125I3 ⇒ I3=-12+(75x0.04)/125= -0.072A

Using equation I1=I2+I3⇒ I1=0.04+0.072= -0.032A (-=current going wrong way)

*Stupid question I know but just want to make sure

 

[gneill]

Your method and results look fine to me.

 

[Daniel2244]

Your method and results look fine to me.

Thanks for letting me know .

 

[Daniel2244]

Your method and results look fine to me.

A quick question, If L1 and L2 were going anti clockwase would the equation still be I1=I2+I3? Nvm yes it would. However, If the current was going anti clockwise do we need to the circuit look anti clock wise?

 

[gneill]

However, If the current was going anti clockwise do we need to the circuit look anti clock wise?

Not sure what you're asking there, but the bottom line is that it doesn't make any difference which way you define your initial current directions so long as you are consistent in determining the potential drop directions due to those current. The math will always take care of itself.

 

[Daniel2244]

Not sure what you're asking there, but the bottom line is that it doesn't make any difference which way you define your initial current directions so long as you are consistent in determining the potential drop directions due to those current. The math will always take care of itself.

Ok, does this -12V -125I3 +75I2=0 equal zero becasue of kirchhoff's second law?

 

[gneill]

Ok, does this -12V -125I3 +75I2=0 equal zero becasue of kirchhoff's second law?

Presumably by Kirchhoff's 2nd law you are referring to KVL, so yes, taking your defined current directions from your image that is the case.