Using Kirchhoff's Rules - Calculating Currents

[PeachBanana]

Homework Statement

What currents are provided by the batteries? What is the voltage drop across the 1.2 Ω
resistor?

Homework Equations

V = IR
Junction Rule = sum of current at a junction = 0
Loop Rule = sum of changes in potential around any closed path of a circuit must be 0

The Attempt at a Solution

I hope what I've drawn is clear enough. I 1 only goes around the small rectangle whereas I 2 encompasses the entire rectangle.

The current provided by the 12 V battery using the loop rule:

12 V + (3.9 Ω)(I1)+(1.2 Ω)(I1) + (9.8 Ω)(I1) = 0

Does that make sense? I'm unsure when it comes to these rules.

 

[gneill]

When you "walk" around a loop doing a KVL sum, be sure to pay attention to the polarities of the potential changes. If you pass through a resistance in the same direction as the current, does the potential rise or drop? Also, it sounds as though you've specified that I2 also passes through the 3.9 and 9.8 Ohm resistors, so it must also contribute to the total potential change in those resistors.

 

[PeachBanana]

Okay I think I understand better. If the resistance is in the same direction as the current, that would be considered a voltage drop.

So really, I have two unknowns.

12 V - (3.9 Ω)(I1) - (1.2 Ω)(I1) - (9.8 Ω)(I1) - (3.9 Ω)(I2) - (9.8 Ω)(I2)= 0

So then maybe applying this same loop rule to I2 I would have:

(3.9 Ω)(I2) - (3.9 Ω)(I1) - (6.7 Ω)(I2) - 9V - (9.8 Ω)(I2) - (9.8 Ω)(I1) + 12 V = 0

 

[gneill]

Okay I think I understand better. If the resistance is in the same direction as the current, that would be considered a voltage drop.

So really, I have two unknowns.

12 V - (3.9 Ω)(I1) - (1.2 Ω)(I1) - (9.8 Ω)(I1) - (3.9 Ω)(I2) - (9.8 Ω)(I2)= 0

So then maybe applying this same loop rule to I2 I would have:

(3.9 Ω)(I2) - (3.9 Ω)(I1) - (6.7 Ω)(I2) - 9V - (9.8 Ω)(I2) - (9.8 Ω)(I1) + 12 V = 0

Yup. Just verify the sign of the first term in the loop 2 equation. Otherwise, nicely done.