Using Kirchhoff's laws to find current in a circuit

[november1992]

Homework Statement

Use KCL and KVL to determine the currents I1 to I3 in the circuit below:

Homework Equations

Ʃ$v_{n}$ = 0
Ʃ$i_{n}$ = 0
v=ir

The Attempt at a Solution

I got two different answers for $I_{1}$ using the two different laws. I'm not sure if I'm doing it right.

KVL @ loop 1: -18 + 6 + 8$I_{1}$ = 0
8 $I_{1}$ = 12
$I_{1}$ = 1.5AKCL @ node a: 3 = $I_{1}$ + $I_{2}$ + 1
KCL @ node b: $I_{3}$ = 1 + $I_{2}$
0 = $I_{1}$ + $I_{2}$ -2
0 = -1 - $I_{2}$ + $I_{3}$
1 = $I_{3}$
0 = $I_{2}$
2 = $I_{1}$

 

[berkeman]

Homework Statement

Use KCL and KVL to determine the currents I1 to I3 in the circuit below:

Homework Equations

Ʃ$v_{n}$ = 0
Ʃ$i_{n}$ = 0
v=ir

The Attempt at a Solution

I got two different answers for $I_{1}$ using the two different laws. I'm not sure if I'm doing it right.

KVL @ loop 1: -18 + 6 + 8$I_{1}$ = 0
8 $I_{1}$ = 12
$I_{1}$ = 1.5A


KCL @ node a: 3 = $I_{1}$ + $I_{2}$ + 1
KCL @ node b: $I_{3}$ = 1 + $I_{2}$



0 = $I_{1}$ + $I_{2}$ -2
0 = -1 - $I_{2}$ + $I_{3}$
1 = $I_{3}$
0 = $I_{2}$
2 = $I_{1}$

Your KVL solution for I1 is correct. For the KCL solution, you have 3 unknowns but you have only written 2 equations. That isn't sufficient to solve for I1 yet...

 

[november1992]

Could I substitute the value that I got for I1 from the KVL equation into KCL?
I wasn't sure if I could do that because I thought the question was asking me to find the values separately using the two different laws.

 

[berkeman]

Could I substitute the value that I got for I1 from the KVL equation into KCL?
I wasn't sure if I could do that because I thought the question was asking me to find the values separately using the two different laws.

It does sound like they want you to solve the circuit with each technique separately. It's a pretty easy circuit to solve by inspection anyway, so they must want you to go through all the steps of each technique separately for practice.

So can you see another node where you could write a 3rd KCL equation?

 

[november1992]

The one between the 8 ohm and 4 ohm resistor?

 

[berkeman]

The one between the 8 ohm and 4 ohm resistor?

That might work...

 

[november1992]

nvm I figured it out, thanks for the help.