Have the Currents Been Labelled Correctly?

[Addem]

Homework Statement

Linked below is a rough picture of the circuit for which I need to find the currents.

https://docs.google.com/drawings/pub?id=1jY6r1yjOf9OG5VixjZKcQbG2rj-ZQ7LZzaybGzkjva0&w=960&h=720

Homework Equations

Kichhoff's Laws, Ohm's Law.

The Attempt at a Solution

I'm doing a self-study using a book that I think may be riddled with errata, so I keep finding that I'm getting a result which disagrees with the answer the back and cannot see why my answer differs. I get the system of equations

$11 - 3i_{3} + 1 = 0$

$11 + 6i_{2} = 0$

$-3i_{3} + 1 -6i_{2} = 0$

$i_{3}=i_{1}+i_{2}$

I believe my choice of direction of flow is consistent and cannot affect my solution up to a change of sign. The book produces the result that $i_{2}=13/6$ when I obtain 11/6.

 

[SammyS]

Homework Statement

Linked below is a rough picture of the circuit for which I need to find the currents.

https://docs.google.com/drawings/pub?id=1jY6r1yjOf9OG5VixjZKcQbG2rj-ZQ7LZzaybGzkjva0&w=960&h=720

Homework Equations

Kichhoff's Laws, Ohm's Law.

The Attempt at a Solution

I'm doing a self-study using a book that I think may be riddled with errata, so I keep finding that I'm getting a result which disagrees with the answer the back and cannot see why my answer differs. I get the system of equations

$11 - 3i_{3} + 1 = 0$

$11 + 6i_{2} = 0$

$-3i_{3} + 1 -6i_{2} = 0$

$i_{3}=i_{1}+i_{2}$

I believe my choice of direction of flow is consistent and cannot affect my solution up to a change of sign. The book produces the result that $i_{2}=13/6$ when I obtain 11/6.

Added in Edit:

If i₂ = 11/6 A, then the voltage drop across the 6Ω resistor is 11 V, which makes perfect sense. So, it looks like you are correct !

 

[CWatters]

I agree. The voltage across the 6 ohm resistor is fixed at 11V so the current is simply 11/6 amps.

Have they/you labelled the currents correctly?