Multibranched Circuit

Kirchhoff's Voltage Rule and Junction Rule will make quick work of this. The sum of the voltage drops/gains around any closed loop must equal 0, and the total current coming into a junction must equal the total current coming out:

When defining voltage drops/gains, as well as current direction, it doesn't really matter as long as we're consistent. I will define coming out of the positive end of the battery as a voltage gain. For current directions, I will use the topmost junction as a reference point;

• $I_{1}$ is the current flowing across the $3\Omega$ resistor, and is going into the topmost junction.
• $I_{2}$ is the current flowing across the $2\Omega$ resistor, and is flowing out of the topmost junction.
• $I_{3}$ is the current flowing across both the $1\Omega$ and the $4\Omega$ resistor, and is flowing out of the topmost junction.

Using the topmost junction as a starting point, we can construct two closed loops going out the sides in both directions and returning through the middle going up, netting us the following two equations: (note that the voltage across a resistor is $V=IR$

$\begin{array}{c}13-3I_{1}-2I_{2}-10=0 \\ 13-3I_{1}+14-4I_{3}-3.5-I_{3}=0 \end{array}$

And since the top junction has current flowing in and out, we know that:

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

This is a system of three equations, which can be simplified into the following and solved using RREF:

$\begin{array}{c}3I_{1}+2I_{2}+0I_{3}=3 \\ 3I_{1}+0I_{2}+5I_{3}=23.5 \\ I_{1}-I_{2}-I_{3}=0 \end{array}$

Which results in $I_{1}=\boxed{2A}$