In the following circuit, the magnitude of current $$I_1$$, is ______ A.

Take the three branch currents as in your equations:

• right branch current $$I_1$$ (required)
• middle branch current $$I_2$$​
• top branch current $$I_3$$

Using KVL for the three loops:

First loop (right loop):

$$I_1+I_3-I_2=2$$

Second loop:

$$I_1+4I_2+I_3=5$$

Third loop:

$$I_1+I_2+4I_3=5$$

Now solve.

Subtract first from second:

$$(I_1+4I_2+I_3)-(I_1+I_3-I_2)=5-2$$

$$5I_2=3$$

$$I_2=\frac{3}{5}$$

Subtract first from third:

$$(I_1+I_2+4I_3)-(I_1+I_3-I_2)=5-2$$

$$2I_2+3I_3=3$$

Substitute

$$I_2=\frac{3}{5}$$

$$\frac{6}{5}+3I_3=3$$

$$I_3=\frac{9}{5}$$​

$$I_3=\frac{3}{5}$$

Now use first equation:

$$I_1+\frac{3}{5}-\frac{3}{5}=2$$

$$I_1=2$$