In the following circuit, the magnitude of current I _ 1 , is...

The figure above shows the current and voltage at different points in circuit.

Applying Kirchhoff's Junction law at $A$,

$\dfrac{x - ( y + 5 )}{1} + \dfrac{x - 2}{2} + \dfrac{x - 0}{2} = 0 . . . \left(1\right)$

Applying Kirchhoff's Junction law at $B$,

$\dfrac{y + 5 - x}{1} + \dfrac{y - 0}{1} + \dfrac{y - 2}{1} = 0 . . . \left(2\right)$

Solving equation (1) and (2), we get

$x = 3$ and $y = 0$

Now, at $D$ junction

${I}_{1} = {i}_{1} + {i}_{2}$

$⇒ {I}_{1} = \dfrac{y - 0}{1} + \dfrac{x - 0}{2}$

$= \dfrac{0 - 0}{1} + \dfrac{3 - 0}{2}$

$⇒ {I}_{1} = 1 . 5A$.