In the given circuit, if the barrier potential across the diode is 0.5 V , the current in the circuit is:

The circuit consists of a diode $D$, a resistor $R = 75 \Omega$, and a voltage source $V = 5 \text{ V}$ connected in series.

The diode is forward-biased as the positive terminal of the battery is connected to the p-side of the diode.

The barrier potential $V_{b}$ across the diode is given as $0.5 \text{ V}$.

Applying Kirchhoff's Voltage Law (KVL) in the loop:
$V - V_{b} - I \times R = 0$

Substituting the given values:
$5 \text{ V} - 0.5 \text{ V} = I \times 75 \Omega$
$4.5 \text{ V} = I \times 75 \Omega$

Solving for the current $I$:
$I = \dfrac{4.5}{75} \text{ A}$
$I = \dfrac{45}{750} \text{ A}$
$I = \dfrac{3}{50} \text{ A}$
$I = 0.06 \text{ A}$

Converting to milliamperes (mA):
$I = 0.06 \times 1000 \text{ mA} = 60 \text{ mA}$

Answer: 60mA