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

Step-by-step solution

The circuit consists of a diode D , a resistor R = 75 , and a voltage source V = 5 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 V . Applying Kirchhoff's Voltage Law (KVL) in the loop: V - V_ b - I R = 0 Substituting the given values: 5 V - 0.5 V = I 75 4.5 V = I 75 Solving for the current I : I = 4.5 75 A I = 45 750 A I = 3 50 A I = 0.06 A Converting to milliamperes (mA)