Physics - Semiconductors Question with Solution | TestHub

In the given circuit, the positive terminal of the $5\text{ V}$ battery is connected to the p-side of the Silicon diode, so the diode is forward-biased.

The forward voltage drop across a Silicon diode is typically $V_D=0.7\text{ V}$.

Applying Kirchhoff's Voltage Law to the loop, we get:

$V-V_D-IR=0$

$5-0.7-I\times 200=0$

$4.3=200I$

$I=\frac{4.3}{200}\text{ A}$

$I=\frac{4.3}{200}\times 1000\text{ mA}$

$I=21.5\text{ mA}$

The reading in the ammeter is $21.5\text{ mA}$.

Answer: 21.5 mA