Semiconductor Electronics Part 1

Given, forward resistance, R1=50Ω
Reverse resistance, R2= infinity
Battery voltage =6V
According to circuit diagram, In this case, diode D1 is forward biased, whereas diode D2 is reverse biased.
So, D2 will act as open circuit.
By using Kirchhoff's voltage law,
⇒‌‌6−50I−130I−120I‌‌=0
⇒‌‌6‌‌=300I
⇒‌‌I‌‌=‌

6

300

=‌

1

50

=‌

2

100

=0.02A=20mA
Hence, current through 120Ω=20mA