Question

In Germanium diode the forward current
increases from 100 mA to to 200 mA when
forward voltage changes from 5V to 10V, The
forward resistance of the diode is
(A) 20
(B) 50
(C) 100
(D) 500​

Answer 1

Solution:

Forward Current increases from 100 mA to 200 mA

Change in Forward Current,

ΔI = Final Current - Initial Current = 200 - 100 = 100 mA or 0.1 A

Forward Voltage increases from 5 V to 10 V

Change in Forward Voltage,

ΔV = Final Voltage - Initial Voltage = 10 - 5 = 5 V

Forward Resistance of the diode, $r_{f} = \frac{Change \ in \ Voltage(\Delta V)}{Change \ in \ Current(\Delta I )} = \frac{5}{0.1} = 50 \ \Omega$

(B) 50 is the correct answer.

The Forward Resistance( $r_{f}$ ) of the diode is 50 ohm.