Home
Class 12
PHYSICS
In a reverse biased diode, when the appl...

In a reverse biased diode, when the applied voltage changes by `1V`, the current is found to change by `0.5muA`. The reversebiase resistance of the diode is

A

`2xx10^(5)Omega`

B

`2xx10^(6)Omega`

C

`200 Omega`

D

`2 Omega`

Text Solution

Verified by Experts

The correct Answer is:
B

Reverse resistance = `(DeltaV)/(DeltaI)=(1)/(0.5xx10^(-6))=2xx10^(6)Omega`
Promotional Banner

Similar Questions

Explore conceptually related problems

In silicon diode, the reverse current increases from 10A to 20 muA . When the reverse voltage change from 2V to 4V . Find the reverse a.c. resistance of the diode

In a silicon diode,the forward current changes by 2.5mA when the voltage changes from 0.08 to 0.09V then the forward dynamic ac resistance of the diode is

If the forward voltage in a semiconductor diode is chaged form 0.5 V to 2 V, then the forward current changed by 1.5 mA. The forward resistance of diode will be-

The reverse bias in a junction diode is changed from 5V to 15 V then the value of current changes from 38 mu A to 88 mu A . The resistance of junction diode will be.

If the forward voltage in a semiconductor diode is changed from 0.5V to 0.7 V, then the forward current changes by 1.0 mA. The forward resistance of diode junction will be

When the reverse potential in a semiconductor diode are 10 V and 30 V, thent he corresponding reverse currents are 25muA . And 50muA respectively. The reverse resistance of junction diode will be-