In a p-n junction diode, the increase in forward voltage of 0.19 V increases the forward current by 37.6 mA. The dynamic resistance of the junction is

Question

In a p-n junction diode, the increase in forward voltage of 0.19 V increases the forward current by 37.6 mA. The dynamic resistance of the junction is (A) 2 ohm (B) 3 ohm (C) 4 ohm (D) 5 ohm.

Tardigrade Admin · Accepted Answer

ra . c=(Δ V/Δ I)=(0.19 V/37.6 m A)

Q. In a p-n junction diode, the increase in forward voltage of 0.19 V increases the forward current by 37.6 mA. The dynamic resistance of the junction is

2 ohm

3 ohm

4 ohm

5 ohm.

$r_{a . c} = \frac{Δ V}{Δ I} = \frac{0.19 V}{37.6 m A}$
$= \frac{0.19 V}{37.6 \times 1 0 ^{- 3} A}$
$= 5 o hm$

Q. In a p-n junction diode, the increase in forward voltage of 0.19 V increases the forward current by 37.6 mA. The dynamic resistance of the junction is

2 ohm

3 ohm

4 ohm

5 ohm.

ra.c​=ΔIΔV​=37.6mA0.19V​=37.6×10−3A0.19V​ =5ohm

$r_{a . c} = \frac{Δ V}{Δ I} = \frac{0.19 V}{37.6 m A}$
$= \frac{0.19 V}{37.6 \times 1 0 ^{- 3} A}$
$= 5 o hm$