CUET Preparation Today
When the reverse potential in a semiconductor diode are 10V and 20V, then the corresponding reverse currents are 25μ A and 50μ A respectively. The reverse resistance of junction diode will be: |
40 $4 × 10^5 Ω$ 40KΩ $4 × 10^{–5} Ω$ |
$4 × 10^5 Ω$ |
$r_r=\frac{V_{r_2}-V_{r_2}}{I_{r_1}-I_{r_1}}=\frac{20-10}{(50-25)×10^{-6}}=4 × 10^5 Ω$ |
