Question bank/JEE Main/Physics

In a common-base configuration of transistor, α = 0.98, I_B =
0.02mA, R_L = 5 kΩ. Output voltage across load is

A3.2 V (B) 4.9 VB5.2C6.2 V

Show answer and solution▾

[IMAGE] I_E 0.98 = I_C
I_E = I_B + I_C
I_E = 1 mA
I_C = 0.98 mA
V_0 = R_L × I_C

Source: Grass-G / P-02 Physics Paper-02_Update.docx