In a triode value for a grid voltage `V_g=-1.2V` , the phase current `I_p` and plate voltage `V_p` are given `I_p=(-5+0.1 V_0),I_p` in `mA and V_p` in volt . When grid voltage is changed to - 3.2 V at constant plate voltage of 150 V , a plate current of 5 mA is observed , then
Plate resistance `R_p` is

Options

A. `10^2 Omega`
B. `10^3 Omega`
C. `10^4 Omega`
D. `10^5 Omega`

Answer

C

Explanation

`V_g=-1.2 V "Now " i_p=(-50+0.1V_p)10^(-3)A " "...(i)`
then `((Deltai_p)/(Deltai_p))=0.1xx10^(-3)AV^(-1)`
then plate resistance
`((Deltai_p)/(Deltai_p))=1/ (0.1xx10^(-3))=10^(4)Omega`