1.

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

Answer»

`10^2 Omega`
`10^3 Omega`
`10^4 Omega`
`10^5 Omega`

SOLUTION :`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`


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