In a triode plate current is 2.5 mA and plate resistance 20 k Ω, amplification of triode is 10, the load resistance will be :

Question

In a triode plate current is 2.5 mA and plate resistance 20 k Ω, amplification of triode is 10, the load resistance will be : (A) 5 k Ω (B) 25 k Ω (C) 20 k Ω (D) 50 k Ω

Tardigrade Admin · Accepted Answer

ip=2.5 mA =2.5 × 10-3 A, rp=20 × 103 Ω A=10, RL=? Vp=ip × rp=50 gm=(ip/Vp)=(2.5 × 10-3/50)=5 × 10-5 μ=r dotp × gm=20 × 103 × 5 × 10-5=1 A=(μ RL/RP+RL) solving RL=20 k Ω

Q. In a triode plate current is $2.5 m A$ and plate resistance $20 k Ω$ , amplification of triode is $10$ , the load resistance will be :

$5 k Ω$

$25 k Ω$

$20 k Ω$

$50 k Ω$

Q. In a triode plate current is 2.5mA and plate resistance 20kΩ, amplification of triode is 10, the load resistance will be :

5kΩ

25kΩ

20kΩ

50kΩ

ip​=2.5mA=2.5×10−3A,rp​=20×103Ω A=10,RL​=? Vp​=ip​×rp​=50 gm​=Vp​ip​​=502.5×10−3​=5×10−5 μ=rp˙​​×gm​=20×103×5×10−5=1 A=RP​+RL​μRL​​ solving RL​=20kΩ

Q. In a triode plate current is $2.5 m A$ and plate resistance $20 k Ω$ , amplification of triode is $10$ , the load resistance will be :

$5 k Ω$

$25 k Ω$

$20 k Ω$

$50 k Ω$