The triodes P and Q have the same amplification factor 40. Their plate resistances are `4kOmega` and `8kOmega`, respectively. If an amplifier circuit is designed using anyone of them and a load resistance is of `8kOmega`, the ratio of the voltage gain obtained from them will be

Options

A. `2:3`
B. `4:3`
C. `3:1`
D. `1:2`

Answer

B

Explanation

`mu=40`=amplification factor
`mu_(p) = mu_(Q)=40`
`r_(P)=4kOmega,r_(Q) = 8kOmega`
`R_(L) = 8kOmega`
Voltage gain for triode A `(muR_(L))/(r_(p)+R_(L))`
For triode `P=A_(P) =(40 xx 8)/(4+8)`
`A_(Q) = (40 xx 8)/(4+8)`
`A_(P)/A_(Q) = ((40 xx 8)/(4+8))/((40 xx 8)/(8+8))=16/12 = 4/3`