In a triode, $gm = 2 × 10^{–3} ohm^{–1}; μ = 42$; resistance of load, R = 50 kilo ohm. The voltage amplification obtained from this triode will be
Answer & explanation
Correct answer: option 2
$A_v=\frac{μR}{r_p+R}=\frac{μR}{(μ/g_m)+R}=\frac{42×(50×10^3)}{42/(2×10^{-3})+50×10^3}=29.57$