MCQ
A triode whose mutual conductance is $2.5 m$$ A/volt$ and anode resistance is $20 kilo ohm$, is used as an amplifier whose amplification is $10$. The resistance connected in plate circuit will be.....$k\Omega$
- A$1 $
- ✓$5$
- C$10$
- D$20 $
✓
Answer
Correct option: B.
$5$
b
(b)$\mu = {r_P} \times {g_m} = 20 \times 2.5 = 50$
From $A = \frac{{\mu \,{R_L}}}{{{r_P} + {R_L}}}$==> ${r_P} + {R_L} = \frac{{\mu \,{R_L}}}{A} = \frac{{50{R_L}}}{{10}} = 5{R_L}$
==> $4{R_L} = {r_p} \Rightarrow {R_L} = \frac{{{r_p}}}{4} = \frac{{20}}{4} = 5k\Omega $
(b)$\mu = {r_P} \times {g_m} = 20 \times 2.5 = 50$
From $A = \frac{{\mu \,{R_L}}}{{{r_P} + {R_L}}}$==> ${r_P} + {R_L} = \frac{{\mu \,{R_L}}}{A} = \frac{{50{R_L}}}{{10}} = 5{R_L}$
==> $4{R_L} = {r_p} \Rightarrow {R_L} = \frac{{{r_p}}}{4} = \frac{{20}}{4} = 5k\Omega $
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