SATHEE: Electronic Devices Question 262

Electronic Devices Question 262

Question: In the grid circuit of a triode a signal $E = 2 \sqrt{2} \cos ω t$ is applied. If m = 14 and rp =10 kW then root mean square current flowing through $R_{L} = 12 k Ω$ will be

Options:

A) 1.27 mA

B) 10 mA

C) 1.5 mA

D) 12.4 mA

Show Answer

Answer:

Correct Answer: A

Solution:

$A = \frac{μ R_{L}}{r_{p} + R_{L}} = \frac{14 \times 12}{10 + 12} = \frac{84}{11}$ .

Peak value of output signal $V_{0} = \frac{84}{11} \times 2 \sqrt{2} V$

Therefore $V_{r m s} = \frac{V_{0}}{\sqrt{2}} = \frac{84 \times 2}{11} V$

Therefore r.m.s.

value of current through the load $= \frac{84 \times 2}{11 \times 12 \times 10^{3}} A = 1.27 m A$

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Electronic Devices Question 262

Question: In the grid circuit of a triode a signal E=22cos⁡ωt is applied. If m = 14 and rp =10 kW then root mean square current flowing through RL=12kΩ will be

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Question: In the grid circuit of a triode a signal $E = 2 \sqrt{2} \cos ω t$ is applied. If m = 14 and rp =10 kW then root mean square current flowing through $R_{L} = 12 k Ω$ will be