SATHEE: Semiconducting Devices Question 59

Semiconducting Devices Question 59

Question: The relation between Ip and Vp for a triode is $I_{p} = (0.125 V_{p} - 7.5) m A$ Keeping the grid potential constant at 1V, the value of rp will be

Options:

A) 8 kW

B) 4 kW

C) 2 kW

D) 8 kW

Show Answer

Answer:

Correct Answer: D

Solution:

$i_{p} = [0.125 V_{p} - 7.5] \times 10^{- 3} a m p$

Differentiating this equation w.r.t.

Vp $\frac{Δ i_{p}}{Δ V_{p}} = 0.125 \times 10^{- 3}$ or $\frac{1}{r_{p}} = 0.125 \times 10^{- 3}$

Therefore $r_{p} = 8 k Ω$

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Semiconducting Devices Question 59

Question: The relation between Ip and Vp for a triode is $I_{p} = (0.125 V_{p} - 7.5) m A$ Keeping the grid potential constant at 1V, the value of rp will be

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Semiconducting Devices Question 59

Question: The relation between Ip and Vp for a triode is Ip=(0.125Vp−7.5)mA Keeping the grid potential constant at 1V, the value of rp will be

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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Question: The relation between Ip and Vp for a triode is $I_{p} = (0.125 V_{p} - 7.5) m A$ Keeping the grid potential constant at 1V, the value of rp will be