JIPMER 2014
Exam JIPMER
Year 2014
Subject Physics
Chapter Semiconductors JIPMER 2014 · Physics · Semiconductors
When plate voltage in diode valve is increased from 100 V to 150 V....
When plate voltage in diode valve is increased from 100 V to 150 V. Then, plate current increases from 7.5 mA to 12 mA, then dynamic plate resistance will be
Options A 10 k Ω 10 \mathrm{k} \Omega 10 k Ω B 11 k Ω 11 \mathrm{k} \Omega 11 k Ω C 15 k Ω 15 \mathrm{k} \Omega 15 k Ω D 11.1 k Ω 11.1 \mathrm{k} \Omega 11.1 k Ω Correct answer D. 11.1 k Ω 11.1 \mathrm{k} \Omega 11.1 k Ω
Step-by-step solution
Plate resistance = Δ V P 2 − Δ V P 1 Δ j P 2 − Δ j R = 150 − 100 ( 12 − 7.5 ) × 1 0 − 3 = 50 × 1 0 3 4.5 \begin{aligned} \text{ Plate resistance } & =\frac{\Delta V_{P_2}-\Delta V_{P_1}}{\Delta j_{P_2}-\Delta j_R} \\ & =\frac{150-100}{(12-7.5) \times 10^{-3}}=\frac{50 \times 10^3}{4.5}\end{aligned} Plate resistance = Δ j P 2 − Δ j R Δ V P 2 − Δ V P 1 = ( 12 − 7.5 ) × 1 0 − 3 150 − 100 = 4.5 50 × 1 0 3 =11.1 \mathrm{k} \Omega$
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