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JEE Main 201411 AprilOnline
Exam
JEE Main
Year
2014
Date
11 April
Mode
Online
Subject
Physics
Chapter
Semiconductors

JEE Main 2014 · 11 April · Online · Physics · Semiconductors

A Zener diode is connected to a battery and a load as show below:...

A Zener diode is connected to a battery and a load as show below: A Zener diode is connected to a battery and a load as show below: The currents, I, I _ Z and I_ L are respectively.
The currents, I,\mathrm{I},\mathrm{I}Z_{\mathrm{Z}} and IL\mathrm{I}_{\mathrm{L}} are respectively.

Options

  1. A15 mA,5 mA,10 mA15 \mathrm{~mA}, 5 \mathrm{~mA}, 10 \mathrm{~mA}
  2. B15 mA,7.5 mA,7.5 mA15 \mathrm{~mA}, 7.5 \mathrm{~mA}, 7.5 \mathrm{~mA}
  3. C12.5 mA,5 mA,7.5 mA12.5 \mathrm{~mA}, 5 \mathrm{~mA}, 7.5 \mathrm{~mA}
  4. D12.5 mA,7.5 mA,5 mA12.5 \mathrm{~mA}, 7.5 \mathrm{~mA}, 5 \mathrm{~mA}

Correct answer

D. 12.5 mA,7.5 mA,5 mA12.5 \mathrm{~mA}, 7.5 \mathrm{~mA}, 5 \mathrm{~mA}

Step-by-step solution

Here, R=4kΩ=4×103ΩR=4 \mathrm{k} \Omega=4 \times 10^3 \Omega V_i=60  V\mathrm{~V}Zener voltage Vz=10 VRL=2kΩ=2×103ΩV_z=10 \mathrm{~V}R_L=2 \mathrm{k} \Omega=2 \times 10^3 \Omega Load current, IL=VZRL=102×103=5 mAI_{\mathrm{L}}=\frac{V_Z}{R_L}=\frac{10}{2 \times 10^3}=5 \mathrm{~mA} Current through R,I=Vi−VZR=60−104×103=504×103=12.5 mAR, I=\frac{V_i-V_Z}{R}=\frac{60-10}{4 \times 10^3}=\frac{50}{4 \times 10^3}=12.5 \mathrm{~mA} Fom circuit diagram, I=IZ+IL⇒12.5=IZ+5⇒I7=12.5−5=7.5 mA\begin{aligned}& I=I_Z+I_L \\ \Rightarrow & 12.5=I_Z+5 \\ \Rightarrow & I_7=12.5-5=7.5 \mathrm{~mA} \end{aligned}
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