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A Zener diode is connected to battery and a load resistance as shown below
The currents $I, I_Z$ and $I_L$ respectively are
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The currents $I, I_Z$ and $I_L$ respectively are
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Verified Answer
The correct answer is:
$12.5 \mathrm{~mA}, 7.5 \mathrm{~mA}, 5 \mathrm{~mA}$
The circuit given is as shown
Potential drop across $R_L=V_L=V_Z=10 \mathrm{~V}$ Potential drop across $4 \mathrm{k} \Omega$ resistance is
$V_R=60-10=50 \mathrm{~V}$
Current through $4 \mathrm{k} \Omega$ resistance is
$I=\frac{V_R}{R}=\frac{50}{4 \times 10^3}=12.5 \times 10^{-3} \mathrm{~A}=12.5 \mathrm{~mA}$
Current through load resistance $=I_L$
$=\frac{V_L}{R_L}=\frac{10}{2 \times 10^3}=5 \mathrm{~mA}$
Also, $I_Z=I-I_L=12.5-5=7.5 \mathrm{~mA}$
Potential drop across $R_L=V_L=V_Z=10 \mathrm{~V}$ Potential drop across $4 \mathrm{k} \Omega$ resistance is
$V_R=60-10=50 \mathrm{~V}$
Current through $4 \mathrm{k} \Omega$ resistance is
$I=\frac{V_R}{R}=\frac{50}{4 \times 10^3}=12.5 \times 10^{-3} \mathrm{~A}=12.5 \mathrm{~mA}$
Current through load resistance $=I_L$
$=\frac{V_L}{R_L}=\frac{10}{2 \times 10^3}=5 \mathrm{~mA}$
Also, $I_Z=I-I_L=12.5-5=7.5 \mathrm{~mA}$
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