A 8,V Zener diode along with a series resistance R is connected across a 20,V supply (as shown in fi

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14.Semiconductor Electronics

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A $8\,V$ Zener diode along with a series resistance $R$ is connected across a $20\,V$ supply (as shown in the figure). If the maximum Zener current is $25\,mA$, then the minimum value of $R$ will be $\Omega .$

A

$480$

B

$441$

C

$420$

D

$460$

(JEE MAIN-2022)

Solution

$\varepsilon- IR – V _{z}=0$

$20- IR -6=0$

$IR =12$

$25 \times 10^{-3} R =12$

$R =\frac{12}{25 \times 10^{-3}}=480\,\Omega$

Standard 12

Physics

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