JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 26)
The circuit diagram used to study the characteristic curve of a zener diode is connected to variable power supply (0 $$-$$ 15 V) as shown in figure. A zener diode with maximum potential Vz = 10 V and maximum power dissipation of 0.4 W is connected across a potential divider arrangement. The value of resistance RP connected in series with the zener diode to protect it from the damage is ________________ $$\Omega$$.
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Answer
125
Explanation
Power $\mathrm{P}_{\mathrm{zm}}=\mathrm{I}_{\mathrm{zm}} \mathrm{V}_{z}$
$$ \begin{aligned} & \mathrm{I}_{\mathrm{ZM}}=\frac{P_{z M}}{V_{z}}=\frac{0.4}{10}=0.04 \mathrm{~A} \\\\ & \text { So, } R_{P}=\frac{E_{\max }-V_{z}}{I_{Z M}}=\frac{15-10}{0.04}=\frac{5}{0.04}=125 \Omega \end{aligned} $$
$$ \begin{aligned} & \mathrm{I}_{\mathrm{ZM}}=\frac{P_{z M}}{V_{z}}=\frac{0.4}{10}=0.04 \mathrm{~A} \\\\ & \text { So, } R_{P}=\frac{E_{\max }-V_{z}}{I_{Z M}}=\frac{15-10}{0.04}=\frac{5}{0.04}=125 \Omega \end{aligned} $$
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