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JEE Main & Advanced Physics Semiconducting Devices Question Bank Self Evaluation Test - Semiconductor Electronics: Materials, Devices and Simple Circuits

  • question_answer
    A zener diode of voltage \[{{V}_{Z}}(=6V)\] is used to maintain a constant voltage across a load resistance \[{{R}_{L}}(=1000\,\,\Omega )\] by using a series resistance \[{{R}_{S}}(=100\,\Omega )\]. If the e.m.f. of source is \[E(=9\,V),\] what is the power being dissipated in Zener diode?

    A)  \[0.144\]watt

    B)  \[0.324\]watt

    C)  \[0.244\]watt

    D)  \[0.544\]watt

    Correct Answer: A

    Solution :

    [a] Here, \[E=9V;\]  \[{{V}_{Z}}=6;\]  \[{{R}_{L}}=1000\,\Omega \] and \[{{R}_{S}}=100\,\Omega ,\] Potential drop across series resistor \[V=E-{{V}_{Z}}=9-6=3V\] Current through series resistance \[{{R}_{S}}\] is \[I=\frac{V}{R}=\frac{3}{100}=0.03A\] Current through load resistance \[{{R}_{L}}\] is \[{{I}_{L}}=\frac{{{V}_{Z}}}{{{R}_{L}}}=\frac{6}{1000}=0.006\,A\] Current through Zener diode is \[{{I}_{L}}=\frac{{{V}_{Z}}}{{{R}_{L}}}=\frac{6}{1000}=0.006\,A\] amp. Power dissipated in Zener diode is \[{{P}_{Z}}={{V}_{Z}}{{I}_{Z}}=6\times 0.024=0.144\] Watt


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