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A zener diode of voltage $\mathrm{V}_{\mathrm{Z}}(=6 \mathrm{~V})$ is used to maintain a constant voltage across a load resistance $\mathrm{R}_{\mathrm{L}}(=1000 \Omega)$ by using a series resistance $R_{\mathrm{s}}(=100 \Omega)$. If the e.m.f. of source is $\mathrm{E}(=9 \mathrm{~V})$, what is the power being dissipated in Zener diode?
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$0.144$ watt
Here, $E=9 \mathrm{~V} ; \mathrm{V}_{\mathrm{z}}=6 ; \mathrm{R}_{\mathrm{L}}=1000 \Omega$ and $\mathrm{R}_{\mathrm{s}}=100 \Omega$,
Potential drop across series resistor $\mathrm{V}=\mathrm{E}-\mathrm{V}_{\mathrm{Z}}=9-6=3 \mathrm{~V}$
Current through series resistance $R_{S}$ is $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{3}{100}=0.03 \mathrm{~A}$
Current through load resistance $R_{L}$ is $\mathrm{I}_{\mathrm{L}}=\frac{\mathrm{V}_{\mathrm{Z}}}{\mathrm{R}_{\mathrm{L}}}=\frac{6}{1000}=0.006 \mathrm{~A}$
Current through Zener diode is $\mathrm{I}_{\mathrm{Z}}=\mathrm{I}-\mathrm{I}_{\mathrm{L}}=0.03-0.006=0.024 \mathrm{amp} .$ Power dissipated in Zener diode is $\mathrm{P}_{\mathrm{Z}}=\mathrm{V}_{\mathrm{Z}} \mathrm{I}_{\mathrm{Z}}=6 \times 0.024=0.144 \mathrm{Watt}$
Potential drop across series resistor $\mathrm{V}=\mathrm{E}-\mathrm{V}_{\mathrm{Z}}=9-6=3 \mathrm{~V}$
Current through series resistance $R_{S}$ is $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{3}{100}=0.03 \mathrm{~A}$
Current through load resistance $R_{L}$ is $\mathrm{I}_{\mathrm{L}}=\frac{\mathrm{V}_{\mathrm{Z}}}{\mathrm{R}_{\mathrm{L}}}=\frac{6}{1000}=0.006 \mathrm{~A}$
Current through Zener diode is $\mathrm{I}_{\mathrm{Z}}=\mathrm{I}-\mathrm{I}_{\mathrm{L}}=0.03-0.006=0.024 \mathrm{amp} .$ Power dissipated in Zener diode is $\mathrm{P}_{\mathrm{Z}}=\mathrm{V}_{\mathrm{Z}} \mathrm{I}_{\mathrm{Z}}=6 \times 0.024=0.144 \mathrm{Watt}$
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