A zener diode of voltage V_ Z (=6 V) is used to maintain a constant...

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}$_{\mathrm{Z}}=6 \times 0.024=0.144 \mathrm{Watt}$