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 \[{{P}_{Z}}={{V}_{Z}}{{I}_{Z}}=6\times 0.024=0.144\] WattYou need to login to perform this action.
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