In the given circuit,the value of I_{L} (in m | vedclass

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(A) The Zener diode is connected in parallel with the load resistor $R_{L} = 2 \ k\Omega = 2000 \ \Omega$.Since the Zener diode is in the breakdown re

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(A) The Zener diode is connected in parallel with the load resistor $R_{L} = 2 \ k\Omega = 2000 \ \Omega$.Since the Zener diode is in the breakdown region,the voltage across the load resistor $R_{L}$ is equal to the Zener voltage $V_{Z} = 5 \ V$.Using Ohm's law,the current through the load resistor $I_{L}$ is given by:$I_{L} = \frac{V_{Z}}{R_{L}}$$I_{L} = \frac{5 \ V}{2000 \ \Omega} = 0.0025 \ A$Converting to milliamperes $(mA)$:$I_{L} = 0.0025 	imes 1000 \ mA = 2.5 \ mA$.

In the given circuit,the value of $I_{L}$ (in $mA$) will be: (in $.5$)

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