What is the current flowing through the Zener | vedclass

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(D) The Zener diode is in parallel with the load resistor of $1 \, k\Omega$. Since the Zener voltage is $5 \, V$,the voltage across the load resistor

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$5$

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(D) The Zener diode is in parallel with the load resistor of $1 \, k\Omega$. Since the Zener voltage is $5 \, V$,the voltage across the load resistor is also $5 \, V$.The current through the load resistor is $I_L = \frac{V_Z}{R_L} = \frac{5 \, V}{1 \, k\Omega} = 5 \, mA$.The total current $I$ flowing from the source through the $500 \, \Omega$ resistor is $I = \frac{V_{source} - V_Z}{R} = \frac{10 \, V - 5 \, V}{500 \, \Omega} = \frac{5 \, V}{500 \, \Omega} = 0.01 \, A = 10 \, mA$.Applying Kirchhoff's Current Law at the node,the current through the Zener diode $I_Z$ (or $I_1$) is $I_Z = I - I_L = 10 \, mA - 5 \, mA = 5 \, mA$.

What is the current flowing through the Zener diode in the given circuit in $mA$?

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