A Zener diode with a breakdown voltage of 15 | vedclass

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(A) The current flowing through the load resistor $R_L$ is given by:${I_L} = \frac{{{V_Z}}}{{{R_L}}} = \frac{{15}}{{1 	imes {{10}^3}}} = 15 \, mA$App

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$5$

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(A) The current flowing through the load resistor $R_L$ is given by:${I_L} = \frac{{{V_Z}}}{{{R_L}}} = \frac{{15}}{{1 	imes {{10}^3}}} = 15 \, mA$Applying Kirchhoff's voltage law to the input loop:${V_i} = I \cdot R + {V_Z}$Since the total current $I = {I_Z} + {I_L}$,we have:${V_i} = ({I_Z} + {I_L})R + {V_Z}$Rearranging to solve for the Zener current ${I_Z}$:${I_Z} = \frac{{{V_i} - {V_Z}}}{R} - {I_L}$Substituting the given values (${V_i} = 20 \, V$,${V_Z} = 15 \, V$,$R = 250 \, \Omega$,${I_L} = 15 \, mA$):${I_Z} = \frac{{20 - 15}}{{250}} - 15 	imes {10^{ - 3}}$${I_Z} = \frac{5}{{250}} - 0.015$${I_Z} = 0.02 - 0.015 = 0.005 \, A = 5 \, mA$

$A$ Zener diode with a breakdown voltage of $15 \ V$ is used as a voltage regulator. In the circuit shown,the current flowing through the Zener diode is ...... $mA$.

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