A Zener diode,having a breakdown voltage equa | vedclass

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(A) The total current $I$ supplied by the source is given by the voltage drop across the series resistor $R_S = 250\ \Omega$.$I = \frac{E - V_Z}{R_S}

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$5$

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(A) The total current $I$ supplied by the source is given by the voltage drop across the series resistor $R_S = 250\ \Omega$.$I = \frac{E - V_Z}{R_S} = \frac{20\ V - 15\ V}{250\ \Omega} = \frac{5\ V}{250\ \Omega} = 0.02\ A = 20\ mA$.The load current $I_L$ flowing through the load resistor $R_L = 1\ k\Omega = 1000\ \Omega$ is determined by the Zener voltage $V_Z = 15\ V$.$I_L = \frac{V_Z}{R_L} = \frac{15\ V}{1000\ \Omega} = 0.015\ A = 15\ mA$.Applying Kirchhoff's Current Law at the node,the current through the Zener diode $I_Z$ is:$I_Z = I - I_L = 20\ mA - 15\ mA = 5\ mA$.

$A$ Zener diode,having a breakdown voltage equal to $15\ V$,is used in a voltage regulator circuit as shown in the figure. The current through the Zener diode is $.....\ mA$.

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