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(A) The input voltage is $V_{in} = 20\, V$. The series resistance is $R_s = 250\, \Omega$. The Zener breakdown voltage is $V_z = 15\, V$. The load res

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$5$

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(A) The input voltage is $V_{in} = 20\, V$. The series resistance is $R_s = 250\, \Omega$. The Zener breakdown voltage is $V_z = 15\, V$. The load resistance is $R_L = 1\, k\Omega = 1000\, \Omega$.$1$. Calculate the total current $(I)$ flowing through the series resistor $R_s$:$I = \frac{V_{in} - V_z}{R_s} = \frac{20\, V - 15\, V}{250\, \Omega} = \frac{5\, V}{250\, \Omega} = 0.02\, A = 20\, mA$.$2$. Calculate the current $(I_L)$ flowing through the load resistor $R_L$:$I_L = \frac{V_z}{R_L} = \frac{15\, V}{1000\, \Omega} = 0.015\, A = 15\, mA$.$3$. The current through the Zener diode $(I_z)$ is given by:$I_z = I - I_L = 20\, mA - 15\, mA = 5\, mA$.

$A$ 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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