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(B) The voltage across the Zener diode is constant at $15 \, V$ because it is in the breakdown region.First,calculate the current through the load res

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$5$

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(B) The voltage across the Zener diode is constant at $15 \, V$ because it is in the breakdown region.First,calculate the current through the load resistor $(R_L = 1 \, k\Omega)$:$I_L = \frac{V_Z}{R_L} = \frac{15 \, V}{1 	imes 10^3 \, \Omega} = 15 \, mA$.Next,calculate the total current $(I)$ flowing through the series resistor $(R_S = 250 \, \Omega)$:The voltage drop across the series resistor is $V_S = V_{in} - V_Z = 20 \, V - 15 \, V = 5 \, V$.$I = \frac{V_S}{R_S} = \frac{5 \, V}{250 \, \Omega} = 0.02 \, A = 20 \, mA$.Finally,apply Kirchhoff's Current Law at the node:$I = I_Z + I_L$$I_Z = I - I_L = 20 \, mA - 15 \, mA = 5 \, mA$.Therefore,the current through the Zener diode is $5 \, mA$.

$A$ 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 diode is.....$mA$.

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