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(B) The voltage across the Zener diode remains constant at $15 \, V$ when it is in the breakdown region.The current through the load resistor $(R_L =

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$5$

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(B) The voltage across the Zener diode remains constant at $15 \, V$ when it is in the breakdown region.The current through the load resistor $(R_L = 1 \, k\Omega)$ is:$i_0 = \frac{V_z}{R_L} = \frac{15 \, V}{1000 \, \Omega} = 0.015 \, A = 15 \, mA$The total current $(i)$ supplied by the source is determined by the voltage drop across the series resistor $(R = 250 \, \Omega)$:$i = \frac{V_{in} - V_z}{R} = \frac{20 \, V - 15 \, V}{250 \, \Omega} = \frac{5 \, V}{250 \, \Omega} = 0.02 \, A = 20 \, mA$Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $(i_z)$ is:$i_z = i - i_0 = 20 \, mA - 15 \, mA = 5 \, mA$

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

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