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(B) $1$. The input voltage is $V_{in} = 20\,V$ and the series resistance is $R_s = 250\,\Omega$.$2$. The Zener diode is in parallel with the load resi

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$5$

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(B) $1$. The input voltage is $V_{in} = 20\,V$ and the series resistance is $R_s = 250\,\Omega$.$2$. The Zener diode is in parallel with the load resistor $R_L = 1\,k\Omega = 1000\,\Omega$. Since the Zener diode is in breakdown,the voltage across the load resistor is $V_L = V_Z = 15\,V$.$3$. The current through the load resistor is $I_L = \frac{V_L}{R_L} = \frac{15\,V}{1000\,\Omega} = 0.015\,A = 15\,mA$.$4$. The total current supplied by the source through the series resistor $R_s$ is $I_s = \frac{V_{in} - V_Z}{R_s} = \frac{20\,V - 15\,V}{250\,\Omega} = \frac{5\,V}{250\,\Omega} = 0.02\,A = 20\,mA$.$5$. Applying Kirchhoff's current law at the node,the current through the Zener diode is $I_Z = I_s - I_L = 20\,mA - 15\,mA = 5\,mA$.

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

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