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(B) The voltage across the load resistor $(R_L = 1 \, k\Omega)$ is equal to the Zener breakdown voltage,$V_Z = 5 \, V$.The current through the load re

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$95$

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(B) The voltage across the load resistor $(R_L = 1 \, k\Omega)$ is equal to the Zener breakdown voltage,$V_Z = 5 \, V$.The current through the load resistor is:$I_L = \frac{V_Z}{R_L} = \frac{5 \, V}{1 \, k\Omega} = 5 \, mA$.The total current supplied by the $20 \, V$ source through the series resistor $(R = 150 \, \Omega)$ is:$I = \frac{V_{in} - V_Z}{R} = \frac{20 \, V - 5 \, V}{150 \, \Omega} = \frac{15 \, V}{150 \, \Omega} = 0.1 \, A = 100 \, mA$.The current through the Zener diode $(I_Z)$ is the difference between the total current and the load current:$I_Z = I - I_L = 100 \, mA - 5 \, mA = 95 \, mA$.

$A$ Zener diode having a breakdown voltage equal to $5 \, V$ is used in the voltage regulator circuit shown in the figure. The current through the diode is ........ $mA$.

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