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(A) The voltage drop across the $1\,k\Omega$ resistor is equal to the Zener breakdown voltage,$V_Z = 15\,V$.The current through the $1\,k\Omega$ resis

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$5$

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(A) The voltage drop across the $1\,k\Omega$ resistor is equal to the Zener breakdown voltage,$V_Z = 15\,V$.The current through the $1\,k\Omega$ resistor $(I')$ is calculated as:$I' = \frac{V_Z}{R_L} = \frac{15\,V}{1 	imes 10^3\,\Omega} = 15 	imes 10^{-3}\,A = 15\,mA$.The voltage drop across the $250\,\Omega$ series resistor is the difference between the input voltage and the Zener voltage:$V_R = 20\,V - 15\,V = 5\,V$.The total current $(I)$ flowing through the $250\,\Omega$ resistor is:$I = \frac{V_R}{R} = \frac{5\,V}{250\,\Omega} = 0.02\,A = 20\,mA$.Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $(I_Z)$ is:$I_Z = I - I' = 20\,mA - 15\,mA = 5\,mA$.

$A$ 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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