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(D) Given: Breakdown voltage of Zener diode $V_Z = 6\,V,$ load resistance $R_L = 4\,k\Omega,$ series resistance $R_i = 1\,k\Omega.$ The current throug

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$0.5\,mA;\,8.5\,mA$

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(D) Given: Breakdown voltage of Zener diode $V_Z = 6\,V,$ load resistance $R_L = 4\,k\Omega,$ series resistance $R_i = 1\,k\Omega.$ The current through the load resistance is constant because the Zener diode maintains a constant voltage $V_Z$ across it: $I_L = \frac{V_Z}{R_L} = \frac{6\,V}{4\,k\Omega} = 1.5\,mA.$ The total current $I_i$ supplied by the battery through the series resistor $R_i$ is given by $I_i = \frac{V_B - V_Z}{R_i}.$ For minimum battery voltage $V_B = 8\,V$: $I_{i,min} = \frac{8\,V - 6\,V}{1\,k\Omega} = \frac{2\,V}{1\,k\Omega} = 2\,mA.$ The Zener current $I_Z = I_i - I_L = 2\,mA - 1.5\,mA = 0.5\,mA.$ For maximum battery voltage $V_B = 16\,V$: $I_{i,max} = \frac{16\,V - 6\,V}{1\,k\Omega} = \frac{10\,V}{1\,k\Omega} = 10\,mA.$ The Zener current $I_Z = I_i - I_L = 10\,mA - 1.5\,mA = 8.5\,mA.$ Thus,the minimum and maximum currents through the Zener diode are $0.5\,mA$ and $8.5\,mA$ respectively.

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