The voltage of a 9V battery falls from 9V when it is new to about 7V near the end of its life. To get reasonable life out of a 9V battery, any circuit powered from it should be able to operate at 7V. To prevent the performance from changing as the voltage of the battery falls, design a circuit, that draws 10mA, to run off a 6V, and to control the voltage supplied to it with a 6V zener diode, which can operate with as little as 2mA through it. R = 82Ω When the battery is new, what current flows from it? Attempt: I tried to work out the resistance of the zener: Rz = Vz / Iz = 6 / 0.002 = 3000Ω I = Iz + Iload Vin max = 9V I = Vin max / R + Rz = 9 / 82 + 3000 = 2.92mA The correct answer was 36.6mA, but I cant see what I did wrong.