Discussion
The current through Zener diode in the following circuit is :
Answer: C
The Zener current is about 19.6 mA.
Without the Zener the load would see \(18 \times \frac{1000}{1270} \approx 14.2\) V, more than 10 V, so the Zener breaks down and holds the load at 10 V. Then:
\[ \begin{aligned} I_{S} &= \frac{18 - 10}{270} \approx 29.6\ \text{mA} \\ I_{L} &= \frac{10}{1000} = 10\ \text{mA} \\ I_{Z} &= 29.6 - 10 = 19.6\ \text{mA} \end{aligned} \]
29.6 mA is the current in the series resistor, which forgets that 10 mA of it goes to the load. 14.8 mA and 39.6 mA do not follow from the circuit.
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