Discussion
In the following zener diode circuit
In the circuit, a 16 V battery feeds, through a 60 Ω series resistor, a 10 V zener diode connected across a 250 Ω load resistor.
the current flowing through the zener diode is :
Answer: B
The zener current is 60 mA.
The zener holds the load at 10 V, so the series resistor drops the remaining 6 V. The currents are:
\[ \begin{aligned} I_{s} &= \frac{16 - 10}{60} = 0.1\ \text{A} = 100\ \text{mA} \\ I_{L} &= \frac{10}{250} = 0.04\ \text{A} = 40\ \text{mA} \\ I_{Z} &= I_{s} - I_{L} = 60\ \text{mA} \end{aligned} \]
The option 40 mA is the load current and 100 mA the total current from the battery; the zener carries the difference. 80 mA does not follow from the circuit values.
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