A Zener diode has a specified Zener voltage of 10 V and conducts a current of 20 mA in regulation. What is the power dissipation in the device?

Introduction / Context:
Zener diodes are used for voltage regulation by operating in reverse breakdown at approximately constant voltage. Knowing the Zener voltage and current allows direct computation of power dissipation, which is critical for selecting an adequate power rating and heat management.

Given Data / Assumptions:

• Zener voltage V_Z = 10 V.
• Zener current I_Z = 20 mA = 0.02 A.
• Steady-state operation within safe breakdown region.

Concept / Approach:
Device power dissipation is P = V * I. For a Zener in regulation, the voltage across the device is approximately V_Z while the current is I_Z. Multiply to find the dissipation in watts, then convert to milliwatts if convenient.

Step-by-Step Solution:
Use formula: P = V_Z * I_Z.Substitute: P = 10 V * 0.02 A.Compute: P = 0.2 W.Convert: 0.2 W = 200 mW.

Verification / Alternative check:
Quick mental math: 10 * 2 = 20; place decimal for 0.02 A gives 0.2 W = 200 mW, consistent.

Why Other Options Are Wrong:
300 mW: would require 30 mA at 10 V.20 mW or 10 mW: far too low for the stated voltage and current.None of the above: incorrect because 200 mW is exact.

Common Pitfalls:
Confusing milliamps and amps when multiplying; ignoring the device power rating (must exceed calculated dissipation with margin).

Final Answer:
200 mW