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An unregulated DC power supply of 12 V is connected across a series combination of a resistor of 1 k and a Zener diode. The Zener diode has a breakdown voltage of 5 V and is connected in reverse bias. Assuming no load resistor is connected across the Zener diode, what is the current flowing through the series resistor?

Options

  1. A12 mA
  2. B17 mA
  3. C7 mA
  4. D0 mA

Correct answer

C. 7 mA

Step-by-step solution

The Zener diode is connected in reverse bias and the input voltage ( 12 V ) is greater than the Zener breakdown voltage ( 5 V ). Therefore, the Zener diode operates in the breakdown region and maintains a constant voltage of 5 V across itself. The voltage drop across the series resistor is the difference between the input voltage and the Zener voltage: V_R = V_ in - V_Z = 12 V - 5 V = 7 V Using Ohm's law, the current flowing through the series resistor is: I = V_R R = 7 V 1 k = 7 mA Answer: 7 mA

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