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A Zener diode with a breakdown voltage of 6 V is used in a voltage regulator circuit. It is connected in parallel with a load resistor of 1 k , and a series resistor of 500 is connected between the unregulated DC input and the Zener diode. If the current flowing through the Zener diode is 8 mA , what is the value of the unregulated input voltage?

Options

  1. A13 V
  2. B10 V
  3. C9 V
  4. D7 V

Correct answer

A. 13 V

Step-by-step solution

The voltage across the load resistor is clamped by the Zener diode, so V_L = V_z = 6 V . The current through the load resistor is: I_L = V_L R_L = 6 1000 = 6 mA According to Kirchhoff's Current Law, the total current through the series resistor is the sum of the Zener current and the load current: I_s = I_z + I_L = 8 mA + 6 mA = 14 mA The voltage drop across the series resistor is: V_s = I_s R_s = (14 10⁻³) 500 = 7 V The unregulated input voltage is the sum of the voltage drop across the series resistor and the Zen

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