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 A shunt regulator utilizing a zener diode with an incremental resistance of 5 Ω is fed through an 82-Ω resistor. If the raw supply changes by 1.0 V, what is the corresponding change in the regulated output voltage?
  • a)
    72.7 mV
  • b)
    73.7 mV
  • c)
    74.7 mV
  • d)
    75.7 mV
Correct answer is option 'C'. Can you explain this answer?
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Ohms is connected to a 12V power supply. If the load current is 500mA, what is the power dissipated by the zener diode?

We can use the formula for power dissipation:

P = V × I

where P is power, V is voltage, and I is current.

First, we need to calculate the voltage across the zener diode. Since it is a shunt regulator, the voltage across the load will be constant at 12V. Therefore, the voltage across the zener diode will be:

Vz = 12V - Vload

where Vload is the voltage across the load, which is also 12V.

Vz = 12V - 12V = 0V

This means that the zener diode is operating in its breakdown region, maintaining a constant voltage of 0V across itself.

Next, we can calculate the power dissipated by the zener diode:

P = Vz × I

P = 0V × 500mA

P = 0

Since the voltage across the zener diode is 0V, there is no power dissipated by the diode. However, the power dissipated by the load would be:

Pload = Vload × I

Pload = 12V × 500mA

Pload = 6W

Therefore, the power dissipated by the zener diode is 0W, and the power dissipated by the load is 6W.
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A shunt regulator utilizing a zener diode with an incremental resistance of 5 Ω is fed through an 82-Ω resistor. If the raw supply changes by 1.0 V, what is the corresponding change in the regulated output voltage?a)72.7 mVb)73.7 mVc)74.7 mVd)75.7 mVCorrect answer is option 'C'. Can you explain this answer?
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