## Compute the values of V_{OUT} & P_{1} in a variable feedback regulator circuit with Vin = 20V, V_{z} = 12V & R_{L} = 2K if the transistor is of silicon and the wiper of R_{3} is adjusted in half-way configuration.

a. 16.8 V & 137.76 mW

b. 10.7 V & 80.09 mW

c. 6.7 V & 43.89 mW

d. 4.7 V & 35.79 mW

Correct Answer: c. 6.7 V & 43.89 mW

**Explanation** From the configuration of feedback regulator circuit,

V

_{OUT} = Voltage at wiper + V

_{BE2}Voltage at wiper must be equal to half the Zener voltage since wiper is adjusted in half way configuration.

V

_{OUT} = (12 / 2) + 0.7

= 6.7 V

P1 = V

_{CE1} x I

_{E1} But, V

_{CE1} = V

_{in} - V

_{out} = 20 – 6.7

= 13.3 V

I

_{E1} = I

_{L} = V

_{out} / R

_{L} = 6.7 /2k = 3.3 mA

Therefore, P1 = 13.3 x 3.3 = 43.89 mW