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The three-terminal linear voltage regulator is connected to a 10 Ω load resistor as shown in the figure given below. If Vin is 10 V, what is the power dissipated in the transistor?
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- 0.6 W
- 2.4 W
- 4.2 W
- 5.4 W
Correct Option: B
BJT operates in active region.
| Current in RL = | = 0.59 amp | |
| 10 |
| Current in 1 kΩ = | = 0.7 M amp | |
| 1 × 1000 |
∴ Total current = 0.59 + 0.0007 amp
Power dissipated in transistor = (0.59) × (3.4 + 0.7)
= 0.59 × (4.1) = 2.4 W
