The Q point value Ic for the circuit is [Figure description: An NPN BJT with emitter grounded, collector connected through RC = 10 kΩ to +12 V, and a base resistor RB = 100 kΩ connected from the collector to the base (collector-to-base feedback bias). βdc = 100.]
- A. 10.18 Micro Amps
- B. 102.5 Milli Amps
- C. 1.018 Milli Amps
- D. 6.325 Micro Amps
Answer: C. 1.018 Milli Amps
Explanation: 12 = IC(10k) + (IC + IB)... with IB = IC/100: 12 − 0.7 = IC(10.1k + 1k) so IC ≈ 1.018 mA.