Voc +9 V Re 560 90v VBB +3 V RB www 2N3904 56 k 3.0v Figure 7 39) Refer to Figure 7. Assume a single problem accounts for the readings. From the readings, you can conclude that A) the base-collector junction is open C) RC is shorted 40) Refer to Figure 7. The transistor is A) in the active region C) cutoff D) Rg is open 39) B) the base-emitter junction is open 40) B) saturated D) in the breakdown region + Voc=15 V BDC = 100 VBB 4.0 V RB ww Re≥ 1.5 k Figure 4 26) Refer to Figure 4. Assume RB 22 k. The collector current will be approximately 130 A) 10 mA B) 5 mA C) 20 mA 27) Refer to Figure 4. The largest base resistor that will still saturate the transistor is approximately Α) 33 ΚΩ Β) 330 ΚΩ 28) Refer to Figure 4. If RB = 100 k2, Vc is approximately A) 12.5 V B) 10 V €) 40 ΚΩ C) 7.5 V D) 27 mA 27) D) 400 kc 28) D) 5V 163 29) 29) Refer to Figure 4. If you wanted to change the transistor to a pnp type, you need to A) increase the base resistor B) increase the collector resistor C) change both dc sources to negative supplies D) all of the above SH 6V Figure 1 R₁ ww 33 ΚΩ R180 Ve 15 V 12) Refer to Figure 1. Assume ẞoc 200. The expected value of ig is A) 161 A B) 182 A C) 200 μA 13) Refer to Figure 1. Assume BDC 200. The value of VCE is approximately A) 5.8 V B) 9.2 V 14) Refer to Figure 1. If VCC is increased to 18 V. IB will A) increase B) decrease 2 C) 7.5 V 12) D) 80 A 13) D) 12.4 V 14) C) not change