Question

6.13 In the circuit shown in Fig. E6.13, the voltage at the emitter was measured and found to be -0.7 V. If B = 50, find lg,

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Answer #1

1E =-07-W) en 0.93mar 9.3 = 0.93 mA (ОК Pok te Te = x 0-93 = 0-911460A = 0.9 1176mA = 18-2352qut. so Vc = 10-5K XIC =(0-5x0.9

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