:Р iL(0+) = iL(0-) = 220/ 20 = 11Р iL() = 220/ 12 = 18.33Р = L/R2 = 0.6 = 0.05Р iL(t) = 18.33 + (11 - 18.33) e - t/0.05Р10.电路如图所示,求支路电流I1, I2 ,I3。Р Р解:用节点分析法,弥尔曼定理Р Uab = (4/10 + 2/10 - 0.1)/ (1/10 + 1/20 + 1/10) = 2VР I2 = Uab / 20 = 0.1AР Uab = -10 * I1 + 4 Р I1 = (4 - Uab) / 10 = 0.2AР I3 = I1 - I2 - 0.1 = 0A Р11.如图所示电路中,开关S原来在“1”位置, 在=0瞬间换接到“2”位置,试求及。Р答:开关换接瞬间,根据换路定律Р根据三要素法求,得Р12.晶体管放大电路如图所示,已知=12V, =3kΩ, =240kΩ,晶体管=50,=0.7V。(其中进行计算)Р(1)试用直流通路估算静态值,,;РIBQ = Vcc-UBEQ/Rb = 12 - 0.7/240 = 0.05mAРICQ = *IBQ = 50 * 0.05 = 2.5 mAРUCEQ = VCC- ICQRc = 12 - 2.5 * 3 = 4.5VРUCEQ >> UBEQ, 即VC< VB<VE ,晶体管工作在作在放大区。Р静态时(=0),、上的电压各为多少?并标出极性。Р C1C2极性如图,C1的电压为零,C2为6V ,ui=0的等效电路如下:Р Р(3)画出微变等效电路,当=6 kΩ求电压放大倍数Р解:微变等效电路Р Р = 200+50* 26/2.5 = 2.54 kΩР = Uo/UI = -(RC//RL) / (Rb+rbe)Р = -50*3||6 / 2.54Р = -39
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