华南理工大学《电工学》考试真题5

华南理工大学《电工学》考试真题5华南理工大学期末考试《电路与电子技术》（双语）试卷B考试时间：120 minutesD a t e ：J a n u a r y , 15, 2013注意事项：1. 考前请将密封线内填写清楚；2. 所有答案请直接答在答题纸上； 3．考试形式：闭卷；1. consider the circuit shown in Fig.1, find current I 1, I 2 and I 3 in KCL.（6 scores ）Fig.1I 1=1－3=－2A 2 I 3=－31266+⨯=－1A 4I 2=－I 3－1=1－1=0 62. Consider the circuit shown in Fig.2, assuming that the diodes are ideal. Find the value of I 1 and I 2. （8 scores ）Fig.2由D 1导通，D 2截止 得I 2＝0，I 1＝()1233+mA ＝5mA3．Consider the circuit shown in Fig.3 given that U S =9V ，I S =3A ，R 1=3Ω，R 2=6Ω. Determine current I 1，I 2 in superposition, which elements are supplying power? （8 scores ）22Fig.3I R 22' 原图=I R 22"I S 单独作用时：'=+⨯=I R R R I 12122S AI 2'=1A2U S 单独作用时：''=''=+=I I U R R 12121SA4叠加得：I 1=I 1'+I 1"=3A I 2=I 2'－I 2"=0P U I P R I I U I ====S S W 122270()P I R P R R 1 2W ===121270U S 是电源4.Consider the circuit shown in Fig.4 given that U S1=10V ，U S2=20V ，R 1=R 2=R 3=R 4=R 5=2Ω. Determine current I in thevenin theorem. （10 scores ）U S 25Fig.4R 3U R R R U R R R U 02124455==V S1S2+⨯-+⨯-4R 0=R 1//R 2+R 4//R 5=2Ω7 I U R R ==A 003125+-.105.As shown in following circuit Fig.5, R 1=60Ω, R 2=120Ω, R 3=40Ω, L =4H, U S V 124=，U S V 220=. the switch S has been open for a long time ，and then closed at t =0, please find the i t L ()after S is closed. (10 scores)S 2Fig.5i i U R L L ()().000523+-===S A2i U R U R L ().∞=+=S S A 112309(叠加原理)4τ==LR R R 12302////. s6i t i i i L L L L t()()()()=∞+-∞+-0e τ=--09045..e A t10 6.Consider the sinusoidal circuit shown in Fig.6 given that: V t v )10000cos(2318=，R 1=R 2=1Ω，L =0.1mH ，C =100μF. Determine: i(t), the power factor and average power of the total circuit. （10 scores ）R 2Fig.6V V ︒∠=02318Z R L R CR C=++--=∠︒122159184j j /j /..ωωωΩ3A ZVI ︒-∠==4.1822005 A t i )4.1810000cos(2200︒-=6 Power factor:λ==cos ..184095o8 Average power:42.604.18cos =︒=rms rms I V P kW107.In the balanced three-phase system with a 380V line voltage of fig. 7 givenZ=24+18j Ω. Find the line current of I a A , I a B ，I a C , and plot phasor diagram. Assume that the phase of V ab is zero (10 scores)Z ZZ设： A B =3800 V =j =3036.9L U Z R X ︒︒∠+∠Ω 2A BA B 12.6636.9 A U I Z︒==∠- 4A =312.6(36.930)=21.9266.9 A I ︒︒︒⨯∠--∠-6因为：负载为三相对称负载所以：B C 21.92173.1 A 21.9253.1 A I I ︒︒=∠=∠8相量图I .U .BI .AI .C120120120A Booo60o8. Consider the circuit shown in Fig. given that: R 1=R 2=R 3 R F1=R F2=R F3，find the relationship between u o and u I .（scores 10）'=-u u u O O I由u u O O =-'得u u u O O I =-+故u u u O I I 2==129.Transistors are operating in the active region in Fig.9，assume that: β=40，r π=1kΩ，determine ：（1）plot small signal equivalent circuit ；（2）calculate input resistor r i and output resistor r o ；（3）voltage gainA u 。

(10 scores).oU CCFig.9(1)(2)()()[]()()[]Ω=Ω++=++=k 1.31k 5.1//34011//120//1//L E be B iR R r R r β(3)()()()()1//1//1L E be L E ≈+++=R R r R R A u ββ10. Consider the common-source amplifier illustrated in Figure 10，(1) plot small signal equivalent circuit; (2) Compute the voltage gain, input resistance, and output resistance.（10 scores ）.(t vFigure 10R L'(t v R LG4We denote the equivalent resistance byL D d L R R r R /1/1/11'++=The output voltage is the product of the current from the controlled source and the equivalent resistance:L gs m R v g v ')(0-= The input voltage and the gate-to-source voltage are equal:gsin v v =we obtain the voltage gain:L m inov R g v v A '-==6Input Resistance21R R R i v R G ininin ===8Output ResistanceoFigure 3.19 Circuit used to find R O .d D r R R /1/110+=1011. In Fig.11，write the logic formula and make simplification in De Morgan ’s law, analyze its logic function.(8 scores)FFig.11F ABAB ABAB AB A B ===+()=++=+++=+()()A B A B AA BB A B AB A B AB是“异或”门。