?15РR2 = XL = | Р?= √× √Ω= 7.5ΩР√2?2?2Р4.5.4Р在图9(a)中,I1 = I2 = 10A,U = 100V ,u与i同相,试求I ,R,XCР及XL 。Р[解]Р作相量图:以U˙ R 为参考相量,I˙2与U˙ R 同相,I˙1超前于U˙ R?90◦;I˙= I˙1 + I˙2 =Р图 9: 习题4.5.4图Р10√2∠45◦A,U˙ L 超前于I˙Р?90◦;U˙与I˙同相;U˙Р?= U˙ L + U˙ R 。于是得出РI?= 10√2A = 14.1AРR?=?URРI2Р?√2UР=РI2Р?√2?100Р=Р×Р10Р?РΩ= 14.1ΩРXC?=Р?UR?√2?100Р×Р=?Ω= 14.1ΩРI1 10РUL?UР?100РXL?=Р4.5.12Р1Р在图14所示的电路中,已知Uab = Ubc ,R = 10Ω,XC =РωCР感性负载。试求U˙和I˙同相时Zab 等于多少?Р[解]Р令Zab = R1 + jXLР图 14: 习题4.5.12图Р?Р= 10Ω,Zab 为电Р又求得Р?РZbc = −Р?РjRXCР?Р= −j10 × 10Р?РΩ= (5 − j5)ΩРR − jXCР所以Р?10 − j10РU˙= U˙ ab + U˙ bc = (Zab + Zbc )I˙Р= [(R1 + jXL ) + (5 − j5)]I˙= [(R1 + 5) + j(XL − 5)]I˙Р若U˙和I˙同相,则上式的虚部必为零,即РXL − 5 = 0?XL = 5ΩР又因Uab = Ubc ,则|Zab | = |Zbc |РqR2Р?2 √ 2 2Р1 + XL =Р?5 + 5Р解之得?R1 = 5ΩР于是?Zab = (5 + j5)ΩР?Р=?= Ω= 7.07ΩР√РI?I?10 2Р?Р1Р0.8Р?
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