例13-5电路如图13-16(a)所示。已知us (t)= 10 /2 cos10t V试求: (l) i(t),i,(t) ;(2）1.6Q2负载电阻吸收的功率。j220.2H12120.42i,0.422us10/0°j32j221·620·3H0.2H1.62福(b)(a)图13-16
例13－5 电路如图13-16(a)所示。已知 uS (t) =10 2 cos10t V 图13－16 试求：(l) i 1 (t),i 2 (t)； (2) 1.6负载电阻吸收的功率
j220.2Hi1212i10.420.42i210/0°vus0.2Hj3230.3H31-62j221-6Q(b)(a)解：画出相量模型，如图(b)所示。求出反映阻抗24M2=(1- jl)QrefW2+i222求出输入阻抗Z, = Z + Zref= (1+ j3 +1- j1)2 =(2 + j2)求出初级电流100°A= 2.5V2Z- 45°A2 + j2
解：画出相量模型，如图(b)所示。求出反映阻抗 = − + = = (1 j1) 2 j2 2 2 2 2 2 2 ref Z M Z 求出输入阻抗 = + + − = + = + (1 j3 1 j1) (2 j2) Zi Z1 1 Zref 求出初级电流 A 2.5 2 45 A 2 2 10 0 i S 1 = − + = = Z j U I