《电路》（英文版）9-2 Average power

89-2 Average power P、1 nn c0s9+-"1"cos(2Ot+9)t 2 2 1mmcs9)7+0≈1mn1p cos 9=VI cos 9 2 The pure resistor: 9=0 c0S 9=1 mm=V=RI=V/ R 2 The pure inductor(capacitor): 9=90(90)COS 9=0 P=O

§ 9-2 Average power cos ) 0 2 ( 1 = T + V I T m m   = T pdt T P 0 1 The pure resistor:  = 0 cos = 1 VI RI V R V I P m m R / 2 2 2 = = = = PX = 0 The pure inductor(capacitor):  = 90 (−90 ) cos = 0   t dt V I V I T T m m m m  = + + 0 cos(2 )] 2 cos 2 [ 1    cos cos 2 VI Vm I m = =

元 Example 1: If 0=42 cos tV, Z=2/60Q2 Find p(t)and P S D,l, p olution 4∠0 z2∠60 =2∠-60°A i(t)=2√2cos(2t-60°)A 6 p()=42cos"tx2√2cos(t-60°) 6 cos60+ cos(t-60°) 16 3 2 =4+8c0s(t-60°)4P=4×2cos60°=4W 3

Example 1: If , 2 60 . ( ) . 6 = 4 2 cos t V Z =   Find p t and P    Solution: A Z V I =  −    = = • • 2 60 2 60 4 0   i t t 60 )A 6 ( ) = 2 2 cos( −   t 60 )A 3 = 4 + 8cos( −   60 ) 6 2 2 cos( 6 p(t) = 4 2 cos t  t −    2 60 ) 3 cos 60 cos( 16 + −  =  t   P = 42cos 60 = 4W i ,i, p t  p

R1 VOFF =0 3.3 VAMPL =5.656 FREQ=0.0803 0 10 口V(V1:+)◇W(R1)V工(R1) Time

Time 12s 14s 16s 18s 20s 22s 24s V(V1:+) W(R1) I(R1) -10 0 10 R 1 1 V1 FREQ = 0.0803 VAMPL = 5.656 VOFF = 0 W 0 I L1 3.3 1 2 V