题目,所示电路中,当R为何值时获得最大功率?
2个回答
展开全部
干路电流为
i = 12/[4R/(4+R)+3]
电阻R两端电压为
U = 12 - 3i = 12 - 36/[4R/首碰孝(4+R)+3] = [48R/(4+R)]/[4R/(4+R)+3] = 48R/[4R+3(4+R)] = 48R/(7R+12)
于是电阻R功率为
P = U2/R = 2304R/(7R+12)2 = 2304/(49R+168+144/R)
当R = 12/7Ω时,Pmax = 2304/者稿(84+168+84) = 48*48/84*4 = 48*3/21 = 48/吵裤7 W
i = 12/[4R/(4+R)+3]
电阻R两端电压为
U = 12 - 3i = 12 - 36/[4R/首碰孝(4+R)+3] = [48R/(4+R)]/[4R/(4+R)+3] = 48R/[4R+3(4+R)] = 48R/(7R+12)
于是电阻R功率为
P = U2/R = 2304R/(7R+12)2 = 2304/(49R+168+144/R)
当R = 12/7Ω时,Pmax = 2304/者稿(84+168+84) = 48*48/84*4 = 48*3/21 = 48/吵裤7 W
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询