52.答案(1)每盏灯电流：
每台电阻炉的电流
(2)P=10PA+2PB=3000W (3)W=Pt=3?3kWh=9kWh
53.答案(a)I=0，P=0(b)I=1A，供出100W(c)I=?10A，吸收100W 54.答案US=Uab+IR0US=6VR0=0.2?
I=2AUab=5.6V，I=3AUab=5.4V，I=4AUab=5.2V
55.答案P=IR I=30A
2
R0=0.1?P0=IR0=30?0.1W=90W
56.答案(a)U=0P=0 (b)U=50V供出500W (c)U=10V供出100W 60.答案I=1AU=(?1?20+100?60)V=20V 61.答案
22
I1=1AI2=1AI3=4A
I4=5AI5=6AI6=5A
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62.答案UBC=?2V UAD=5V UAB=4V IAB=0.5A
IBC=?1A IAD=0.25A ICA=0.75A ICD=?1.75A IDB=?1.5A 63.答案U=?5V
64.答案 US=200V
65.答案KVL：I2R2=6+UbeI2=0.335mA KVL：I1R1+Ube=24
I1=0.376mA Ib=I1?I2=0.041mA 66.答案U=(12+2?2)V=16V U2=?4V
67.答案
U=(3?2+1?8+2)V=16V
68.答案W=IUt,120=0.1?U?0.5?60U=40V (方向从下至上)
(方向从下至上)
I=(0.1+0.08?0.06)A=0.12A
69.答案P=UI?2IRl即10=220I?0.4II=50A和I=500A
2
4
2
I=50AU=(220?2?50?0.2)V=200V 70.答案I=2AUab=35A(a)US=35V
(b)IS=2A(c)
71.答案(1)U1=(9+4)V=13VU2=(2?3+9)V=15VI=4(2)9V电压源反接时，则
U1=?5VU2=?3,I=?14A
72.答案I1=5AI2=0U1=?8V电流源供出功率为：6W 73.答案应用KVL5I1=10I2+10I5故I5=?0.5A
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应用KCLI3=I1+I5=2.5AI4=I2?I5=2.5A 应用KVLI4R4=10I5+6I3故R4=4?
US=5I1+6I3=30V 74.答案I5=1AI3=1A
应用KVL6I1+Uab=I2?14故I2=1AI4=I2+I5=2A应用KVL 6I3=2I5+R4I4R4=2?
应用KVL9I6=6I1+6I3故I6=2A IS=I1+I2+I6=5A 75.答案S断开时
I3=I4=0S闭合时
不变
I5=?I3?I1=?2.12A
I6=I3?I2=7.89A
76.答案I4=I1+I3=0.04A
77.答案电流源的电压
(上“?”，下“?”）
=(10?4)mA=6
由KVL
故
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