推荐回答（2个）
一楼回答不详细 我用我的说法来告诉您吧
第一题： 因为1500r/KW.h 说明电能表转1500转 消耗1KW.h=3.6X10六次方J电能
当转了10转时 则消耗了10分之1500 X3.6X10六次方J=24000J
所以4分钟消耗24000J电能 即：4分之60小时消耗24000J电能
1分之15小时消耗24000J电能 则1小时消耗 15X24000J=360000J （360000J=0.1KW.h）
答：消耗电能是36000J（或者说是0.1KW.h）。
第二题： （1）因为电路并联 所以R1中的电压=电源电压=6V 因为R1的阻值=20欧 则可用欧姆定律I=U/R 得I（R1）=U（R1）/R1=6V/20欧=0.3A
（2）因为电流表接在干路上 所以总电流=0.5A 因为电路并联 所以总电压=电源电压=6V 再根据欧姆定律 变换得 R总=U总/I总=6V/0.5A=12欧 再依并联电路的公式：1分之R总=1分之R1+1分之R2 所以1/12欧=1/20欧+1/R2欧 解出R2=30欧
（3）因为I总=0.5A I（R1）=0.3A（第一问已解出）电路并联
所以I（R2）=0.5-0.3=0.2A U（R2）=U总=6V
根据公式P=UI 得P（R2）=6V.0.2A=1.2W
第三题：因为串联 则两只灯泡通过的电流都相同
再依题可知L1的额定电压为220V 额定电功率为40W L2的额定电压为110V 额定电功率为100W 因为P=U²/R 所以R=U²/P 所以R（L1）=220V²/40W=1210欧 R（L2）=110V²/100W=121欧 根据P=UI 可以解出I（L1）=P/U=40W/220U= 2/11 A（所以L1的额定电流为2/11A）I（L2）=P/U=100W/110V=10/11A（所以L2的额定电流为10/11A） 因为电流都相同 L1的电流又不能大于2/11A 所以这两个灯泡通过的电流最大也就只能为2/11A（如果超过这个电流值 灯泡L1就会烧坏） 则此时P总=P（L1）+P（L2）= I²R（L1）+I²R（L2）=I²（R1+R2）=4/121（1210+121）=44W 答：总功率最大不得超过44瓦。
第四题： 依题可知电器的额定电压为220V 额定电功率为140W
根据P=UI 所以I=140W/220V=7/11A（额定电流） 所以正常工作时他的电流为7/11A
根据W=UIt得 W=220V.7/11A.2.60.60S=1.008X10六次方J 1.008X10六次方J=0.28KW.h 因为1KW.h=1度电 所以正常工作两小时的电能为0.28度。
LZ啊，我写这些不容易啊 求采纳了！
1、先算灯泡功率W电=10/1500 KW.H =24000J P= W电/t=100w =0.1kw 所以后面的电功
W=pt=0.1kw*1h=0.1kwh
2、I1=U/R1=0.3A，I2=0.5-0.3=0.2，R2=U/I2=30欧，P2=UI2=6*0.2=1.2w
3、串联时考虑两灯各自允许通过的最大电流I1=40/220 ，I2=100/110 ，取两者中小的故取I1，然后总功率P=（R1+R2）*I1；并联时，能加的最大电压取两者中小的110V，然后自己求出求出总功率
4、自己用基本公式算
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