【单选题】
<img title=无标题.jpg alt=无标题.jpg src=https://huaweicloudobs.ahjxjy.cn/FEA630E4A29C8297FA52F7F8C1541FFB.jpg/>
①
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/A69A8A8E03B543F42881BE1FECF1D22F.png data-tex=I=14.14mA,i=20sin(1000t+{ 45 }^{ 0 })mA/>
②
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/61A79C8AA26A0B2C1204D4815BE900D6.png data-tex=I=14.14mA,i=20sin(1000t-{ 45 }^{ 0 })mA/>
③
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/76C2FE44ED64E707C4F6D00421095DAE.png data-tex=I=20mA.i=20sin(1000t+{ 45 }^{ 0 })mA/>
④
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/C45DCC8EC5DFF29788D659A0E3D310BC.png data-tex=I=14.14mA,i=14.14sin(1000t+{ 45 }^{ 0 })mA/>
【单选题】
<img title=无标题516.jpg alt=无标题516.jpg src=https://huaweicloudobs.ahjxjy.cn/BED698B3285A85529515F13CAB5AFBB3.jpg/>
①
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/47B6A2CAB0FB3169C091C0DD95BD9C76.png data-tex={ i }_{ 1 }=6.74\sqrt { 2 } sin(314t+{ 47.6 }^{ 0 })A\quad \\ \quad { i }_{ 2 }=67.4\sqrt { 2 } sin(314t+{ 47.6 }^{ 0 })A\quad \\ \quad { u }_{ 2 }=2.1\sqrt { 2 } sin(314t+{ 47.6 }^{ 0 })V/>
②
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/D7322B6290F75C65E60F173FF2DE7209.png data-tex={ i }_{ 1 }=6.74sin(314t+{ 47.6 }^{ 0 })A\quad \\ \quad { i }_{ 2 }=67.4sin(314t+{ 47.6 }^{ 0 })A\quad \\ \quad { u }_{ 2 }=2.1sin(314t+{ 47.6 }^{ 0 })V/>
③
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/5D9278CB3B1F41ECEA0BBD5364925DB1.png data-tex={ i }_{ 1 }=6.74\sqrt { 2 } sin(50t+{ 47.6 }^{ 0 })A\quad \\ \quad { i }_{ 2 }=67.4\sqrt { 2 } sin(50t+{ 47.6 }^{ 0 })A\quad \\ \quad { u }_{ 2 }=2.1\sqrt { 2 } sin(50t+{ 47.6 }^{ 0 })V/>
④
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/FE7BE4B6478D54C2EC3A8769D0E689EF.png data-tex={ i }_{ 1 }=6.74\sqrt { 2 } sin(314t+{ 25 }^{ 0 })A\quad \\ \quad { i }_{ 2 }=67.4\sqrt { 2 } sin(314t+{ 25 }^{ 0 })A\quad \\ \quad { u }_{ 2 }=2.1\sqrt { 2 } sin(314t+{ 25 }^{ 0 })V/>
【单选题】
<img title=无标题421.jpg alt=无标题421.jpg src=https://huaweicloudobs.ahjxjy.cn/054A15CAB30331320A7A62518FECCB6C.jpg/>
①
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/5F931CE6FE3F3703B2BFEF357DAD2B21.png data-tex=I=2.24A,{ U }_{ ab }=20V,\hat { S } =(20-j40)VA/>
②
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/DB777B73C24D41B286D5E026FC00C8D6.png data-tex=I=2.24A,{ U }_{ ab }=20V,\hat { S } =(20+j40)VA/>
③
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/819522447EA95D8DB3A83DC72EC3A361.png data-tex=I=2.24A,{ U }_{ ab }=20V,\hat { S } =(40-j20)VA/>
④
<img class=jc-formula style=vertical-align: middle; src=https://huaweicloudobs.ahjxjy.cn/D9CE7A72F664633AC68008A7A39F9B30.png data-tex=I=2.24A,{ U }_{ ab }=20V,\hat { S } =(40+j20)VA/>