🎯题库

{"bank":[{"analysis":"电气安全的 4 种危害是**触电、电弧、静电、雷击**：触电源于接触带电体，电弧源于短路/过载的高温火花，静电源于干燥摩擦的积累放电，雷击分直接雷与感应雷。对应防范分别是绝缘/接地/隔离、保护/熔断、接地/加湿、避雷针/接地。","answer":"A","chapter":"电工·电气安全","difficulty":1,"explain":{"B":"短路、过载是产生**电弧**的成因，漏电是**触电**的隐患，都不是危害分类本身。","C":"**感应**只是雷击的一种形式，**接地**是防范措施，二者不能与 4 种危害并列。","D":"**过载**是电弧的成因之一，**潮湿**是静电与漏电发生的条件，均不属于危害类别。"},"id":"tz-01-001","options":["触电、电弧、静电、雷击","触电、短路、过载、漏电","电弧、静电、感应、接地","雷击、触电、过载、潮湿"],"source":"chapter","stem":"教材把电气作业的主要危害归纳为 4 种。下列哪一组与这 4 种危害完全对应？","subject":"特种作业","type":"single"},{"analysis":"带电作业是**高危红线**：除持证并履行审批的带电作业外，一律**严禁带电作业**。正常检修必须先**停电、验电、挂接地线、挂牌**，危险场所还要坚持一人操作一人监护。","answer":"B","chapter":"电工·电气安全","difficulty":2,"explain":{"A":"绝缘手套只是辅助安全用具，**不能替代资格与审批**，无证带电作业属严重违章。","C":"危险作业必须**一人操作一人监护**，赶时间不是免除监护的理由。","D":"380V 属于危险电压，**不存在「低于 380V 就安全」**的说法，安全电压等级另有规定。"},"id":"tz-01-002","options":["只要戴上绝缘手套，任何人都可以带电作业","未取得相应资格、未经批准严禁带电作业；确需带电作业必须持证并按规程设专人监护","带电作业只要动作快就可以不设监护人","线路电压低于 380V 就属于安全电压，可以随意带电操作"],"source":"chapter","stem":"关于带电作业，下列说法正确的是？","subject":"特种作业","type":"single"},{"analysis":"停送电的顺序是**先停电，后验电，再接地**：断开电源后必须用相应电压等级的验电器确认无电，再装设接地线，顺序不能乱。","answer":"D","chapter":"电工·电气安全","difficulty":1,"explain":{"A":"未停电就装接地线相当于**带电挂接地线**，会直接造成短路与电弧伤人。","B":"设备尚未停电就验电，**验电器无法确认是否真正断电**，不符合规程。","C":"停电后未验电就接地，可能在**实际仍带电**的情况下挂接地线，属严重违章。"},"id":"tz-01-003","options":["装设接地线 → 验电 → 停电","验电 → 停电 → 装设接地线","停电 → 装设接地线 → 验电","停电 → 验电 → 装设接地线"],"source":"chapter","stem":"对停电设备进行检修前的安全操作，正确顺序是？","subject":"特种作业","type":"single"},{"analysis":"接地线的装设顺序是**先接接地端、后接导体端**，拆除顺序**完全相反**：**先拆导体端，后拆接地端**，以免作业人员先失去接地保护而触电。","answer":"A","chapter":"电工·电气安全","difficulty":2,"explain":{"B":"先拆接地端会让导体端**暂时失去接地保护**，设备一旦意外来电即发生触电。","C":"两端同时拆除无法保证**始终有一端可靠接地**，不符合安全顺序要求。","D":"接地线未拆除就送电等于**带接地线合闸**，会形成短路并损坏设备。"},"id":"tz-01-004","options":["先拆导体端，后拆接地端","先拆接地端，后拆导体端","两端同时拆除","无需拆除，可直接送电"],"source":"chapter","stem":"拆除检修作业中已装设的接地线，正确顺序是？","subject":"特种作业","type":"single"},{"analysis":"触电急救 4 步：**①脱离电源 → ②判断意识 → ③呼救（120）→ ④心肺复苏（按压与吹气 30:2）**。第一步永远是**脱离电源**，之后才谈判断、呼救与抢救。","answer":"D","chapter":"电工·电气安全","difficulty":2,"explain":{"A":"徒手拉人会让施救者**一起触电**，必须借助绝缘工具或直接断电。","B":"未脱离电源前任何施救动作都无效且有风险，喂水掐人中还会延误抢救。","C":"胸外按压是第 4 步，必须在脱离电源并判断意识、呼救之后进行。"},"id":"tz-01-005","options":["直接用手把触电者从带电体上拉开","先给触电者喂水并掐人中","立即对触电者进行胸外按压","迅速使触电者脱离电源，如切断开关或用干燥绝缘物挑开电线"],"source":"chapter","stem":"发现有人触电倒在地上，作为急救者首先应当做的是？","subject":"特种作业","type":"single"},{"analysis":"电气火灾 3 步：**①先断电 → ②用干粉/CO₂ 灭火 → ③呼救（119）**。核心是**先断电**，灭火剂必须用不导电的干粉或二氧化碳，**严禁用水**。","answer":"B","chapter":"电工·电气安全","difficulty":1,"explain":{"A":"未断电就灭火，设备残留电压仍会伤人，且可能复燃。","C":"水导电，带电泼水等于**为自己制造触电通道**，是电气火灾的大忌。","D":"泡沫灭火器含水导电，**不能用于带电电气火灾**。"},"id":"tz-01-006","options":["先用干粉灭火器灭火，再切断电源","先切断电源，再用干粉或二氧化碳灭火器灭火并报警","立即用水扑救，水能快速降温","用泡沫灭火器带电灭火，边灭火边等待救援"],"source":"chapter","stem":"电气设备着火时，正确的处置顺序是？","subject":"特种作业","type":"single"},{"analysis":"《安全生产法》的方针是**安全第一、预防为主、综合治理**，其中**安全第一**是核心，任何情况下都不能以生产进度牺牲安全。","answer":"A","chapter":"电工·安全法规","difficulty":1,"explain":{"B":"把生产置于安全之前，与**安全第一**的方针直接冲突。","C":"这是管理责任划分的表述，不是法定的安全生产方针。","D":"**效率优先**与安全第一相矛盾，属于典型的错误表述。"},"id":"tz-02-001","options":["安全第一、预防为主、综合治理","生产第一、安全第二、奖惩结合","谁主管、谁负责、谁受益","以人为本、效率优先、兼顾安全"],"source":"chapter","stem":"《安全生产法》确立的安全生产方针（12 字方针）是？","subject":"特种作业","type":"single"},{"analysis":"**生产经营单位（企业）是安全生产的责任主体**，必须建立安全生产责任制、保障安全投入、组织教育培训并落实规章制度。","answer":"B","chapter":"电工·安全法规","difficulty":2,"explain":{"A":"从业人员承担遵章守纪、接受培训的义务，不是主体责任方。","C":"监管部门承担的是**监督责任**，不能代替企业的主体责任。","D":"供应商只对设备产品质量负责，不对使用单位的安全生产负责。"},"id":"tz-02-002","options":["一线从业人员","生产经营单位（企业）","当地应急管理部门","设备供应商"],"source":"chapter","stem":"依据《安全生产法》，对本单位安全生产工作承担主体责任的是？","subject":"特种作业","type":"single"},{"analysis":"从业人员享有**知情权、拒绝权、紧急避险权**三种权利。**拒绝违章指挥和强令冒险作业**属于**拒绝权**，行使该权利不得被降低待遇或解除劳动合同。","answer":"C","chapter":"电工·安全法规","difficulty":2,"explain":{"A":"知情权是了解作业场所危险因素、防范措施与事故应急预案的权利。","B":"紧急避险权用于险情危及人身安全的紧急情况，可停止作业撤离。","D":"建议权是提出安全改进意见的权利，与拒绝违章指挥不是同一项权利。"},"id":"tz-02-003","options":["知情权","紧急避险权","拒绝权","建议权"],"source":"chapter","stem":"班组长要求电工在未停电、未验电的情况下直接检修配电柜，电工依据《安全生产法》可以行使的权利是？","subject":"特种作业","type":"single"},{"analysis":"电工、焊工、高处作业等属特种作业，必须持**特种作业操作证**方可上岗，**没有证就不能上岗**，安排无证人员作业属违法行为。","answer":"B","chapter":"电工·安全法规","difficulty":2,"explain":{"A":"持证是法定前置条件，**不存在先上岗后补证**的做法。","C":"他人监护不能替代本人持证资格，无证作业本身即违规。","D":"企业证明不能替代法定的特种作业操作证。"},"id":"tz-02-004","options":["只要技术熟练，可以先上岗后补证","无证不得上岗，该安排违规，作业人员应拒绝","有老师傅在旁边看着就可以无证作业","只要企业出具证明就可以无证上岗"],"source":"chapter","stem":"某企业安排一名未取得特种作业操作证的人员独立从事电气检修，下列说法正确的是？","subject":"特种作业","type":"single"},{"analysis":"**两票**指工作票与操作票。**停送电、倒闸操作必须使用操作票**，票面写明操作步骤、操作人和监护人，按步执行并核对。","answer":"B","chapter":"电工·安全法规","difficulty":2,"explain":{"A":"工作票用于检修/施工，明确工作任务、安全措施与工作负责人。","C":"领料单是物资凭证，与操作安全无关。","D":"交接班记录属于三制内容，不能替代操作票。"},"id":"tz-02-005","options":["工作票","操作票","领料单","交接班记录"],"source":"chapter","stem":"对电气设备进行停送电和倒闸操作时，必须使用的票据是？","subject":"特种作业","type":"single"},{"analysis":"**两票 = 工作票 + 操作票**；**三制 = 交接班制、巡回检查制、设备定期试验轮换制**，共 5 项制度，是电气运行管理的固定口径。","answer":"A","chapter":"电工·安全法规","difficulty":3,"explain":{"B":"工作票、操作票属于**两票**，监护是安全措施而非三制内容。","C":"停送电、验电、接地属于《电气安全规程》的操作要求。","D":"教育培训与奖惩属于企业安全管理制度，不是两票三制所指。"},"id":"tz-02-006","options":["交接班制、巡回检查制、设备定期试验轮换制","工作票制、操作票制、监护制","停送电制、验电制、接地制","安全教育制、考核制、奖惩制"],"source":"chapter","stem":"「两票三制」中的「三制」指的是？","subject":"特种作业","type":"single"},{"analysis":"按欧姆定律的变形式 **I = U / R**，I = 12 ÷ 4 = **3A**。欧姆定律基本式为 U = I × R。","answer":"B","chapter":"电工·电路基础","difficulty":1,"explain":{"A":"0.33A 是把 R/U 当成了 I，公式方向写反。","C":"8A 是用 U − R 相减得到的，欧姆定律不含减法关系。","D":"48A 是用 U × R 相乘得到的，混淆了乘法与除法。"},"id":"jg-01-001","options":["0.33A","3A","8A","48A"],"source":"chapter","stem":"某电阻两端电压为 12V，阻值为 4Ω，则通过它的电流是？","subject":"职业资格","type":"single"},{"analysis":"**KCL 针对节点**：ΣI = 0，即**流入 = 流出**。其典型应用是并联电路——**总电流等于各支路电流之和**。","answer":"B","chapter":"电工·电路基础","difficulty":2,"explain":{"A":"这是**KVL**（电压定律）的内容，针对的是回路。","C":"串联电路中各电阻电流**相等**，与阻值无关。","D":"并联总电阻的倒数等于各支路电阻倒数之和，不是各电阻直接相加。"},"id":"jg-01-002","options":["沿任一闭合回路，各段电压升之和等于电压降之和","对电路中任一节点，流入该节点的电流之和等于流出该节点的电流之和","串联电路中各电阻的电流与其阻值成反比","并联电路的总电阻等于各支路电阻之和"],"source":"chapter","stem":"基尔霍夫电流定律（KCL）的内容是？","subject":"职业资格","type":"single"},{"analysis":"**KVL 针对回路**：ΣU = 0，即**电压升 = 电压降**。其典型应用是串联电路——**总电压等于各电阻电压之和**。","answer":"C","chapter":"电工·电路基础","difficulty":2,"explain":{"A":"这是 **KCL**（电流定律）的内容，针对的是节点。","B":"并联分流按电阻**反比**分配，电阻越大分得的电流越小。","D":"总电压是各段电压**之和**，不是乘积。"},"id":"jg-01-003","options":["对任一节点，流入电流等于流出电流","并联各支路电流与电阻成正比","沿任一闭合回路，各段电压升之和等于电压降之和","串联电路的总电压等于各电阻电压之积"],"source":"chapter","stem":"关于基尔霍夫电压定律（KVL），下列说法正确的是？","subject":"职业资格","type":"single"},{"analysis":"串联总电阻 R = 10 + 20 = 30Ω，电流 I = 30 ÷ 30 = 1A，U1 = I × R1 = 1 × 10 = **10V**。串分压：**大电阻分大电压**。","answer":"B","chapter":"电工·电路基础","difficulty":2,"explain":{"A":"5V 没有对应的计算依据，属猜测值。","C":"15V 是按电阻个数均分 30V 的误算，串联按阻值比例分配。","D":"20V 是 R2 两端的电压，不是 R1 的。"},"id":"jg-01-004","options":["5V","10V","15V","20V"],"source":"chapter","stem":"R1=10Ω 与 R2=20Ω 串联后接在 30V 电源上，R1 两端的电压为？","subject":"职业资格","type":"single"},{"analysis":"并联各支路电压相等：I1 = 20 ÷ 10 = 2A，I2 = 20 ÷ 20 = 1A，**总电流 = 2 + 1 = 3A**。并分流：大电阻分小电流。","answer":"D","chapter":"电工·电路基础","difficulty":2,"explain":{"A":"0.67A 是把总电阻倒数误当成电流的算法。","B":"1A 只是 R2 支路的电流。","C":"2A 只是 R1 支路的电流。"},"id":"jg-01-005","options":["0.67A","1A","2A","3A"],"source":"chapter","stem":"R1=10Ω 与 R2=20Ω 并联后接在 20V 电源上，干路总电流为？","subject":"职业资格","type":"single"},{"analysis":"有功功率 **P = U × I × cosφ** = 220 × 5 × 1 = **1100W**。功率因数为 1 时视在功率与有功功率相等。","answer":"C","chapter":"电工·电路基础","difficulty":3,"explain":{"A":"44W 是把 220 与 5 相减后的误算。","B":"225W 是把 220 与 5 相加后的误算。","D":"2200W 是把结果多乘了 2，属于重复计算。"},"id":"jg-01-006","options":["44W","225W","1100W","2200W"],"source":"chapter","stem":"单相交流电路中电压为 220V、电流为 5A，功率因数为 1，则其有功功率约为？","subject":"职业资格","type":"single"},{"analysis":"变比 **K = U1/U2 = N1/N2**，故 U2 = U1 × N2/N1 = 220 × 100 ÷ 1000 = **22V**。此处 K = 10 \u003e 1，属降压变压器。","answer":"B","chapter":"电工·电机与控制","difficulty":2,"explain":{"A":"2.2V 多除了一个 10，匝数比方向也算反了。","C":"220V 忽略了变比，误以为二次电压与一次相同。","D":"2200V 把升降压关系弄反，当成了升压变压器。"},"id":"jg-02-001","options":["2.2V","22V","220V","2200V"],"source":"chapter","stem":"某变压器一次侧电压 220V、一次绕组 1000 匝、二次绕组 100 匝，二次侧空载电压约为？","subject":"职业资格","type":"single"},{"analysis":"教材列出两种组别：**Y/Y** 用于低压配电，**Y/Δ** 用于中压配电。斜杠**前为一次侧、后为二次侧**，故 Y/Δ 是一次星形、二次三角形。","answer":"A","chapter":"电工·电机与控制","difficulty":2,"explain":{"B":"顺序颠倒，先三角后星形应写作 Δ/Y。","C":"两侧均为星形是 Y/Y 组别。","D":"两侧均为三角形是 Δ/Δ 组别。"},"id":"jg-02-002","options":["一次侧星形接线、二次侧三角形接线","一次侧三角形接线、二次侧星形接线","一次侧与二次侧均为星形接线","一次侧与二次侧均为三角形接线"],"source":"chapter","stem":"变压器接线组别 Y/Δ 指的是？","subject":"职业资格","type":"single"},{"analysis":"同步转速 **n = 60f / p**。4 极对应磁极对数 p = 2，n = 60 × 50 ÷ 2 = **1500 r/min**。教材的 4 档转速为 3000/1500/1000/750，对应 2/4/6/8 极，**极数越多转速越低**。","answer":"C","chapter":"电工·电机与控制","difficulty":2,"explain":{"A":"750 r/min 对应 8 极电机。","B":"1000 r/min 对应 6 极电机。","D":"3000 r/min 对应 2 极电机。"},"id":"jg-02-003","options":["750 r/min","1000 r/min","1500 r/min","3000 r/min"],"source":"chapter","stem":"工频 50Hz 下，三相异步电动机 4 极的同步转速约为？","subject":"职业资格","type":"single"},{"analysis":"启动方式三种：**11kW 以下直接启动，11~75kW 采用星三角启动以降低启动电流，75kW 以上考虑软启动**。30kW 落在 11~75kW 区间，故选星三角启动。","answer":"B","chapter":"电工·电机与控制","difficulty":2,"explain":{"A":"直接启动冲击电流大，教材口径适用于 11kW 以下小电机。","C":"软启动一般用于 75kW 以上、需要平滑启动的场合。","D":"大功率电机直接合闸会造成电网电压跌落，并可能损坏设备。"},"id":"jg-02-004","options":["直接（全压）启动","星三角（Y-Δ）启动","软启动","不设启动装置，直接合闸"],"source":"chapter","stem":"一台 30kW 三相异步电动机，按教材口径宜采用的启动方式是？","subject":"职业资格","type":"single"},{"analysis":"**热继电器**靠**双金属片**受热弯曲动作，切断控制回路，实现电动机过载保护。","answer":"C","chapter":"电工·电机与控制","difficulty":2,"explain":{"A":"接触器用于通断主回路，本身不承担过载保护功能。","B":"中间继电器用于信号放大与隔离，不做过载保护。","D":"按钮只发出启动/停止指令，不检测电流。"},"id":"jg-02-005","options":["接触器","中间继电器","热继电器","按钮"],"source":"chapter","stem":"电动机控制回路中，利用双金属片实现过载保护的元件是？","subject":"职业资格","type":"single"},{"analysis":"自锁靠与启动按钮并联的**接触器辅助常开触点**：按下启动按钮线圈得电，辅助常开触点闭合后即使松手线圈仍保持得电。口诀：**启动常开，停止常闭，自锁靠辅助**。","answer":"B","chapter":"电工·电机与控制","difficulty":3,"explain":{"A":"启动按钮常开触点只在按下瞬间闭合，松手即断开，不能维持得电。","C":"停止按钮常闭触点串联在回路中，作用是在按下时断开回路。","D":"热继电器常闭触点在过载时断开，起保护作用，不负责自锁。"},"id":"jg-02-006","options":["启动按钮的常开触点","接触器自身的辅助常开触点","停止按钮的常闭触点","热继电器的常闭触点"],"source":"chapter","stem":"电机启停自锁电路中，实现自锁的是？","subject":"职业资格","type":"single"},{"analysis":"相线用**黄/绿/红**，零线用**淡蓝**，地线**必须是黄绿双色且不得挪用**。设备外壳接 PE，**接地电阻一般不大于 4Ω**。","answer":"A","chapter":"电工·实操排故","difficulty":2,"explain":{"B":"把外壳接零线属于错误接法，一旦零线断开外壳就会带电。","C":"接地电阻越大，故障时外壳对地电压越高，保护效果越差。","D":"PE 线必须独立可靠接地，不能与相线共用端子。"},"id":"jg-03-001","options":["设备外壳应接黄绿双色 PE 线，接地电阻一般不大于 4Ω","设备外壳接淡蓝色零线即可，无需另设 PE 线","接地电阻越大，保护效果越好","PE 线可以与相线共用同一端子"],"source":"chapter","stem":"关于保护接地（PE），下列说法正确的是？","subject":"职业资格","type":"single"},{"analysis":"口诀：**电压并、电流串、测电阻先断电**。电流档**严禁带电换档**，否则会烧表并引发触电。","answer":"A","chapter":"电工·实操排故","difficulty":2,"explain":{"B":"电流档并联相当于把电流表直接跨接在电源两端，会造成短路并烧毁万用表。","C":"带电换档会产生电弧并损坏仪表，必须先断电或断开表笔。","D":"电阻档由表内电池供电，带电测量结果错误且不安全。"},"id":"jg-03-002","options":["测电压时表笔并联接入，测电流时串联接入，且严禁带电换档","测电流时把表笔并联在被测电路两端","测量过程中直接旋转旋钮切换档位","用电阻档测量时保持线路带电"],"source":"chapter","stem":"使用万用表时，下列操作正确的是？","subject":"职业资格","type":"single"},{"analysis":"**30mA 是人身保护级**，家用及潮湿场所必须安装。100~300mA 属于工业线路或设备保护级，**等级不能混淆**。","answer":"B","chapter":"电工·实操排故","difficulty":2,"explain":{"A":"10mA 过于灵敏，容易误动作，教材口径的人身保护级为 30mA。","C":"100mA 属线路/设备保护级，对人身触电保护不足。","D":"300mA 主要用于工业线路保护，不能替代人身保护。"},"id":"jg-03-003","options":["10mA","30mA","100mA","300mA"],"source":"chapter","stem":"家用及潮湿场所用于人身保护的漏电保护器，其动作电流一般为？","subject":"职业资格","type":"single"},{"analysis":"短路排故五步：①隔离（先断总开关）②目测烧焦/鼓包/异味 ③电阻档从电源到负载逐段测量 ④定位 ⑤修复复测送电。**电阻为 0 处即为短路点**。","answer":"A","chapter":"电工·实操排故","difficulty":2,"explain":{"B":"电阻接近无穷大说明该段**断路**，不是短路。","C":"断电状态下测电压没有意义，测电压也不是查找短路点的方法。","D":"断电后线路本就没有电流，不能据此定位故障。"},"id":"jg-03-004","options":["测得某段电阻接近 0","测得某段电阻接近无穷大","测得某段电压接近 0","测得某段电流为零"],"source":"chapter","stem":"断电后用万用表电阻档逐段排查短路点，判定短路点的依据是？","subject":"职业资格","type":"single"},{"analysis":"断路的特征是**一送就断**。可先用通断档分段测，或送电后用**分压法**测各点电压，**电压突变处就是断点**；定位后换线或紧固，复测再送电。","answer":"B","chapter":"电工·实操排故","difficulty":3,"explain":{"A":"电阻接近 0 是**短路**的特征，不是断路。","C":"兆欧表用于绝缘/漏电检测，且必须**停电**操作，带电摇测十分危险。","D":"盲目全部换线既浪费材料，又无法确认故障点是否真正排除。"},"id":"jg-03-005","options":["用电阻档判断哪一段电阻接近 0","送电后用分压法测各点电压，电压突变处即断点","带电用兆欧表逐段摇测","直接更换全部导线"],"source":"chapter","stem":"排查线路断路时，较有效的定位方法是？","subject":"职业资格","type":"single"},{"analysis":"漏电排故先断总开关，再**逐路送电**定位问题回路，最后用**兆欧表分段摇测**：**绝缘电阻低于 0.5MΩ 视为绝缘不良**，多为受潮或破皮，须停电后摇测。","answer":"B","chapter":"电工·实操排故","difficulty":3,"explain":{"A":"0.5Ω 是普通电阻量级，混淆了绝缘电阻的单位量级。","C":"5MΩ 远高于判废标准，属于绝缘良好。","D":"50MΩ 属绝缘良好，不可能判为漏电隐患。"},"id":"jg-03-006","options":["0.5Ω","0.5MΩ","5MΩ","50MΩ"],"source":"chapter","stem":"用兆欧表（摇表）检测线路绝缘时，通常把绝缘电阻低于多少视为绝缘不良（漏电隐患）？","subject":"职业资格","type":"single"}],"cert":"diangong","chapters":[{"name":"电工·电气安全","title":"第 1 章 · 电气安全：4 种危害与 3 种操作","url":"/diangong/books/tezhong/tz-01-anquan/"},{"name":"电工·电路基础","title":"第 1 章 · 电路基础：欧姆定律、基尔霍夫、串并联","url":"/diangong/books/jigui/jg-01-dianlu/"},{"name":"电工·安全法规","title":"第 2 章 · 安全生产法规：《安全生产法》与《电气安全规程》","url":"/diangong/books/tezhong/tz-02-fagui/"},{"name":"电工·电机与控制","title":"第 2 章 · 电机与电气控制：变压器、电动机、接触器","url":"/diangong/books/jigui/jg-02-dianji/"},{"name":"电工·实操排故","title":"第 3 章 · 实操技能：安装、调试、排故","url":"/diangong/books/jigui/jg-03-shicao/"}],"practiceUrl":"/diangong/practice/"}