8.3.5power
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This document discusses power calculations in electrical circuits. It shows that total power (PT) in a circuit equals the sum of individual powers (P1 + P2 + P3...). It then calculates the power of two resistors in a circuit, which add up to 3kW total. It demonstrates that total power can be calculated from current and total resistance, or from total voltage and current. The document concludes by instructing the reader to complete review questions on power calculations.
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9.2.2percentage
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4.1.10 Light Demand 2
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1.4.2cosinelaw
The cosine law describes the relationship between illuminance (E), luminous intensity (l), distance from the light source (d), and the angle of incidence (θ). Specifically, it states that illuminance (E) equals the luminous intensity (l) divided by the distance from the light source squared (d2), multiplied by the cosine of the angle of incidence (Cos θ).
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This short document does not provide any substantive information to summarize in 3 sentences or less. It only notes that an explanation is not contained in a workbook, but provides no other context or details.
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The document outlines the demand calculations for 19 different circuit load groups across 3 phases. It lists the load description, current allowance calculation method, and resulting demand current for each phase. The total demand current per phase is calculated at the bottom, with values of 143.8 amps for phase 1, 153.25 amps for phase 2, and 145.05 amps for phase 3.
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The document contains a table that calculates the current demand per phase for various circuit load groups in an electrical system. It lists 19 load groups categorized by letters A through D, describing each load. It shows the current allowance calculation method and resulting demand in amps for each phase. The total demand current calculated per phase is 143.8 amps for L1, 153.25 amps for L2, and 145.05 amps for L3.
4.1.10 Light Demand 2
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4.1.9 Light Demand 1
The document describes the functions of 19 electrical circuits in a building. It lists the types of equipment connected to each circuit such as fluorescent lighting, outlets, motors, and appliances. It also indicates which of the 3 electrical phases (L1, L2, L3) each circuit is connected to.
4.1.7 Example Domestic 3
The document calculates the electrical load of communal services in an apartment building. It shows that 24 lighting points will draw 240 watts and 6 10A sockets will draw up to 12A, for a total demand of 18A per phase. The total demand current per phase for communal services is 18A.
4.1.6 Example Domestic 2
The document calculates the electrical load and demand for 6 living units per phase. It lists the types of loads in each unit, the quantity and allowance per unit, and uses this to calculate the total demand current for phases L1, L2 and L3, which is 154.4 amps for each phase. The key loads included are lighting, power outlets, cooking ranges, air conditioners and hot water systems.
4.1.5 Example Domestic 1
This document discusses the number of living units per phase of a project. It calculates that for 18 total living units divided into 3 phases, there would be 6 living units per phase.
4.1.4 Step 2
This document summarizes the electrical load calculations for 11 circuits in a home. It lists the load type and description for each circuit, the current allowance per unit, and calculates the demand in amps for circuits 1-10. The largest demands are 17.7 amps for an air conditioner and 15 amps for an off-peak hot water system. The total calculated demand current for each phase is 40.7, 39.5, and 46.5 amps respectively.
4.1.3 Example Solution Three
The document calculates the electrical load and demand current for 11 circuits in a home. It groups the loads into categories like lighting, outlets, appliances, and assigns each a description, allowance, and demand calculation. The total demand current per phase is summarized at the bottom, with the highest draw of 46.5 amps on phase L3.
4.1.2 Example Three
The document outlines the functions and ratings of 12 electrical circuits. Circuit 1 provides power for 13 indoor lighting points. Circuit 10 powers an air conditioner rated at 23.6 amps per phase and can draw power from circuits L1, L2, and L3. Circuits 11a and 11b each power a controlled load hot water unit rated at 3.6 kW.
4.1.1 Example Single
This document calculates the total demand current for an electrical installation consisting of:
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- 15 single sockets contributing 10A
- A 6kW oven contributing 0.5A
- A 2.4kW water heater contributing 0.33A
The total demand current calculated is 45.83A.
5.4.4 LR Circuits
The document discusses the time constant and final current value for an RL circuit. It states that:
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14.5.4 Example 1 Part 1
This document calculates the apparent power, power factor, and phase angle for a circuit. It finds that the apparent power is 2.308 kVA by multiplying the current of 9.615 by the voltage of 240. This apparent power is larger than the actual power of 1.5 kW, indicating a poor power factor of 0.65 or a 49.46 degree phase angle between the current and voltage.
14.1.2 TRA Power
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14.5.8 Example 1 Part 5
This document calculates the true power, apparent power, and total current for a circuit. It determines that the true power is 1.5 kW, the apparent power is 1.5009 kVA, and the power factor is 0.99994. It then calculates that with an apparent power of 1.5009 kVA at 240 Volts, the total current would be 6.25 amps.
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二、回国进私企、外企、自己做生意的情况，这些单位是不查询毕业证真伪的，而且国内没有渠道去查询国外文凭的真假，也不需要提供真实教育部认证。鉴于此，办理一份毕业证【微信：A575476】即可
三、进国企，银行，事业单位，考公务员等等，这些单位是必需要提供真实教育部认证的，办理教育部认证所需资料众多且烦琐，所有材料您都必须提供原件，我们凭借丰富的经验，快捷的绿色通道帮您快速整合材料，让您少走弯路。
留信网认证的作用:
1:该专业认证可证明留学生真实身份
2:同时对留学生所学专业登记给予评定
3:国家专业人才认证中心颁发入库证书
4:这个认证书并且可以归档倒地方
5:凡事获得留信网入网的信息将会逐步更新到个人身份内，将在公安局网内查询个人身份证信息后，同步读取人才网入库信息
6:个人职称评审加20分
7:个人信誉贷款加10分
8:在国家人才网主办的国家网络招聘大会中纳入资料，供国家高端企业选择人才
→ 【关于价格问题（保证一手价格）
我们所定的价格是非常合理的，而且我们现在做得单子大多数都是代理和回头客户介绍的所以一般现在有新的单子 我给客户的都是第一手的代理价格，因为我想坦诚对待大家 不想跟大家在价格方面浪费时间
对于老客户或者被老客户介绍过来的朋友，我们都会适当给一些优惠。
选择实体注册公司办理，更放心，更安全！我们的承诺：可来公司面谈，可签订合同，会陪同客户一起到教育部认证窗口递交认证材料，客户在教育部官方认证查询网站查询到认证通过结果后付款，不成功不收费！
How are Lilac French Bulldogs Beauty Charming the World and Capturing Hearts....
“After being the most listed dog breed in the United States for 31
years in a row, the Labrador Retriever has dropped to second place
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popular canines. The French Bulldog is the new top dog in the
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- 1. PT = P1 + P 2 + P3 . . .P
- 2. P= I2 R or P = V I
- 3. R1 = (10)2x 10 = 1000 W = 1 kW
- 4. R2 = (10)2x 20 = 2000 W = 2kW
- 5. Pt = PR1 + PR2 = 1000 + 2000 = 3000 W = 3kW
- 6. Rt = 10 + 20 = 30 Ω Pt = I2 Rt = (10) 2 x 30 = 3000 W
- 7. P t = Vt I = (100 + 200) x 10 = 3000W
- 8. Complete the Review Questions on page 70 of Electrical Trade Principles by Hampson & Hanssen.