作业帮英语翻译1.For capacitor charge it is necessary to take resistor
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英语翻译
1.For capacitor charge it is necessary to take resistor with resistance 47 Ohm and 1Watt power.
2.Measure voltage of stabilized power supply,what will be used to charge capacitor (desirable power supply voltage is 12-14V).
3.Multiply measured (received) power supply voltage by 0.67,for example – 14 V *0.67 = 9.38 V.Round up and have 9.4 V – it is voltage,to which capacitor has to be charged by resistor.
4.Multiply declared capacitance of capacitor by resistor resistance and as a result have charge equivalent time.For example:it is declared that capacitor volume (capacitance) is 1 Farad,then 47 Ohm * 1 Farad = 47 seconds.This capacitor has to be charged to 9,4 V in 47 seconds with the use of 47 Ohm resistor!
5.If it takes less time (for example,30 seconds) to charge capacitor to 9,4 V with the use of resistor,so its capacitance is lower than it was declared!Percentage of received capacitance can be calculated by proportion,where 47 seconds – 100%,30 seconds – X%.Totally 30*100/47 = 63.8% from declared capacitance.
1.For capacitor charge it is necessary to take resistor with resistance 47 Ohm and 1Watt power.
2.Measure voltage of stabilized power supply,what will be used to charge capacitor (desirable power supply voltage is 12-14V).
3.Multiply measured (received) power supply voltage by 0.67,for example – 14 V *0.67 = 9.38 V.Round up and have 9.4 V – it is voltage,to which capacitor has to be charged by resistor.
4.Multiply declared capacitance of capacitor by resistor resistance and as a result have charge equivalent time.For example:it is declared that capacitor volume (capacitance) is 1 Farad,then 47 Ohm * 1 Farad = 47 seconds.This capacitor has to be charged to 9,4 V in 47 seconds with the use of 47 Ohm resistor!
5.If it takes less time (for example,30 seconds) to charge capacitor to 9,4 V with the use of resistor,so its capacitance is lower than it was declared!Percentage of received capacitance can be calculated by proportion,where 47 seconds – 100%,30 seconds – X%.Totally 30*100/47 = 63.8% from declared capacitance.
1.电容器充电必须用阻力为47欧姆和1瓦的电阻器
2.测试将要用於电容器充电的 稳定的电源 的电压(需要的电源电压为12伏~14伏)
3.将测试所得的电源电压乘以0.67,例如-14V*0.67=9.38V. 四舍五入得出9.4V- 这就是电容器充电要求的电阻器电压
4.用列出的电阻力乘以电容器电流容量,得出充电的等效时段.例如,电容器电流容量若为1法拉,那麼就是47欧姆*1法拉=47秒. 那麼这个电容器就必须在9.4伏电压情况下用电阻未47欧姆的电阻器充电47秒
5.假如只用了更少的时间(例如30秒)去充电,那麼他的电容量就低於它原本所陈述的.接收的电容量百分比可用比例计算, 47秒-100%,30秒-%,得出30*100/47=63.8%
(这里是数学的比例计算,我就不具体翻译了 应该懂吧)
大概意思就是这样 我不熟悉电容测试 但还是能翻译 望采纳
2.测试将要用於电容器充电的 稳定的电源 的电压(需要的电源电压为12伏~14伏)
3.将测试所得的电源电压乘以0.67,例如-14V*0.67=9.38V. 四舍五入得出9.4V- 这就是电容器充电要求的电阻器电压
4.用列出的电阻力乘以电容器电流容量,得出充电的等效时段.例如,电容器电流容量若为1法拉,那麼就是47欧姆*1法拉=47秒. 那麼这个电容器就必须在9.4伏电压情况下用电阻未47欧姆的电阻器充电47秒
5.假如只用了更少的时间(例如30秒)去充电,那麼他的电容量就低於它原本所陈述的.接收的电容量百分比可用比例计算, 47秒-100%,30秒-%,得出30*100/47=63.8%
(这里是数学的比例计算,我就不具体翻译了 应该懂吧)
大概意思就是这样 我不熟悉电容测试 但还是能翻译 望采纳
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