An expression for the voltage drop between inverter and PUC
[tex]V D \rightarrow \frac{I(\Omega / k f t) d}{2770}$[/tex]
This is further explained below.
Generally, the equation for Voltage drop is mathematically given as
[tex]V=\frac{I R}{V}[/tex]
Therefore
[tex]\%$ Voltage drop $=\frac{I R}{V} * 100\\\\$\rightarrow \frac{100 I(\Omega / 15 f t) d}{1000 V_{p}}\\\\$V_{P}=277 \mathrm{~V}$[/tex]
[tex]V D=\frac{I(\Omega / \mathrm{kft}) d}{10 * 2770} \rightarrow \frac{I(\Omega / \mathrm{kft}) d}{2770} \mathrm{~V}\\\\$\% V D \rightarrow \frac{I(\Omega / k f t) d}{2770}$[/tex]
In conclusion, the expression for the voltage drop between inverter and PUC
[tex]V D \rightarrow \frac{I(\Omega / k f t) d}{2770}$[/tex]
CQ
An inverter has a balanced 3-phase, 277/480 V output and is installed a distance, d, ft from the point of utility connection. The DC/AC converter is shown below in the inverter. And the distance d is the physical separation between this DC/AC converter and the PCC.
Derive an expression for the voltage drop between inverter and PUC in terms of d using the basic equation %VD =100IR/V.
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