# why is the the Line voltage drop formula is $\Delta V=2I_1 (Rcos\theta+Xsin\theta)$,not $\Delta V=2I_1(\sqrt{R^2+X^2})$?

There is a single-phase two-wire system,if the resistor $$\(R)\$$ of line is 1Ω and the reactance $$\(X)\$$ of line is 2Ω,and the power factor is $$\\theta\$$ .The Line voltage drop formula should be

$$\Delta V=2I_1 \times(Rcos\theta+Xsin\theta),\Delta V=VS-VR$$ simulate this circuit – Schematic created using CircuitLab

I want to ask why is the the Line voltage drop formula is $$\\Delta V=2I_1 (Rcos\theta+Xsin\theta)\$$,because i think the according to $$\V=IZ\$$,the line of impedance should be $$\\sqrt{R^2+X^2}\$$,the line of impedance should be $$\\Delta V=2I_1(\sqrt{R^2+X^2})\$$

• ΔV - what do you mean precisely by ΔV? – Andy aka Mar 30 at 9:13
• That formula is an approximation, powerline calculations for small voltage drops are full of them, the approximations were developed in the era before electronic calculators.Today computers have made those simplifications less important, but old texts evaporate slowly. – user287001 Mar 30 at 9:22
• @Andyaka sorry,i forget to mention that,i have edited the question already – shineele Mar 30 at 9:52
• "appears to be based on an assumption that the voltage drop will be relatively small." when small resistances assumed it yields that approximation instead of R+jX: electronics.stackexchange.com/questions/188378/… – user1245 Mar 30 at 9:54
• @user1245 The website you provide didn't mention any sin or cos,so although i know $(Rcos\theta+Xsin\theta)$ is a approximation of $\sqrt{R^2+X^2}$ when $R$ and $X$ are both very small,but i still don't know how do we approximate $\sqrt{R^2+X^2}$ or R+jX – shineele Mar 30 at 10:58