# Approaching to Integral from Limit of Sum ->Finding voltage drop due to uniformly distributed load current on a distributor

As shown in the picture above, I want to calculate the voltage drop due to a distributed load.

l is the length of the distributor;

n is simply no of segments, n=l/Δx;

I have written down the sum to determine the voltage drop. But what actually I want to do is to find the "LIMIT OF SUM" of this integral is Δx → 0.

How can I carry out the solution ?

• Why you didn't title your question "Distributed load volt drops" beats me. Jul 14, 2017 at 15:52
• I'm sorry I didn't used that title, because that was not the context of this question. Jul 14, 2017 at 16:06

I can't say that I feel I completely understood your question. But if I gather it correctly, the value of $i$ doesn't change as you move from a finite number of taps to an infinite number of them. This, as you will see, creates a problem that won't be resolvable with a finite answer.

I'm going to label each tap, left to right, as $k$ starting with $k=1$ on the left and $k=n$ at the rightmost tap. You have $n$ taps. The current removed at each tap is a constant $i$. I'm going to call the entire length, $L$ (and not $l$.) But the length of each segment of $L$ I'll call $l_k=\frac{L}{n}$. Length $L$ exhibits a certain resistance per unit length $r$. The resistance of any segment length is $r_k=r\cdot l_k=r\frac{L}{n}$.

The current entering from the left must be the sum of all the branch currents, or $I=n\cdot i$. This current must pass through the first segment $l_1$ of length $L$ and drop a voltage $v_1=n\cdot i\cdot r\cdot\frac{L}{n}$. The next segment will drop a voltage $v_2=\left(n-1\right)\cdot i\cdot r\cdot\frac{L}{n}$. Etc.

So the question you are asking, "what is the voltage drop across length $L$ as $n\to\infty$," can be written out this way:

\begin{align*} \Delta V=V-V^{'}&= \lim_{n\to\infty} \sum_{k=1}^n \left(n-k+1\right)\cdot i\cdot r\cdot \frac{L}{n}\\\\ &=i\cdot r\cdot L\:\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \left(n-k+1\right)\\\\ &=i\cdot r\cdot L\:\lim_{n\to\infty} \frac{1}{n}\left[\left(n+1\right)\sum_{k=1}^n 1-\sum_{k=1}^n k\right]\\\\ &=i\cdot r\cdot L\:\lim_{n\to\infty} \frac{1}{n}\left[\left(n+1\right)n-\frac{\left(n+1\right)n}{2}\right]\\\\ &=i\cdot r\cdot L\:\lim_{n\to\infty} \left[\frac{1}{n}\frac{\left(n+1\right)n}{2}\right]\\\\ &=i\cdot r\cdot L\:\lim_{n\to\infty} \left[\frac{n+1}{2}\right]\\\\ &=\frac{i\cdot r\cdot L}{2}\:\left[1+\lim_{n\to\infty} n\right] \end{align*}

And I don't think I need to tell you how to figure out that limit.

Of course, if you also diminished $i$ as a function of $n$ then perhaps there might be a different answer. But I didn't see that behavior in your question.