# Thread: double sum designation question..

1. ## double sum designation question..

the telegan law basically states that the total sum of power is zero.
my prof proved lik this:

we choose a node (a point where more then one currents come together)

and decide that the voltage on that node to be zero.
we designate the voltages on the nodes to be $e_k$
$J_k$ is the current.

$v_kJ_k=(e_a-e_b)J_{ab}$
$v_kJ_k=\frac{1}{2}[(e_b-e_a)J_{ab}+(e_a-e_b)J_{ab}]$
$n_t$ is the number of nodes.[/tex]
$B$ is the number of branches.[/tex]
$\sum_{k}^{B}v_kJ_k=\frac{1}{2}\sum_{a=1}^{n_t}\sum _{b=1}^{n_t}(e_a-e_b)J_{ab}$
each J that does not exist in the graph will be zero.
$\sum_{k}^{B}v_kJ_k=\frac{1}{2}\sum_{a=1}^{n_t}e_a\ sum_{b=1}^{n_t}J_{ab}-\frac{1}{2}\sum_{a=1}^{n_t}e_b\sum_{b=1}^{n_t}J_{a b}=0$

because by kcl
$\sum_{b=1}^{n_t}J_{ab}=0$

my problem iswhen he sums for all nodes
he uses
$\sum\sum$ sign which by me represents multiplication
of the sums

why not $\sum+\sum$,thus we can know that ist the sum of many similar equations.

but how he did it doesnt represent a sum

2. The double sum in this case represents a nested sum (if you're familiar with programming, it's like a nested for loop).
$\frac{1}{2}\sum_{a=1}^{n_t}\sum_{b=1}^{n_t}(e_a-e_b)J_{ab}$
means for every iteration of a, we go through all $n_t$ iterations of b. I like to think of it as sort of a series exponentiation.
Formally if we define: $F = \sum_{i=0}^{n} a_{i}$ and $G = \sum_{j=0}^{n} b_{j}$
$F * G = \sum_{i=0}^{n} a_{i}b_{n-i}$
$F º G = \sum_{i=0}^{n}\sum_{j=0}^{n} a_{i}b_{j}