# Double sum sign question

1. Nov 22, 2009

### lom

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.
$$\sum_{k}^{B}v_kJ_k=\frac{1}{2}\sum_{a=1}^{n_t}e_a\sum_{b=1}^{n_t}J_{ab}-\frac{1}{2}\sum_{a=1}^{n_t}e_b\sum_{b=1}^{n_t}J_{ab}=0$$

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. Nov 22, 2009

### rickz02

$$\sum_{a=1}^{n_t}\sum_{b=1}^{n_t}$$

is not multiplication of summations in this context. It means that you take the sum of the powers at all nodes. For example say you have 2 total nodes, so the summation would look like

$$\frac{1}{2}\sum_{a=1}^{2}\sum_{b=1}^{2}(e_a-e_b)J_{ab} = \frac{1}{2}[(e_1-e_1)J_{11}+(e_1-e_2)J_{12}+(e_2-e_1)J_{21}+(e_2-e_2)J_{22}].$$

You'll say that it's multiplication when there is already something in between the two summation terms as in the second summation equation.

(An analogy is by considering a loop with in a loop in programming. The outer loop will start say from 1, so while the first loop is at 1, the inner loop will continue looping until it satisfies a certain condition. After the inner loop stopped, the outer loop will proceed to the 2nd iteration and so the inner loop will loop again and so on... until both conditions in the inner and outer loop is satisfied.)

Last edited: Nov 22, 2009