Need some help with an electricity query...

The formula for power losses in a wire is

\(\displaystyle P_L = I^2 R\)

where I is the current passing through the wire and R is the resistance. If the power transmitted along the cable is P and the voltage level is V, then the losses can be written as

\(\displaystyle P_L = \frac{RP^2}{V^2}\)

In an A.C. circuit, V is not constant, so can substitute

\(\displaystyle V = |V|\cos\phi\)

where \(\displaystyle \cos\phi\) is called the 'power factor' and is a parameter of the generator. The formula can therefore be rewritten as

\(\displaystyle P_L = \frac{R}{|V|^2 \cos^2 \phi } \cdot P^2 = B P^2\)

where B is called the 'B-loss coefficient', which is

\(\displaystyle B = \frac{R}{|V|^2 \cos^2 \phi }\)

Now... for the case of two generators connected to a single node (aka bus) via two separate wires, we can rewrite the loss formula to be

\(\displaystyle P_{L,i} = \frac{R_i}{|V_i|^2 \cos \phi_i }\)

I would've thought that

\(\displaystyle P_L = P_{L,1} + P_{L,2}\)

\(\displaystyle P_L = \frac{R_1}{|V_1|^2 \cos \phi_1 } P_1^2 + \frac{R_2}{|V_2|^2 \cos \phi_2 } P_2^2\)

or, if we define

\(\displaystyle B_i = \frac{R_i}{|V_i|^2 \cos^2 \phi_i }\)

we get

\(\displaystyle P_L = B_1 P_1^2 + B_2 P_2^2\)

However, according to an online journal article, (http://www.arpapress.com/volumes/vol12issue2/ijrras_12_2_20.pdf) they seem to define B something like this (not sure about this one):

\(\displaystyle B_{ij} = \frac{R}{|V_i||V_j| \cos \phi_i \cos \phi_j}\)

and then quote the following formula as the solution:

\(\displaystyle P_L = B_{11} P_1^2 + 2B_{12} P_1 P_2 + B_{22} P_2^2\)

My questions are:

1. Why isn't the power loss just the sum of the power losses across each cable? What is the significance of the additional \(\displaystyle 2 B_{12} P_1 P_2\) term?

2. What is R in the definition of \(\displaystyle B_{ij}\)? Is it the total resistance calculated from \(\displaystyle \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\)? If so, how come we need to know the total resistance of a circuit to calculate the power losses in a single wire? Is it because of the dependence of power injection across the other wire?

The formula for power losses in a wire is

\(\displaystyle P_L = I^2 R\)

where I is the current passing through the wire and R is the resistance. If the power transmitted along the cable is P and the voltage level is V, then the losses can be written as

\(\displaystyle P_L = \frac{RP^2}{V^2}\)

In an A.C. circuit, V is not constant, so can substitute

\(\displaystyle V = |V|\cos\phi\)

where \(\displaystyle \cos\phi\) is called the 'power factor' and is a parameter of the generator. The formula can therefore be rewritten as

\(\displaystyle P_L = \frac{R}{|V|^2 \cos^2 \phi } \cdot P^2 = B P^2\)

where B is called the 'B-loss coefficient', which is

\(\displaystyle B = \frac{R}{|V|^2 \cos^2 \phi }\)

Now... for the case of two generators connected to a single node (aka bus) via two separate wires, we can rewrite the loss formula to be

\(\displaystyle P_{L,i} = \frac{R_i}{|V_i|^2 \cos \phi_i }\)

I would've thought that

\(\displaystyle P_L = P_{L,1} + P_{L,2}\)

\(\displaystyle P_L = \frac{R_1}{|V_1|^2 \cos \phi_1 } P_1^2 + \frac{R_2}{|V_2|^2 \cos \phi_2 } P_2^2\)

or, if we define

\(\displaystyle B_i = \frac{R_i}{|V_i|^2 \cos^2 \phi_i }\)

we get

\(\displaystyle P_L = B_1 P_1^2 + B_2 P_2^2\)

However, according to an online journal article, (http://www.arpapress.com/volumes/vol12issue2/ijrras_12_2_20.pdf) they seem to define B something like this (not sure about this one):

\(\displaystyle B_{ij} = \frac{R}{|V_i||V_j| \cos \phi_i \cos \phi_j}\)

and then quote the following formula as the solution:

\(\displaystyle P_L = B_{11} P_1^2 + 2B_{12} P_1 P_2 + B_{22} P_2^2\)

My questions are:

1. Why isn't the power loss just the sum of the power losses across each cable? What is the significance of the additional \(\displaystyle 2 B_{12} P_1 P_2\) term?

2. What is R in the definition of \(\displaystyle B_{ij}\)? Is it the total resistance calculated from \(\displaystyle \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\)? If so, how come we need to know the total resistance of a circuit to calculate the power losses in a single wire? Is it because of the dependence of power injection across the other wire?

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