How Do You Calculate Line Current for Different Connections in a 3 Phase Line?

[scorpio_wan1945]

3 x 15Ohm resistor (R) and 3 x 3.83 mH reactors (L) are connected in different ways across a 500V, 3 phase line. without drawing a phasor diagram, calculate the line current for each of the following connections:

a. R and L in series, connected in wye

b. R and L in parallel, connected in delta

i have difficulty in starting to solve...anyone could shed some light here?

thanks

 

[berkeman]

Well, in -a-, you just have a 15 Ohm resistor in series with a 3.83mH inductor from each phase to Neutral, right? It seems that the current in each leg would be identical and independent...

For -b-, you can look first at Wye-Delta transformations,

http://en.wikipedia.org/wiki/Y-delta_transform

or you could solve it using the phase angle differences...

 

[piercas]

for your question. In order to calculate the line current for each of the given connections, we will need to use Ohm's Law and the formula for calculating the impedance of a series and parallel circuit. Let's start by defining some variables:

R = 15Ohm resistor
L = 3.83 mH reactor
V = 500V
I = line current

a. R and L in series, connected in wye:
In this configuration, the total impedance (Z) can be calculated as the sum of the individual impedances:

Z = R + jωL
where j is the imaginary unit and ω is the angular frequency.

Since the components are connected in series, the line current is the same as the total current (I = I1 = I2 = I3).

To calculate the line current, we can use the formula:

I = V/Z

Substituting the values, we get:

I = 500V / (15Ohm + jω*3.83 mH)

To simplify the calculation, we can convert the inductance to ohms by using the formula:

ωL = jXl
where Xl is the inductive reactance.

Substituting the value of Xl, we get:

I = 500V / (15Ohm + j*2π*60Hz*3.83 mH)
= 500V / (15Ohm + j0.457Ohm)
= 500V / (15Ohm - j0.457Ohm)
= 500V / (15.003Ohm ∠-1.73°)

Using the polar form of the impedance, we can calculate the magnitude and phase angle of the impedance as:

|Z| = √(15.003^2 + 0.457^2) = 15.01Ohm
θ = tan^-1(0.457/15.003) = 1.73°

Therefore, the line current can be calculated as:

I = 500V / (15.01Ohm ∠1.73°)
= 33.3A ∠-1.73°

b. R and L in parallel, connected in delta:
In this configuration, the total impedance (Z) can be calculated as:

1/Z = 1/R + 1/jωL

Using the same approach as before, we can