Question

If VBA = 240 < 120° VRMs, positive sequence, determine, in the circuit of the attached figure. A) Neutral displacement voltage Vnn' B) Reading of wattmeters placed in B and C(common A). 5 j1522 12 w -j12 22 HE B n j7 Ω ni с 10 kVA, fp 0.6

Answer 1

Ans)

A) Given

VBA 2402120° V

VAB = -VBA 240Z - 60° V

the line to neutral voltage is given as

phase voltage lags line voltage by 30 degrees

VAB VAN -2 -30° V3 240 L-60° -30° V V3

poistive sequence balanced source other phase voltages are

Van' 138.562 - 90° V

VBA 138.562 - 90° - 120° 120º = 138.562 - 2100 V

$V_{Cn'}=138.56\angle -90^{o}+120^{o}=138.56\angle 30^{o}\: \: V$

now Kirchofff's current law at node 'n' gives

$\frac{V_{nn'}-V_{An'}}{5+j15}+\frac{V_{nn'}-V_{Bn'}}{12-j12}+\frac{V_{nn'}-V_{Cn'}}{9+j7}=0$

VAN VBn' + Ven' 9+17 Vn 12 - 312 9+17 5+j15 5+j15 12 - 312

$0.1495\angle -28.87^{o}V_{nn'}=\frac{138.56\angle -90^{o}}{5+j15}+\frac{138.56\angle -210^{o}}{12-j12}+\frac{138.56\angle 30^{o}}{9+j7}$

$0.1495\angle -28.87^{o}V_{nn'}=7.76\angle -122.44^{o}$

$V_{nn'}=\frac{7.76\angle -122.44^{o}}{0.1495\angle -28.87^{o}}$

is the answer

51.922 – 93.56° V nn'

=================

B)

first we will find current through parallel load connected given S=10kVA fp=0.6 + leading

as load is connected between A and B the voltage is $V_{AB}=240\angle -60^{o}\: \: V$

the current through this load is given as

$I_{AB}=(\frac{S\angle -cos^{-1}f_{p}}{V_{AB}})^{*}=(\frac{10k\angle -cos^{-1}0.6}{240\angle -60^{o}})^{*}$

$I_{AB}=41.67\angle -6.87^{o}\: \: A$

now currents in load impedances connected in star network is

$I_{An}=\frac{V_{An'}-V_{nn'}}{5+j15}=5.49\angle -159.44^{o}\: \: A$

$I_{Bn}=\frac{V_{Bn'}-V_{nn'}}{12-j12}=9.91\angle 178.96^{o}\: \: A$

$I_{Cn}=\frac{V_{Cn'}-V_{nn'}}{9+j7}=15.15\angle 6.62^{o}\: \: A$

now two loads in place using Kirchhoff's current law the currents drawn from source is

$I_{B}=I_{Bn}-I_{AB}=51.54\angle 174.25^{o}\: \: A$

$I_{C}=I_{Cn}=15.15\angle 6.62^{o}\: \: A$ as the load is not connected in this branch

given   

VBA 2402120° V

$V_{AB}=240\angle -60^{o}\: \: V$

$V_{CA}=240\angle -60^{o}+120^{o}\: \: V$

$V_{CA}=240\angle 60^{o}\: \: V$

Now wattmeter reading in B is

$\mathbf{W_{B}=7226.8\: \: W}$

Now wattmeter reading in C is

$\mathbf{W_{C}=2169.6\: \: W}$