(i) Find the Thevenin’s equivalent across a-ba−b in the
circuit in Fig. 7.7-2. (ii) Find the voltage across
a 500Ω500Ω connected across a-ba−b. (iii) Find the value
of capacitive reactance to be connected across a-ba−b if
the magnitude of voltage across the 500Ω500Ω load is to
be raised to 132kV132kV rmsrms. (iv) If a-ba−b gets
shorted what is the current that flows through the short?
Step-by-Step
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Verified Solution
(i) We use superposition principle to obtain the V_{oc}Voc
phasor
\begin{array}{l} V_{o c}=127 \angle 6^{\circ} \times
\frac{1.5+j 9}{3.5+j 21}+127 \angle 5^{\circ} \times
\frac{2+j 12}{3.5+j 21}=127 \angle 5.43^{\circ}
\mathrm{kV} \mathrm{rms} \\ Z_{t h}=(2+j 12) / /(1.5+j
9)=\frac{(2+j 12)(1.5+j 9)}{3.5+j 21}=0.857+j 5.143
\Omega \end{array}Voc=127∠6∘×3.5+j211.5+j9
+127∠5∘×3.5+j212+j12=127∠5.43∘kVrmsZth
=(2+j12)//(1.5+j9)=3.5+j21(2+j12)(1.5+j9)=0.857+j5.143Ω
(ii) The voltage phasor across the load of 500
Ω500Ω
=127 \angle 5.43^{\circ} \times \frac{500}{500.857+j
5.143}=126.77 \angle 4.84^{\circ} \mathrm{kV}
\mathrm{rms}=127∠5.43∘×500.857+j5.143500
=126.77∠4.84∘kVrms
(iii) We first determine a new Thevenin’s equivalent of the circuit
in Fig. 7.7-3 (b) for further load connection across ab.a−b. This equivalent circuit will have an open-circuit voltage
phasor = 126.77∠4.84°=126.77∠4.84° kV rmskVrms and
Thevenin’s equivalent impedance = (0.857 + j5.143)//500 =
0.908 + j5.125 Ω=(0.857+j5.143)//500=0.908+j5.125Ω.
Let –jX_{C}–jXC be the capacitive reactance connected now
across the output of this new equivalent circuit. The magnitude
of voltage phasor across the capacitor has to be 132
kV132kV rms.rms.
\begin{array}{l} \therefore\left|\frac{-j
X_{C}}{0.908+j\left(5.125-
X_{C}\right)}\right|=\frac{132}{126.77}=1.0413 \\ \text
{ i.e. } \frac{X_{C}{ }^{2}}{0.908^{2}+\left(5.125X_{C}\right)^{2}}=1.0413^{2}=1.0843 \end{array}∴∣∣∣∣
0.908+j(5.125−XC)−jXC∣∣∣∣=126.77132=1.0413 i.e. 0.9082+(5.125−XC)2XC2
=1.04132=1.0843
There are two solutions for X_{C}XC. They are 2.7
Ω2.7Ω and 129.1 Ω129.1Ω. The higher value is accepted.
(Why?).
Therefore, the required reactance is –129.1Ω–129.1Ω and the
required capacitive reactance is 129.1Ω129.1Ω. Once we
qualify a reactance by using capacitive we do not have to
include the negative sign.
(iv) The short-circuit current at a-ba−b is = 126.77∠4.84°
÷
(0.908 + j 5.125)=126.77∠4.84°÷(0.908+j5.125) =
126.77∠4.84°÷5.205∠- 80° = 6.26 –
=126.77∠4.84°÷5.205∠−80°=6.26–j23.54
kAj23.54kA rms = 24.36∠-75.16°rms=24.36∠−75.16° kA
rms.kArms.