A balanced 3-ϕ, star connected AC generator generates a phase voltage of magnitude |E_A| = |E_B| = |E_C| = 100 V. If the connection to one of the phases is reversed, what will be the magnitude of the line voltages?

Question

Question

This question was previously asked in

MPPGCL JE Electrical 19 March 2019 Shift 2 Official Paper

Answer 1

Concept

The relationship between the phase voltage (V_P) and line voltage (V_L) in a star connection is:

\(V_L=√{3}× V_{P}\)

Now, the phasors are as shown below:

qImage67d2e458c8064db59457064c

If the connection to one of the phases is reversed:

qImage67d2e459c8064db59457064f

Calculation

From the first phasor diagram, under normal conditions, the magnitudes of the line voltages are:

|EA| = |EB| = |EC| = √ 3 × 100 = 173 V

If one of the phases is reversed as shown in the figure, the line voltage for phase 'A' is:

\(E_{A}=\sqrt{(E_{A})^2+(E_{B})^2+2E_{A}E_{B}cos\theta_{AB}}\)

\(E_{A}=\sqrt{(100)^2+(100)^2+2(100)(100)cos(60)}\)

\(E_{A}=173.2 \space V\)

The line voltage for phase 'B' with respect to phase 'C' or vice versa is:

\(E_{B}=\sqrt{(E_{B})^2+(E_{C})^2+2E_{B}E_{C}cos\theta_{BC}}\)

\(E_{B}=\sqrt{(100)^2+(100)^2+2(100)(100)cos(240)}\)

\(E_{B}=\sqrt{(100)^2+(100)^2+2(100)(100)cos(-0.5)}\)

|EB| = |EC| = 100 V