GATE 2020 EE Q.46: solution and answer

Official answer: 123.00 to 127.00Source: Official answer key, GATE 2020 EE (IIT Delhi)

1. Split the original armature current into its active and reactive parts. The terminal voltage $V$ is constant, so $P=VI\cos \varphi \propto I\cos \varphi$ and $Q=VI\sin \varphi \propto I\sin \varphi$: $$I_p=100\times 0.9=90\text{A},I_q=100\sqrt{1-0.9^2}=100\sqrt{0.19}=43.589\text{A (lagging)}$$

2. An ideal reactor absorbs only lagging reactive power, so the real power stays the same. With $V$ and $P$ both constant, the active component does not change: $I_p^{\prime }=90\text{A}$.

3. The question says the total lagging reactive power (load plus reactor) is now twice what it was. At the same $V$, the reactive component of current doubles too: $$I_q^{\prime }=2\times 43.589=87.178\text{A}$$

4. Add the two components in quadrature. Since $I_q^2=100^2\times 0.19=1900$, we have $I_q^{\prime 2}=4\times 1900=7600$: $$I^{\prime }=\sqrt{I_p^{\prime 2}+I_q^{\prime 2}}=\sqrt{90^2+7600}=\sqrt{8100+7600}=\sqrt{15700}$$

5. $$I^{\prime }=125.30\text{A}$$ The new power factor is $90/125.30=0.718$ lagging. You do not need the machine's reactance or excitation: at a fixed terminal voltage, the generator supplies whatever $P$ and $Q$ the load draws, and the excitation only settles the internal EMF needed to do that.