GATE EE · Power Systems

Power System Stability – GATE EE Power SystemsPractice Questions & PYQs

Generate GATE-level questions on Stability analysis. Focus on: 1. Steady-state stability: Power-angle equation and Synchronizing power coefficient. 2. Transient stability: Equal Area Criterion and Swing Equation. 3. Methods to improve stability (Fast valving, High-speed circuit breakers).

25 questions · 20 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2022NAT

Two generating units rated for 250 MW and 400 MW have governor speed regulations of 6% and 6.4%, respectively, from no load to full load. Both the generating units are operating in parallel to share a load of 500 MW. Assuming free governor action, the load shared in MW, by the 250 MW generating unit is _________. (round off to nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

Let no-load frequency is 50 Hz. Draw the curve : From the curve,

\begin{aligned} \frac{50-f}{3}&=\frac{P_1}{250} \;\;\;..(1) \\ \frac{50-f}{3.2}&=\frac{P_2}{400} \;\;\;..(2) \end{aligned}

From eq. (1) & (2),

\begin{aligned} \frac{3P_1}{250}&= \frac{3.2P_2}{400} \\ 3P_1&=2P_2\;\;\;...(3) \end{aligned}

Given: $P_1+P_2=500$ From eq. (3), $P_1+1.5P_1=500$ $P_1=200MW$

Explanation Figure 1

Q2GATE 2022NAT

A 20 MVA, 11.2 kV, 4-pole, 50 Hz alternator has an inertia constant of 15 MJ/MVA. If the input and output powers of the alternator are 15 MW and 10 MW, respectively, the angular acceleration in mechanical $degree/s^2$ is __________. (round off to nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

We have, swing equation $\frac{2HS}{\omega _s}\frac{d^2\delta }{dt^2}=P_a$ Put the values,

\begin{aligned} \frac{2 \times 15 \times 20}{360 \times 50}\frac{d^2\delta }{dt^2}&=15-10\\ \frac{d^2\delta }{dt^2}&=\frac{5 \times 360 \times 50}{2 \times 15 \times 20}\\ &=150 \text{ electric degree per} sec^2\\ \frac{d^2\delta }{dt^2}&=\alpha =\frac{2}{4} \times 150 \text{ Mech. degree/}sec^2\\ &=75\text{ Mech. degree/}sec^2 \end{aligned}

Q3GATE 2021MCQ

In the figure shown, self-impedances of the two transmission lines are $1.5j\:p.u$ each, and $Z_{m}=0.5j \:p.u$ is the mutual impedance. Bus voltages shown in the figure are in p.u. Given that $\delta>0$ , the maximum steady-state real power that can be transferred in p.u from Bus-1 to Bus-2 is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} L_{e q}&=\frac{L_{1} L_{2}-M^{2}}{\left(L_{1}+L_{2}-2 M\right)} \\ X_{\mathrm{eq}}&=\frac{1.5 \times 1.5-0.5^{2}}{1.5+1.5-2 \times 0.5}=1 \mathrm{p.u.} \end{aligned}

\begin{aligned} P_{\max }&=\frac{|E||V|}{1}\\ P_{\max }&=|E||V| \end{aligned}

Explanation Figure 2

Explanation Figure 3

Q4GATE 2019NAT

In the single machine infinite bus system shown below, the generator is delivering the real power of 0.8pu at 0.8 power factor lagging to the infinite bus. The power angle of the generator in degrees (round off to one decimal place) is _________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} X &=0.25+0.2+0.4||0.4 \\ &=0.45+0.2=0.65pu \\ P&=V_{pu} \times I_{pu} \cos \phi \\ 0.8 &=1 \times I_{pu} \times 0.8 \\ I_{pu} &= 1pu\\ E &= V+jI_aX_s\\ &=1+1\angle -36.86^{\circ} \times j0.65 \\ &= 1.484 \angle 20.51^{\circ}pu\\ \delta &= 20.51^{\circ} \end{aligned}

Q5GATE 2018MCQ

Consider a lossy transmission line with $V_{1} \; and \; V_{2}$ as the sending and receiving end voltages, respectively. Z and X are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be

🚀 Solve in Practice Mode

📖 Explanation

With only x: $P_{max}=\left | \frac{V_1V_2}{x} \right |$ With Lossy Tr, Line $P=\left | \frac{V_1V_2}{z} \right | \cos (\beta -\delta )-\left | \frac{AV_2^2}{z} \right |\cos (\beta -\delta )$ Therefore, with Lossy Line $P_{max} \lt \left | \frac{V_1V_2}{x} \right |$

Explanation Figure 1

Explanation Figure 2

Q6GATE 2017NAT

The figure shows the single line diagram of a power system with a double circuit transmission line. The expression for electrical power is $1.5 sin\delta$ , where $\delta$ is the rotor angle. The system is operating at the stable equilibrium point with mechanical power equal to 1 pu. If one of the transmission line circuits is removed, the maximum value of $\delta$ as the rotor swings, is 1.221 radian. If the expression for electrical power with one transmission line circuit removed is $P_{max} sin\delta$ , the valueof $P_{max}$ , in pu is _________.

🚀 Solve in Practice Mode

📖 Explanation

Using equal area criteria: $A_1=A_2$ $\int_{\delta _0}^{\delta _c}(P_{m0}-P_{max1} \sin \delta )d\delta =\int_{\delta _c}^{\delta _2}(P_{max1} \sin \delta -P_{m0})d\delta$ By solving above integration, $P_{max1}=\frac{P_{m0}(\delta _2-\delta _0)}{\cos \delta _0- \cos \delta _2}$ Given data:

\begin{aligned} P_{m0} &=1pu \\ P_e&= 1.5 \sin (\delta _0)=P_{m0}=1pu\\ \delta _0&41.8^{\circ}=0.7295\; rad \\ \delta _2 &=1.221\; rad =96.95^{\circ} \end{aligned}

Substitute above values in above equation, $P_{max1}=\frac{1(1.221-0.7295)}{\cos 41.8- \cos 69.95}=1.220 p.u.$

Explanation Figure 2

Q7GATE 2017NAT

A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} S&=250 MVA \\ \cos \phi &=0.8 \\ K.E. &=1000MJ \\ P_e&= 60MW\\ \delta _0&=10^{\circ} \\ t&=10\; cycles\\ &=\frac{10}{50}=0.2\; sec \\ t&=5\; cycle =0.1sec \\ \text{Load}&\text{ is removed,} \\ P_e&=0 \\ P_a&=P_m-P_e \\ P_m &=60MW \\ M&=\frac{HS}{180f}=\frac{K.E.}{180f} \\ &= \frac{1000}{180 \times 50}=0.111\\ \frac{d^2 \delta }{dt^2}&=\frac{P_a}{M}=545.45 \text{ ele.degree}/s^2 \\ \text{Integrating}&\text{ the above eq.} \\ \int \left ( \frac{d^2 \delta }{dt^2} \right )dt&=\int \frac{P_a}{M}dt=545.45 \times t\\ \text{Integrating}&\text{ tit once again with dt} \\ &=545.45 \times t(dt)\\ \delta &=\frac{545.45 \times t^2}{2}=2.7^{\circ}\\ \text{So, new}&\text{ value of power angle}\\ &=10^{\circ}+2.7^{\circ}=12.7^{\circ} \end{aligned}

Q8GATE 2015NAT

A 50 Hz generating unit has H-constant of 2 MJ/MVA. The machine is initially operating in steady state at synchronous speed, and producing 1 pu of real power. The initial value of the rotor angle $\delta$ is 5 $^{\circ}$ , when a bolted three phase to ground short circuit fault occurs at the terminal of the generator. Assuming the input mechanical power to remain at 1 pu, the value of $\delta$ in degrees, 0.02 second after the fault is _________.

🚀 Solve in Practice Mode

📖 Explanation

Let $3-\phi$ fault occurs at t=0 $\Rightarrow \; P_e=0\text{ for }t\geq 0^+$ Now, swing equation for $t\geq 0^+$

\begin{aligned} \frac{2H}{\omega _s}\frac{d^2\delta }{dt^2} &=P_m \\ \frac{d^2\delta }{dt^2}&= \frac{P_m \omega _s }{2H}\\ \frac{d\delta }{dt}&= \frac{P_m \omega _s }{2H}t+C\\ C=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right| _{t=0^+}&=\omega (t=0^+)-\omega _s\\ C=\omega _s-\omega _s&=0\;(\omega (t-0^+)=\omega _s) \\ \frac{d\delta }{dt}&=\frac{P_m \omega _s }{2H}t \\ \Rightarrow \; \delta &= \frac{P_m \omega _s }{4H}t^{2+}C'\\ C'&= \delta |_{t=0^+}=\delta _0\\ \delta &= \frac{P_m \omega _s }{4H}t^{2+}\delta _0\\ \text{Now \in problem,}\\ \delta (t=0.02)&\\ &=\frac{1 \times 2 \times 180 \times 50 \times 0.02^2 +5^{\circ}}{4 \times 2}\\ &=5.9^{\circ} \end{aligned}

Q9GATE 2015MCQ

The synchronous generator shown in the figure is supplying active power to an infinite bus via two short, lossless transmission lines, and is initially in steady state. The mechanical power input to the generator and the voltage magnitude E are constant. If one line is tripped at time $t_1$ by opening the circuit breakers at the two ends (although there is no fault), then it is seen that the generator undergoes a stable transient. Which one of the following waveforms of the rotor angle $\delta$ shows the transient correctly?

🚀 Solve in Practice Mode

📖 Explanation

Initial value of $\delta =\delta _0$ at $t =t_1$ with both lines are in service of one of the line circuit breakes opened the $\delta$ is increased for same time and stabilizes to new value of $\delta$ .

Q10GATE 2014NAT

The figure shows the single line diagram of a single machine infinite bus system. The inertia constant of the synchronous generator H = 5 MW-s/MVA. Frequency is 50 Hz. Mechanical power is 1 pu. The system is operating at the stable equilibrium point with rotor angle $\delta$ equal to 30 $^{\circ}$ . A three phase short circuit fault occurs at a certain location on one of the circuits of the double circuit transmission line. During fault, electrical power in pu is $P_{max} \sin \delta$ . If the values of $\delta \; and \; d\delta/dt$ at the instant of fault clearing are 45 $^{\circ}$ and 3.762 radian/s respectively, then $P_{max}$ (in pu) is ______.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \frac{2H}{\omega _s}\frac{d^2\delta }{dt^2}&=(P_m-P_e) \\ \text{Let, } \omega _r&=\frac{d\delta }{dt} \\ \frac{2H}{\omega _s}\frac{d\omega _r }{dt} &=(P_m-P_e) \\ \text{Multiplying with } &\omega _r \text{ \in both side}\\ \frac{H}{\omega _s}\left ( 2\omega _r \frac{d\omega _r }{dt}\right )&= (P_m-P_e)\frac{d\delta }{dt}\\ \Rightarrow \; \frac{H}{\omega _s} 2d(\omega _r)^2&=(P_m-P_e)d\delta \\ \Rightarrow \; \frac{H}{\omega _s}\int_{\omega _{r_1}}^{\omega _{r_2}}d(\omega _r^2) &= \int_{\delta _1}^{\delta _2}(P_m-P_e)d\delta \\ \text{where, } & \\ \omega _{r_1}&=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right|_{\delta =\delta _1} =\omega (\delta =\delta _1)-\omega _s\\ \omega _{r_2}&=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right|_{\delta =\delta _2} =\omega (\delta =\delta _2)-\omega _s\\ \frac{H}{\omega _s}(\omega _{r_2}^2 -\omega _{r_1}^2)&=\int_{\delta _1}^{\delta _2}(P_m-P_e)d\delta \\ \text{In problem, }&\\ \frac{d\delta }{dt}=0, &\frac{d\delta }{dt}=3.762\\ \frac{5}{314}(3.762^{2-}0^2)&=\int_{30^{\circ}}^{45^{\circ}}(1-P_{max} \sin \delta )d\delta \\ \Rightarrow \;\; P_{max}&=0.23p.u. \end{aligned}

Explanation Figure 2

Q11GATE 2014NAT

A synchronous generator is connected to an infinite bus with excitation voltage $E_f= 1.3$ pu. The generator has a synchronous reactance of 1.1 pu and is delivering real power (P) of 0.6 pu to the bus. Assume the infinite bus voltage to be 1.0 pu. Neglect stator resistance. The reactive power (Q) in pu supplied by the generator to the bus under this condition is _____.

🚀 Solve in Practice Mode

📖 Explanation

Given, $E_f=1.3pu, X_s=1.1pu, V=1 pu, P=0.6pu$ Neglecting stator resistance i.e. $\theta _s=90^{\circ}$ Output power delivered,

\begin{aligned} P &=\frac{VE_f}{X_s}\sin \delta \\ \sin \delta &=\frac{PX_s}{VE_f} \\ &= \frac{0.6 \times 1.1}{1 \times 1.3}\approx 0.5\\ \therefore \;\; \delta &=30^{\circ}=\text{Power angle} \\ \therefore \;\; \cos \delta &= \cos 30^{\circ}=\frac{\sqrt{3}}{2} \end{aligned}

Reactive power is given by,

\begin{aligned} Q&=\left[ \frac{VE_f}{X_s} \cos \delta -\frac{V^2}{X_s} \right]\$\\ &=\left[ \frac{1 \times 1.3}{1.1} \times \frac{\sqrt{3}}{2}-\frac{1^2}{1.1} \right]\$\\ Q&=(1.023-0.909)=0.114pu\\ &\approx 0.11pu \end{aligned}

Q12GATE 2014NAT

There are two generators in a power system. No-load frequencies of the generators are 51.5 Hz and 51 Hz, respectively, and both are having droop constant of 1Hz/MW. Total load in the system is 2.5 MW. Assuming that the generators are operating under their respective droop characteristics, the frequency of the power system in Hz in the steady state is ______.

🚀 Solve in Practice Mode

📖 Explanation

Let no-load frequency of generator-1 be 51.5 Hz and no-load frequency of generator-2 be 51 Hz. Given total load = 2.5 Mwand drop of both machines =1 Hz/MW Let machine-1 shares a load of $P_1$ MW them machine-2 will share a load of $(2.5-P_1)$ MW Le the steady state frequency of the system be f.

\begin{aligned} \text{Now, }\frac{51.5-f}{P_1}&=1=droop \\ P_1&= (51.5-f)\;\;...(i)\\ \text{Also, } \left ( \frac{51-f}{2.5-P_1} \right )&=1=droop \\ 51-f &=2.5-P_1 \\ P_1 &=f-48.5 \;\;...(ii) \\ \text{equating eq. }&\text{(i) and (ii), we get,} \\ 51.5-f &=f-48.5 \\ 2f&=100 \\ f&=50Hz \end{aligned}

Explanation Figure 1

Q13GATE 2012MCQ

A cylinder rotor generator delivers 0.5 pu power in the steady-state to an infinite bus through a transmission line of reactance 0.5 pu. The generator no-load voltage is 1.5 pu and the infinite bus voltage is 1 pu. The inertia constant of the generator is 5MW-s/MVA and the generator reactance is 1 pu. The critical clearing angle, in degrees, for a three-phase dead short circuit fault at the generator terminal is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} P_e&= \frac{|E||V|}{(X_d+X_T)}\sin \delta _0\\ 0.5 &=\frac{1.5 \times 1}{(1+0.5)}\sin \delta _0 \\ \delta _0 &=\sin ^{-1}(0.5)=30^{\circ}=\frac{\pi}{6} \end{aligned}

Critical clearing angle

\begin{aligned} \delta _{cr}&=\cos ^{-1}[(\pi-2\delta _0) \sin \delta _0 -\cos \delta _0]\\ &=\cos ^{-1}\left[ \left ( \pi-\frac{\pi}{3} \right ) \times \frac{1}{2} -\frac{\sqrt{3}}{2}\right]\\ &= \cos ^{-1} (0.18)=79.56^{\circ} \end{aligned}

Explanation Figure 1

Q14GATE 2009MCQ

A 500 MW, 21 kV, 50 Hz, 3-phase, 2-pole synchronous generator having a rated p.f = 0.9, has a moment of inertia of $27.5 \times 10^{3} kg-m^{2}$ .The inertia constant (H) will be

🚀 Solve in Practice Mode

📖 Explanation

J= moment of inertia $=27.5 \times 10^3 kg-m^2$ Synchronous speed $N_s=\frac{120f}{P}=\frac{120 \times 50}{2}=300rpm$ Kinetic energy of the rotor $=\frac{1}{2}J\omega _s^2=\frac{1}{2}\times 27.5 \times 10^3 \times (314.16)^2$ K.E.=1375 MJ G= Machine rating $=\frac{500}{pf}=\frac{500}{0.9}=555.55MVA$ Inertia constant $=H=\frac{K.E.}{G}=\frac{1357}{555.55}=2.44 \; MJ/MVA$

Q15GATE 2008MCQ

A loss less single machine infinite bus power system is shown below : The synchronous generator transfers 1.0 per unit of power to the infinite bus. The critical clearing time of circuit breaker is 0.28 s. If another identical synchronous generator is connected in parallel to the existing generator and each generator is scheduled to supply 0.5 per unit of power, then the critical clearing time of the circuit breaker will

🚀 Solve in Practice Mode

📖 Explanation

For such faults, $\delta _{cr}$ is given by $\delta _{cr}=\cos ^{-1} [(\pi-2\delta _0) \sin \delta _0 -\cos \delta _0]\;...(i)$ and

\begin{aligned} t_{cr}&=\text{critical cleaning time } \\ &= \left ( \frac{2H(\delta _{cr}-\delta _0)}{\pi f P_m} \right )^{1/2}\;...(ii) \end{aligned}

CASE-1: When only once generator is connected Mechanical input to the generator $(P_{m_1})=$ Electrical power delivered by the generator $(P_{e_1})$ $\Rightarrow \; P_{m_1}=P_{e_1}=1p.u.$ CASE-2: When two generator connected in parallel Electrical power delivered by each generator $P_{e_2}=0.5p.u.$ Mechanical inout to each generator $(P_{m_2})=P_{e_2}=0.5p.u.=\frac{P_{m_1}}{2}$ Assuming $\delta _0$ in each case same $\delta _{0_1}=\delta _{0_2}=\delta _{0}$ So, $\delta _{cr}$ will also be same in both cases, (from equation (i)),

\begin{aligned} t_{cr}&\propto \frac{1}{\sqrt{P_m}}\\ &(\delta _{cr}, H \text{ and } \delta _0 \text{ are same}) \\ \therefore \; t_{cr_2} &=\sqrt{\frac{P_{m_1}}{P_{m_2}}}\times t_{cr_1} \\ t_{cr_2}&= \sqrt{\frac{P_{m_1}}{P_{m_1}/2}}\times 0.28\\ &= \sqrt{2} \times 0.28=0.396\;sec \end{aligned}

Explanation Figure 2

Q16GATE 2007MCQ

An isolated 50 Hz synchronous generator is rated at 15 MW which is also the maximum continuous power limit of its prime mover. It is equipped with a speed governor with 5% droop. Initially, the generator is feeding three loads of 4 MW each at 50 Hz. One of these loads is programmed to trip permanently if the frequency falls below 48 Hz .If an additional load of 3.5 MW is connected then the frequency will settle down to

🚀 Solve in Practice Mode

📖 Explanation

Here, change in (frequency) w.r.t. to power

\begin{aligned} \left ( \frac{5}{15} \right ) \times 3.5&=1.16\%\\ \therefore \; \Delta f &=0.583\\ \therefore \; f&=50-0.583\\ &=49.417 Hz \end{aligned}

Q17GATE 2007MCQ

Consider a synchronous generator connected to an infinite bus by two identical parallel transmission line. The transient reactance 'x' of the generator is 0.1 pu and the mechanical power input to it is constant at 1.0 pu. Due to some previous disturbance, the rotor angle ( $\delta$ ) is undergoing an undamped oscillation, with the maximum value of $\delta (t)$ equal to 130 $^{\circ}$ .One of the parallel lines trip due to the relay maloperation at an instant when $\delta (t) = 130^{\circ}$ as shown in the figure. The maximum value of the per unit line reactance, x such that the system does not lose synchronism subsequent to this tripping is

🚀 Solve in Practice Mode

📖 Explanation

The curve $P_{e1}$ and $P_{e2}$ represents power before the tripping of one line and after the tripping respectively $P_{e\; max\;2}=\frac{EV}{X_2}$ $X_2=0.1+X$ for max value of X system will not loose synchronization, if at $\delta _2$

\begin{aligned} P_{e2} &=P_m \\ \frac{EV}{X_2} \sin \delta _2&=P_m \\ \frac{1 \times 1}{X_2} \sin 130^{\circ}&=1 \\ X_2&=0.77 \\ X&=0.67 \end{aligned}

Explanation Figure 2

Q18GATE 2006MCQ

A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30 $^\circ$ . The system frequency is 50 Hz. If the initial accelerating power is X pu, the initial acceleration in elect $deg/sec^2$ , and the inertia constant in MJ-sec/elect deg respectively will be

🚀 Solve in Practice Mode

📖 Explanation

Inertia constant (in MJ-s/elec deg) $M=\frac{GH}{180f}$ where, G = Machine rating = 100MVA H = Ineria constant in MJ/MVA = 5 MJ/MVA

\begin{aligned} M&=\frac{GH}{180f}=\frac{5 \times 100}{180 \times 50} \\ &=0.056\; \text{MJ-s/elec deg }\\ M_{\in \; pu}&= \frac{H}{180f}=\frac{5}{180 \times 50}\\ \alpha &= \text{Acceleration \in elect deg/sec}\\ M\alpha &=P_a \\ P_a&= \text{Accelerating power}=X \\ \alpha &= \frac{P_a}{M}=\frac{X}{(5/180 \times 50)}\\ &= 1800 \; \text{X elec deg/sec} \end{aligned}

Q19GATE 2005MCQ

A generator with constant 1.0 p.u. terminal voltage supplies power through a step-up transformer of 0.12 p.u. reactance and a double-circuit line to an infinite bus bar as shown in the figure. The infinite bus voltage is maintained at 1.0 p.u. Neglecting the resistances and suspectances of the system, the steady state stability power limit of the system is 6.25 p.u. If one of the double-circuit is tripped, then resulting steady state stability power limit in p.u. will be

🚀 Solve in Practice Mode

📖 Explanation

Induced emf of the generator $=|t_g|=1p.u.$ Termincal voltage $=|V_t|=1p.u.$ Reactance of transformer $=X_t=0.12p.u.$ When double-circuit line is connected Reactance of line $=X||X=\frac{X}{2}$ Steady state stability power limit

\begin{aligned} P_1&= \frac{|E_g||V_t|}{X_t+\frac{X}{2}}=\frac{|X|}{0.12+\frac{X}{2}}\\ 6.25&= \frac{1}{0.12+\frac{X}{2}}\\ X&= 0.08p.u. \end{aligned}

When one of the double circuit is tripled, steady state stability power limit $=P_2=\frac{|E_g||V_t|}{X_t+X}=\frac{|X|}{0.12+0.08}=5p.u.$

Q20GATE 2005MCQ

An 800 kV transmission line has a maximum power transfer capacity of P. If it is operated at 400 kV with the series reactance unchanged, the new maximum power transfer capacity is approximately

🚀 Solve in Practice Mode

📖 Explanation

Power transfer capacity $\propto V^2$

\begin{aligned} \frac{P_1}{P_2}&=\frac{V_1^2}{V_2^2} \\ P_2&=P_1 \times \left ( \frac{V_2}{V_1} \right )^2 \\ &= P \times \left ( \frac{400}{800} \right )^2\\ &=\frac{P}{4} \end{aligned}

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