I know that in a power system there are no energy storage elements. Due to this, as soon as power is generated by the generators it is transmitted to the end user where the power is utilized.

Say a power system is stable at the moment (energy supply is equal to the energy consumption.) If new equipment is connected to the power system (say 5 MW) it is energized as soon as it is connected.

Where does that additional 5 MW of power come from instantly until the power system frequency goes down and the active power governors speed up the generators? (Obviously there is a time delay.)

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    \$\begingroup\$ One word: inertia. After that, system will sense the tiny drop in frequency and request more power to compensate and the system is once again in balance. \$\endgroup\$
    – winny
    Jun 22, 2021 at 16:44
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    \$\begingroup\$ Off course they are storage elements: Most hydro dams act as storage: pump during low demand and generate high demand. \$\endgroup\$
    – Hilmar
    Jun 23, 2021 at 1:57
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    \$\begingroup\$ Also search on: spinning reserve or operating reserve. This may be fast response conventional such as hydropower including pumped storage and more recently battery storage and flywheels. \$\endgroup\$
    – Russell McMahon
    Jun 23, 2021 at 9:26
  • \$\begingroup\$ @Hilmar, by "storage elements", we're talking about things like batteries or capacitors to bridge the few-second gap between everyone turning on their microwave at halftime and the grid operator opening the intake valves on the hydroelectric turbines to compensate. \$\endgroup\$
    – Mark
    Jun 23, 2021 at 23:45

5 Answers 5


Where that additional 5MW of power came from instantly until the power system frequency go down and active power governors speed up the generators (obviously there is a time delay)?

The voltage on the line sags when loads are switched on (which is dependent on resistance and inductance in physical wires; if they were superconducting wires, the load and source would react almost instantaneously). When large loads are switched on and the grid can't source the power instantly because the wire inductance and resistance prevents it, the voltage sags and can cause a brownout (the same thing happens in your house when you turn on a vacuum and the lights dim for a bit).

Usually in the case of larger loads the power company needs to be notified so that they can be ready to avoid a local brownout. The power generation on the other end must provide more current (to match the current being drawn at the other end by the load) or the voltage will drop and no one likes that.

In the short term, capacitors and DFR's (Distributed Feeder Regulators) and other regulators can make up some of the difference, but they can only cover line regulation for short amounts of time with small amounts of power.

If too much power is drawn, the power generation facility might 'trip' and a blackout will occur until generation is restored.

  • 1
    \$\begingroup\$ "the same thing happens in your house when you turn on a vacuum and the lights dim for a bit" Are you sure this isn't due to the inrush current of the vacuum as the motor gets up to speed? \$\endgroup\$
    – Phil Frost
    Jun 23, 2021 at 17:56
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    \$\begingroup\$ My point is the inrush current is a large load and it dims the lights, same thing with a very large load and the local grid. \$\endgroup\$
    – Voltage Spike
    Jun 23, 2021 at 18:01
  • \$\begingroup\$ If I understand correctly the OP question is less concerned with voltage sags, brownouts or extremely large overloads for a system. This answer talks more about what happens to the electric grid but relatively little about "where does the newly connected load receive its power during the initial moments after being connected". \$\endgroup\$
    – Radu Popa
    Jun 24, 2021 at 22:26
  • \$\begingroup\$ I thought I'd explain the whole picture to give a better understanding for everyone from circuit theory. Apparently, they liked the answer. \$\endgroup\$
    – Voltage Spike
    Jun 24, 2021 at 22:28

When you connect a new load, it becomes part of the larger circuit. The current starts flowing through the new load. This is a transient process and the associated electromagnetic field propagates close to the speed of light through the entire grid - that is, the change will be detected almost immediately in other parts of the Power grid, even at relatively large distances away from where the new load was connected.

Where does that additional 5 MW of power come from instantly until the power system frequency goes down and the active power governors speed up the generators?

If you zoom-in at small time scales (micro- to milli-seconds), as current starts flowing through the new load and before the generators have a chance to adjust their power production, the power that the new load starts drawing is collected by "starving" the other loads as the current/power flow redistributes. You can consider two circuits:

  1. The grid before the new load was connected.
  2. The grid right after the new load was connected. In the time instance immediately after the new load is connected, this circuit has the same power sources as the original circuit (no generator changes).

You can calculate the current flow through each topology and will see that, given the same power sources, the new load will get its power by "starving other loads".

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    \$\begingroup\$ The UK Grid can supply 60GW but normally runs daytime at about 30GW. (grid.iamkate.com). 5MW is < 0.02% and won't be noticed at grid voltages -- I have seen a 6MW electric fire (OK, flash dryer in a paper mill). Come to think of it, there was a 5MW log grinder at the start of the same paper run too. \$\endgroup\$ Jun 23, 2021 at 8:49

We can't simply connect a huge load to a power system without informing the authorities(5 MW is really a huge load)

When such a huge load is connected to the existing balanced grid, the grid will initially try to satisfy the load requirements. As a result, the voltage and frequency of the grid will drop as the system doesn't know about the But the Automatic Voltage regulators (AVR) and automatic frequency controller(AFC) will try to make the system stable by satisfying the increasing torque demand of the generator(By increasing the fuel level, steam level, etc.)

And generally, in a Power plant,

  • Diversity factor must be greater than 1
  • Demand factor must be less than 1

Diversity factor Demand factor

  • \$\begingroup\$ Isn't max_demand another way to say max_load? If so, max_load / connected_load should always be greater than 1 (not less), because you want the connected load to be less than the maximum. Or am I missing something with Demand factor? \$\endgroup\$ Jun 23, 2021 at 6:10
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    \$\begingroup\$ 5MW is not a "huge" load, that's for example two electric locomotives (whose power draw changes quickly and frequently). A lot of industrial equipment is in the same ballpark. "Huge" loads might be represented by things like big electric arc furnaces ("average" ones are in the higher tens of MW, "big" ones draw >100 MW). \$\endgroup\$
    – TooTea
    Jun 23, 2021 at 11:26
  • \$\begingroup\$ @PeterCordes For demand factor - if all the connected load draws power to the maximum or more than that. Then Demand factor will become greater than or equal to 1. eg: Total load connected= 10KW, then maximum demand will be practically less than 10KW \$\endgroup\$
    – raghul
    Jun 23, 2021 at 17:03

There are always small fluctuations in both frequency and voltage in the electrical grid. The size of these fluctuations depends on the power of the generators in the grid as well as the load currently present and the load being added.

Under normal circumstances 5 MW is not a big load compared to the load normally present in the grid. For example a train leaving a train station may easily exceed this. For comparison: In 2019 the power generation capacity in Germany was about 200GW with Germany being part of a huge synchronous grid covering almost the whole of Europe with a generation capacity of moren than 600GW (more information in the wikipedia: https://en.wikipedia.org/wiki/Synchronous_grid_of_Continental_Europe) - in this context 5MW is trivial and the change in frequency or voltage from that load will not be noticable.

The voltage and - actually more important - the frequency are monitored constantly and when they exceed certain thresholds generators will be added or removed from the grid or controllable loads will be added or removed.

Grid operators have many ways to adjust the load and generation in the grid. For example pumped-storage plants can provide or consume a huge power within seconds to keep the frequency stable. Once the frequency drops below a certain value rolling blackouts will occur and cities and wherever possible industrial plants will be shut down to prevent a total blackout.

Big loads of several dozen or hundreds of Megawatts are mostly controllable by the grid operators so they can stabilize the grid. Aluminium huts e.g. have a load of several hundred Megawatts and have to be scheduled with the grid operator to be turned on or even off.

You can actually see periodic drops in frequency of the Swiss train grid every hour at minute 00 since that is when a lot of trains are leaving the stations.


While other answers are excellent and to the point, there is one thing in the question that pokes me:

I know that in a power system there are no energy storage elements. Due to this, as soon as power is generated by the generators it is transmitted to the end user where the power is utilized.

There's no long term energy storage. You can't generate a kWh today, and consume it tomorrow. But there's energy storage: rotating machinery have inertia. If you slow a rotating electrical machine down from 60Hz to 59.9Hz you've consumed a bit of this energy.

With many generators in parallel, a tiny dip in frequency can supply a lot of energy. The dip in frequency can also be used by the regulator to increase the energy flow into the prime mover (water to a hydroelectric plant, fuel into a jet turbine etc).


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