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I am really stuck on this and I couldn't find any references on the internet regarding this. When a load is connected to a generator, why does the current withdrawn from the generator increases. Since the connected load will increase the overall impedance of the system, shouldn't the generator now supply less current ?

I was able to explain this in case of induction motors (When they are loaded, their effective reactance decreases and hence they draw more current), however I'm not able to explain this in cases of heaters and light bulbs.

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  • \$\begingroup\$ How much impedance does an open circuit have? \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 17 '13 at 7:58
  • \$\begingroup\$ An open circuit has infinite impedance \$\endgroup\$ – Kartik Anand Nov 17 '13 at 7:58
  • \$\begingroup\$ So then how does adding a load increase the impedance? \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 17 '13 at 8:00
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    \$\begingroup\$ Perhaps you should be glad if you don't get that job... \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 17 '13 at 8:09
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    \$\begingroup\$ If you put two subcircuits in series with a supply, the supply doesn't have two loads. It has one load whose characteristics are a composite of the characteristics of the two subcircuits. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 17 '13 at 8:17
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It's worth remembering that a generator has an active control system that works to maintain a certain output voltage.

That is, the generator is responsive to changes in the load impedance. A step increase in load power requirement will cause a temporary voltage dip, as the output power of the generator at that instant is less than the amount of power required by the load.

However, within a period of five seconds, the generator's governor and automatic voltage regulator will act to increase the generator's power output (voltage, current) until it matches the amount of power (voltage, current) required by the loads.

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An unloaded generator will generate zero current, since it is open circuit. Hence there is no load angle. The load has infinite Z = no current draw.

In the case of induction motors, the current increase due to loading, is due to increase in the slip between rotor and stator (synch speed-rotor speed).

Heaters and lightbulbs are likely to be fairly resistive, probably slightly inductive.

Adding another element in parallel with a load like this may not change the Cos Phi all that much, and if the elements resistance is already quite high, you may not see any difference in current drawn.

To be any real help, we'd need to know the phase angle of your loads.

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