Calculating Power Delivery of a DC Generator with Varying Load and Speed

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Discussion Overview

The discussion revolves around calculating the power delivered by a DC generator under varying load and speed conditions. Participants explore the relationship between generator speed, open-circuit voltage, and power delivery to a resistive load, addressing both theoretical and practical aspects of the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a problem involving a DC generator with specified parameters, seeking assistance in calculating power delivery at a higher speed.
  • Another participant questions whether the open-circuit voltage of the generator varies linearly with speed, suggesting that this assumption is necessary for further calculations.
  • A participant expresses confusion about the initial steps, mistakenly calculating armature current without considering the open-circuit condition.
  • Clarification is provided that the current through the armature is zero when open-circuit, and the correct approach involves finding the open-circuit voltage at the new speed before calculating current and power.
  • Concerns are raised about the lack of a specified field current, which is necessary to determine the open-circuit voltage at the new speed.
  • One participant proposes a calculation for voltage at the new speed, followed by deriving current and power, which is met with positive feedback from another participant.

Areas of Agreement / Disagreement

Participants do not reach a consensus on how to handle the missing field current in determining the open-circuit voltage, and there are differing views on the assumptions needed for the calculations.

Contextual Notes

The discussion highlights the dependence on assumptions regarding the linearity of voltage with speed and the unspecified field current, which complicates the calculations.

cabellos
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DC generator question please help...

Please could I get some help with the following question:

An open circuit voltage of a d.c. generator driven at 1800rpm with rated field current is 85Volts. The resistance of the armature is 4.3 ohms. If a resistive load of 50 ohms is connected to the armature terminals, what will be the power delivered to this load when the machine is driven at 2400rpm?
 
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This question requires some assumptions. Does the open-circuit DC voltage of the generator vary linearly with its speed?

If so, you just need to solve for the voltage of the generator at 2400 rpm, then find the current through the loop (50 ohms + 4.3 ohms = 54.3 ohms), then find the power dissipated in the load resistor (P = I2R).

- Warren
 
Thanks but I am a bit lost with your explanation. The first step i thought i had to take was to find the armature current. simply using V=IR? is this correct? so it would be 85/4.3 = 19.77A ... but i really am lost as to the next step...
 
If the armature is not connected to anything (it's "open circuit"), then the current through it is zero, not 19.77 A.

What you want to do first is find the open circuit voltage of the generator at speed.

Next, find out how much current would flow when the load resistor is connected. The completed circuit has two resistors in it -- one's 50 ohms and the other is 4.3 ohms.

Next, find out how much power is dissipated by the load resistor alone. The power dissipated by a resistor is given by P = I2R.

Follow these steps exactly. I can't really give you any more hints; I'd just be doing the problem for you.

- Warren
 
thanks but i have one final question and need another push in the right direction if possible. Iv just been looking through some similar examples and to me it seems you are required to know the field current to find the open circuit voltage of the generator at the new speed. In my question no field current value is given?
 
All you need to analyze a DC loop composed of resistors is the voltage and the resistances.

I honestly do not know how you're supposed to find the voltage at the new speed -- as I said in my first post, there problem has some assumptions. I assume that your textbook (or teacher) can help you further.

- Warren
 
I think iv cracked it...does this sound right...
Voltage = 85 x (2400/1800) = 113.33 Volts
Now V=IR 113.33/54.3 = 2.09 Amps
And finally P = (2.09^2)50
= 217.81 Watts
 
Looks right, cabellos!

- Warren
 

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