Help with resistor power dissipation homework

In summary, the question asks for the average power dissipated on a resistor in a circuit with a sinusoidal voltage source, diodes, and a transformer. After some discussion and calculations, it is determined that the voltage across the resistor can be represented by 10*sin(100pi*t) and the average power can be calculated by finding the area under the curve of this voltage waveform. The final solution also takes into account a voltage drop across the diodes.
  • #1
This question is in reference to the diagram here:
http://imageshack.us/f/17/ecediagram.jpg/

Find the average power dissipated on the resistor.
Assume Vth for the diodes is 1V, N1 = 1000, N2 = 100, the transformer is ideal (non-lossy), and V(t) is a purely sinusoidal voltage with frequency of 50 Hz and amplitude of 100V.

I've tried working it out several times, but I can't figure out exactly what to integrate. Thanks for your help!
 
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  • #2
Hi thepcphysicia, welcome to PF.

You'll need to show some attempt at the problem before we can give you any help. Have you determined the waveform of the voltage (or current) that will appear across R1?
 
  • #3
Hi,

Here is my work thus far:
http://imageshack.us/photo/my-images/202/photo2la.jpg/

The waveform of the voltage is given by 10*sin(100pi*t). I think that it is squared, because P=V^2/R.

The waveform is also shifted down two due to the voltage drop across the diodes.

Not sure about the current.
 
  • #4
Anybody? The assignment is due in a few hours...
 
  • #5
You might want to consider the period of the sinewave in terms of angle rather than time, and do the integration symbolically rather than carrying all those decimals around and fiddling with the milliseconds and their fractions. Then the period of a cycle is just [itex] 1/(2\pi) [/itex], and you can call the "turn-on angle" for the the diodes [itex] \phi [/itex], given by:
[tex]10V sin(\phi) = 2V[/tex]

attachment.php?attachmentid=41025&stc=1&d=1321630094.jpg


The voltage that appears across the resistor R1 is shown in the diagram. In your proposed solution you made it [itex] (10 sin(stuff))^2 - 2) [/itex], but that's not quite right; The 2V needs to be taken from the sin() before squaring occurs.
 

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  • #6
Awesome, thank you SOOO much! The diagram was incredibly helpful!
 

1. How do I calculate the power dissipation of a resistor?

To calculate the power dissipation of a resistor, you can use the formula P = V^2 / R, where P is power in watts, V is voltage in volts, and R is resistance in ohms.

2. What factors affect the power dissipation of a resistor?

The power dissipation of a resistor is affected by its resistance value, the voltage applied across it, and the ambient temperature.

3. Can I use the same formula to calculate power dissipation for all types of resistors?

Yes, the formula P = V^2 / R can be used for all types of resistors, as long as the voltage and resistance values are known.

4. How does power dissipation affect the temperature of a resistor?

As power dissipation increases, the temperature of a resistor also increases. This is due to the Joule heating effect, where the energy dissipated by the resistor is converted into heat.

5. Are there any safety concerns with high power dissipation in resistors?

Yes, high power dissipation can lead to overheating and potentially cause damage to the resistor or surrounding components. It is important to choose a resistor with a power rating that can handle the expected power dissipation in the circuit.

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