text: # Calculating power of heating element
  
  1. Jul 14, 2007
  
  ### serverxeon
  
  hello all!
  
  If there is a circuit with only one heating element, how do we calculate the power it requires?
  
  According to the physics formula P=V^2/R, whenever you increase the resistance, the power it requires falls. However, I don't think that's right. Or is it?
  
  When R is increased, so the heating element will consume more power right? So if you substitute a larger value of R in the formula, the P falls. Why is it so?
  
  2. Jul 14, 2007
  
  ### ranger
  
  This is quite simple. However, I will resist the temptation to do an explicit calculation for you. But the power (lets use the term heat from now on) it requires depends on what it is heating. If you consider water, can you calculate the amount of heat (in joules) it requires to bring the water to a boil (100 Celsius degrees) from a room temp. of 20 Celsius degrees?
  
  The amount of heat produced by an electric current flowing in a conductor is given by:
  Q = I2*R*t
  where Q is heat; I is current; R is resistance of the conductor; t is time
  
  Or one can say the the resulting work done by the electric current is heat. This gives rise to a second formula:
  W = V*I*t
  where V is volts and W is work.
  No, when we increase R, we present a smaller load to the circuit. This is because it will draw less current. Think of a load with infinite resistance (open circuit); how much current does it draw?
  A large load is considered to be one that draws a significant amount of current. Think about a load with a trivial amount of resistance - like piece of wire.
  
  3. Jul 14, 2007
  
  ### serverxeon
  
  but allow me to post another approach to the question.
  
  Say I connect a heating element (Nichrome wire?) directly to a 230V power source.
  
  The live wire will be 230V, and the neutral wire will be at 0V (theoretically).
  
  Given this, P.D = 230V
  Power = IV
  Substitute I = V/R into eq
  Power = V^2/R
  
  Given V is a constant, R increase will cause a P decrease.
  
  --------------
  
  The difference in my approach is that I deal with one constant and 2 variables, so values can easily be related.
  
  Your approach is rather based on 'logic', where you are dealing with 3 variables in a single equation.
  
  Hope you get my point.
  
  So, who's right?
  
  4. Jul 14, 2007
  
  ### ranger
  
  What is it exactly you're having trouble with?
  
  Your equations are for power (watts). The ones I posted are give rise to the work done by electric current flowing in a conductor - V*I*t = Watts * sec = Joule. In this case its safe to say that this work is heat.
  
  5. Jul 14, 2007
  
  ### serverxeon
  
  If
  P = V^2/R
  Isn't it the same as
  
  Q = V^2/R * t
  
  same thing, isn't it? t is an experimental constant.
  
  6. Jul 15, 2007
  
  ### ranger
  
  If you're talking about power, you want watts, which V^2/R gives. If you want to find the about of (heat) energy expended, we need to get our units in joules. This is why we multiply watts and seconds, because it is equivalent to joules.
  
  The latter of the formula, Q, represents the about of heat energy. Watt is the measure of power - energy per second. Joule is watt-second. The two terms are not synonymous.
  
  7. Jul 15, 2007
  
  ### ank_gl
  
  is this correct??
  
  8. Jul 15, 2007
  
  ### ranger
  
  Wasn't my reply in the latter part of post #2 sufficient?
  
  9. Jul 15, 2007
  
  ### serverxeon
  
  Ok to me, I still disagree about with the problem lying between the argument of POWER and JOULES. Effectively you still can relate higher power means more heat. (relate loosely by logic)
  
  Ok, so, if given by you Q = I^2 * R * t eq,
  i assume it is okay for me to change to Q = V^2 / R * t?
  And given V and t an experimental constant,
  it does shows that with R INCREASED, Q (total energy output by heating element) drops.
  
  then why, do we know heating elements as something with high resistivity?
  It doesn't makes sense.
  
  Unless you tell me heating elements are low in resistance, nothing is going to make sense.
  
  so, anybody know where the 'paradox' lies?
  
  10. Jul 15, 2007
  
  ### ranger
  
  This goes back to my initial response in post #2. The large load is one that draws a significant amount of current. Therefore if we keep on decreasing the resistance of the element, it will draw more and more current. By virtue of ohms law:
  
  If we have a 120V source with a load resistance of 12ohms. We have a 1.2kW of power. Using a resistance 3 times the original; we get a 400W of power. Therefore, the lower resistance will produce more heat!! It does seem counterintuitive at first, but its true.
  
  Thats not all, all voltage sources have some form internal resistance. This is equivalent to a ideal voltage source with a series resistor. Now if we add a huge resistance to this circuit, almost all the voltage will be dropped across it (load) because it is significantly larger when compared to the internal resistance of the voltage source. But the current will limited which results in little heat.
  If you use a small resistance (when compared to the internal resistance of the source), there will be a larger voltage drop across the internal resistance and little current will flow; which results in little heat.
  
  It can be shown both experimentally and therotically that the maximum power will be transferred to the load when the heating element's resistance is equal to that of the load internal resistance (maximum power transfer theorem). If the internal resistance and the load are not the only components in the circuit, then maximum power transfer will occur when the load resistance is equal to that of the thevenin resistance. You can prove this by setting up a SPICE simulation and vary the
  load resistance with respect to the thevenin\internal resistance and plot the results on a graph. For this you can use SPICE-OPUS or switcherCAD. I actually did this to prove something to one of my peers, but the files are at my computer in work; I'll probably post it on Monday if you're still interested.
  
  Are you still confused?
  
  Last edited: Jul 15, 2007
  11. Jul 15, 2007
  
  ### ank_gl
  
  yea i know, i was asking xeon to rethink it, because xeon is confusing between his concepts
  
  Last edited: Jul 15, 2007
  12. Jul 15, 2007
  
  ### ank_gl
  
  hey why are u confusing so much?? heating elements are low in resistance, (thats why they are thick,remember thick wires have low resistance), so they draw more current and dissipate more heat(I^2 * R * t)
  
  13. Jul 15, 2007
  
  ### serverxeon
  
  ok, so contrary to popular belief, a heating element is actually low in resistivity?
  
  Cos, my fren and I both had this idea that heating element = high resistivity; and that wiki says that heating elements generally have relatively high electrical resistance.
  
  So, the general public's concept is wrong?
  
  14. Jul 16, 2007
  
  ### ranger
  
  serverxeon, the heating elements have a certain amount of power dissipation at the rated potential. If you were to increase the resistance of the heating element, less current will flow and there would be less heat energy. If we decrease the resistance, more current will flow and more heat energy will result.
  
  You are taking "relatively high electrical resistance" out of context. Nichrome is a worse conductor than copper, but that does not mean its resistance is anything phenomenal. When designing these heaters, its not simply about having the highest power dissipation possible, one has to obey the physical limitations of the material.
  
  15. Jul 16, 2007
  
  ### Staff: Mentor
  
  It is a misconception, yes. As ranger's post implies, "high" and "low" are meaningless terms to a scientist/engineer.
  
  16. Jul 16, 2007
  
  ### ank_gl
  
  :surprisedis general public so dumb???:grumpy::grumpy:
  
  17. Jul 16, 2007
  
  ### ank_gl
  
  can you explain which are the variables and constants in your approach???
  as i see it, V is constant, R is variable , I and P are dependent. so only R is actual variable, you can change I or P by changing R
  
  18. Jul 17, 2007
  
  ### mengshuen
  
  I'm following this argument. According to this logical argument, if there is low resistance, there will be a high power rating of the appliance. Power is defined as the rate in which electrical energy is transferred to another form.
  
  In the case of a short circuit, there is little to no resistance, hence we would achieve almost infinite power. Infinite rate of change of electrical energy to heat? True. The wire melts.
  
  Going by the definition of the watt, if 1 volt of potential difference is applied to a resistive load, and a current of 1 ampere flows, then 1 watt of power is dissipated. The units of watt is J/s. If your want to heat something, you would want a high wattage, which is, through P=IV, a high current.
  
  V=IR, R is inversely proportional to I, hence R must be small. That's why heating elements are thick, to reduce resistance. You may ask, why add the heating element at all? It needs to be there to heat up the water (or whatever). If we assume that the heating element has the least resistance in the circuit, most power is converted over there, and most (if not all) of the heating effect occurs at the heating element.
  
  Hope that is a clear explanation. Seems that the public is misled.
  
  19. Jul 17, 2007
  
  ### mengshuen
  
  If we were to view this from a different perspective,
  
  a high resistance means little current flow. By definition, current is defined as the rate of charge moving across the appliance, and having units of Cs^-1. If there is only a few Mr. Coulomb running in the circuit, little charge is delivered, and there is a little change of electrical energy to thermal energy.
  
  20. Jul 17, 2007
  
  ### serverxeon
  
  seems that the general public is misled oh well.
  time to go change wiki! =D
  
  21. Jul 17, 2007
  
  ### Staff: Mentor
  
  It is also worth noting that resistance is required for heat disspiation. No resistance, no heat dissipation (p=ir: p=i*0=0). But the fact that having resistance gives you power dissipation does not imply that increasing the resistance increases the power dissipation. In a constant voltage situation, amperage and power are both dependent on resistance:
  
  i=v/r
  p=vi
  
  The decrease in current caused by increasing the resistance is what drives the power. Graphing (for a 120V power outlet), you get these logrithmic amperage and power curves:
  
  #### Attached Files:
  
  • ###### resistor.jpg
  File size:
  11.8 KB
  Views:
  157
  22. Jul 17, 2007
  
  ### ranger
  
  Russ just reminded me that I forgot to post that graph about power dissipation as it related to an internal source resistance.
  
  This demonstrates what I was talking about earlier about maximum power transfer from [non-ideal] source to load. Maximum power transfer will happen when the load resistance is equal to the internal source resistance or thevenin resistance.
  
  Here is the SPICE netlist (complied using SPICE-OPUS):
  Code (Text):
  
  Max_ power transfer
  
  vin 1 0 dc 50
  vx 3 0 dc 0
  
  rin 1 2 200
  r1 2 3 1
  
  .control
  destroy all
  dc r1 1 600 1
  plot (i(vx)*v(2,3))
  +xlabel rL[Ohm]
  +ylabel p[Watts]
  print i(vx)
  .endc
  .end
  
  The internal source has a resistance of 200ohms. The load is varied in steps of one from 1ohm to 600ohms. Note from the graph how max. power transfer occurs when the load is 200ohms.
  
  #### Attached Files:
  
  • ###### max_pwr_thr.png
  File size:
  7.6 KB
  Views:
  148
  23. Jul 17, 2007
  
  ### Staff: Mentor
  
  Good thinking - yes, there can be no situation where the power source has zero internal resistance, so at low resistance for the load, the internal resistance of the source plays a big role.