Electrical energy converted to heat in coil

In summary: Power is not the same as total energy. Power is the instantaneous energy dissipation with respect to time. How might you find the total energy from the equation?
  • #1
Erik P
33
0

Homework Statement


Given a coil with N turns, a radius of r, a resistance of R', and an induced current i'(t) running through the coil, determine the totale energy converted to heat in the coil for t>=0.EDIT:
BnZkW10.png

The circuit on figure 4 consists of a resistance R and a capacitor with capacitance C. The voltage over the capacitor is V0 for the time t=<0 with the shown polarity. A small coil with N turns and a radius r is placed in the same plan as the large circuit, at a distance d from the wire ab. The coils resistance is R'. At the time t = 0 the switch is turned on. Both circuits are stationary and we assume that ONLY the straight wire ab, which is closest to the coil, generates a magnetic field. We can ignore the self-inductance from both circuits.

a.) Determine the current i(t) through the large circuit at the time t>=0.

\begin{equation}
i(t) = \frac{V_0}{R}e^{-\frac{t}{RC}}
\end{equation}

The wire ab is considered infinitly long and the magnetic field at the coil is assumed to be uniform (d>> r).

b.) Determine the direction and size of the induced current i'(t) in the coil for the time t>= 0.

\begin{equation}
i'(t) = \frac{r^2N\mu_0V_0}{2dR^2CR'}e^{-\frac{t}{RC}}
\end{equation}

c.) Calculate the total energy converted to heat in the coil for t>=0.

this is where the problem arises. The answer should be:

\begin{equation}
U = \frac{\mu_0^2N^2V_0^2r^4}{8R'R^3d^2C}
\end{equation}

Homework Equations


I found the following, but it is for energy stored in a toroidal solenoid so I don't think it is correct:

\begin{equation}
U = \frac{1}{2}Li'(t)^2 = \frac{1}{2}\frac{\mu_0N^2A}{2\pi r}i'(t)^2
\end{equation}

The Attempt at a Solution


Using the above equation:

\begin{equation}
U = \frac{\mu_0N^2r}{4}i'(t)^2
\end{equation}

But according to the answer this isn't correct.
 
Last edited:
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  • #2
Any formula for energy will have to involve time because the longer the duration that current flows, the more energy will be dissipated.

Is there a possibility it may be a trick question? You have a current i'(t) flowing in a resistance R'.
 
  • #3
NascentOxygen said:
Is there a possibility it may be a trick question?

You have a current i'(t) flowing in a resistance R'.
Well actually that formulation is my fault. This is the last problem of a 3 part problem, in the first two parts I found the current i'(t) that would be induced in a coil with a resistance of R', the last part simply asks what the total energy to heat conversion would be in the coil as a function of time.
 
  • #4
If current is i'(t), what exactly is capital I?
 
  • #5
NascentOxygen said:
If current is i'(t), what exactly is capital I?
Sorry, I should have been written as i'(t), will fix it.
 
  • #6
Perhaps remember that an ideal inductor is lossless. So no energy is converted to heat in the ideal part of a real inductor. What about losses in the other parts of a real inductor?
 
  • #7
CWatters said:
Perhaps remember that an ideal inductor is lossless. So no energy is converted to heat in the ideal part of a real inductor. What about losses in the other parts of a real inductor?
Hmm... I think my explanation is causing more confusion than anything. Give me a few minutes to write the full original text in.
 
  • #8
CWatters said:
Perhaps remember that an ideal inductor is lossless. So no energy is converted to heat in the ideal part of a real inductor. What about losses in the other parts of a real inductor?
I updated the question so all the details are there.
 
  • #9
Your equation 4 is for the energy stored in an inductor. Not the energy dissipated in the inductor.

Consider a resistor R with current I(t) flowing through it. What is the simple equation for the instantaneous power dissipated in the resistor.
 
  • #10
CWatters said:
Your equation 4 is for the energy stored in an inductor. Not the energy dissipated in the inductor.

Consider a resistor R with current I(t) flowing through it. What is the simple equation for the instantaneous power dissipated in the resistor.
Hey, thanks for replying. Was away when you sent that. The instantaneous power dissipated in the resistor is the product of the square of the current and the resistance over which the current runs?

\begin{equation}
P = I^2R \Rightarrow P' = i'(t)^2R'
\end{equation}

However this doesn't give the same result as the answer, this gives:

\begin{equation}
P' = \frac{r^4N^2\mu_0^2V_0^2}{4d^2R^4C^2R'}e^{\frac{t^2}{R^2C^2}}
\end{equation}
 
  • #11
Erik P said:
However this doesn't give the same result as the answer, this gives:

\begin{equation}
P' = \frac{r^4N^2\mu_0^2V_0^2}{4d^2R^4C^2R'}e^{\frac{t^2}{R^2C^2}}
\end{equation}
Power is not the same as total energy. Power is the instantaneous energy dissipation with respect to time. How might you find the total energy from the power?
 
  • #12
gneill said:
Power is not the same as total energy. Power is the instantaneous energy dissipation with respect to time. How might you find the total energy from the power?
Intergrating over the power from 0 to t, at which point I get the same equation :) well sort of :( The problem is the integration of e^(x^(2)), I looked it up and it seems to involve 1/2 sqrt(pi) erf which isn't something I recall having had about in calculus. The answer sheet gives the equation without the sqrt(pi) erf part but otherwise seems to be the same as you can see from the OP. Am I making a mistake here?
 
  • #13
I think you'll find that the exponent of e in your equations should be negative. Also, note that ##(e^{-x})^2 \ne e^{(-x)^2}##. So you'll want to check the exponent of e in your power equation for two reasons.
 
  • #14
gneill said:
I think you'll find that the exponent of e in your equations should be negative. Also, note that ##(e^{-x})^2 \ne e^{(-x)^2}##. So you'll want to check the exponent of e in your power equation for two reasons.
I ran it through wolfram alpha to check, and surely you were right about the exponent. I am however a little confused now because I get the following, keep in mind the i had to change the name of some things, like R' is now R_0, V_0 is now v etc.
alw7DhA.png

as you can see I now have a negative value and e is still present, which it shouldn't be.
 
  • #15
You've taken an indefinite integral, so you've ended up with an unknown constant. What are the integration limits for the time interval of interest?
 
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Likes Erik P
  • #16
gneill said:
You've taken an indefinite integral, so you've ended up with an unknown constant. What are the integration limits for the time interval of interest?
I see the problem :D its from 0 to infinity, not 0 to t. Also the t isn't multiplied on the fraction, its only present as an exponent, so t=0 doesn't make the entire thing 0 it just makes the e part equal to 1 and for t = infinity we get e expression becomes 0, which means we have 0 - (above expression without the e part), so double negatives cancel out. Sorry, I guess I'm just tired... Should have caught that mistake.
 

1. What is the process of converting electrical energy to heat in a coil?

Electrical energy is converted to heat in a coil through a phenomenon called Joule heating. This occurs when an electric current passes through a conductor, such as a coil, and encounters resistance. The resistance causes the electrical energy to be converted into heat energy, which increases the temperature of the coil.

2. How is the amount of heat produced in a coil determined?

The amount of heat produced in a coil is determined by the current passing through the coil, the resistance of the coil, and the time it takes for the current to pass through the coil. This can be calculated using the equation Q = I^2RT, where Q is the amount of heat produced, I is the current, R is the resistance, and T is the time.

3. Is there a limit to how much electrical energy can be converted to heat in a coil?

Yes, there is a limit to how much electrical energy can be converted to heat in a coil. This limit is determined by the maximum temperature that the coil can withstand before it becomes damaged. If too much electrical energy is converted to heat, the coil can overheat and potentially melt or burn out.

4. How is the heat generated in a coil used in practical applications?

The heat generated in a coil can be used in a variety of practical applications, such as in electric stoves, water heaters, and electric blankets. It can also be used in industrial processes such as soldering, welding, and heat treating.

5. Can the process of converting electrical energy to heat in a coil be reversed?

Yes, the process of converting electrical energy to heat in a coil can be reversed. This is known as the Seebeck effect, where a temperature difference between two ends of a conductor can create an electric current. This is the principle behind thermoelectric generators, which can convert waste heat into usable electricity.

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