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Joule heating of resistors

  • #1
158
7

Homework Statement


2. Homework Equations
[/B]
Consider the following circuit where:

$$R_A=330 k\Omega$$
$$R_B=220 k\Omega$$
$$C= 2.2 nF $$
$$v_1(t)=6u(t) V$$
$$ i_2(t)=-10u(t-t_2) \mu A$$
$$t_2 = 12 ms $$

NfuQiBL.png


All the currents go clockwise (with the obvious exception of the current source).

The first part of the problem asked for the equations of vc and vb. I had no trouble on getting to the correct answers so I won't post the developed answer:

$$v_C(t) = 0V$$
if $$t<0$$

$$v_C(t) = 6(1-e^{-\frac{t}{1.21 ms}}) V$$
if $$0\leq t <12 ms$$


$$v_C(t) = 2.7+3.3e^{-\frac{t-12ms}{1.21 ms}} V$$
if $$t \geq 12 ms$$

$$v_B(t) = 0V$$
if $$t<0$$

$$v_B(t) = 2.4e^{-\frac{t}{1.21 ms}} V$$
if $$0\leq t <12 ms$$


$$v_B(t) = -1.32e^{-\frac{t-12ms}{1.21 ms}} V$$
if $$t \geq 12 ms$$

Now I have question in the next part of the problem. It asks me the instant of time of the maximum dissipated power by Joule heating, and the value of that power.


3. The Attempt at a Solution


So we have that the power P

$$P= R_A (i_A)^2 + R_B (i_B)^2$$

The equations for iB are pretty straight forward. We only need to take the equations of vB and divide it by the value of the resistance RB

$$i_B(t) = 0\mu A$$
if $$t<0$$

$$i_B(t) = 10.9e^{-\frac{t}{1.21 ms}} \mu A$$
if $$0\leq t <12 ms$$

$$i_B(t) = -6e^{-\frac{t-12ms}{1.21 ms}} \mu A$$
if $$t \geq 12 ms$$

The equations of iA are less obvious.
The first case is direct:
$$i_A(t) = 0\mu A$$
if $$t<0$$

On the second case the currents through both resistors are the same (since the current source is still not working).
$$i_A(t) = 10.9e^{-\frac{t}{1.21 ms}} \mu A$$
if $$0\leq t <12 ms$$

For the third case, if we take KCL applied to the upper node:

$$ -i_A+i_B-(-10\mu A)=0$$

Using that we get to
$$i_A(t) = 10-6e^{-\frac{t-12ms}{1.21 ms}} \mu A$$
if $$t \geq 12 ms$$

Now looking to this equations and by verifying them in key points (0+, 12-, 12+, and infinity), I concluded that the maximum in both sets is in 0+ and therefore the power is (since in this case both currents are equal):

$$P= (R_a+R_B) \times (i_A^2)=65.35 \mu W $$

However my textbook chooses instead the instant 12+ and says that the power is
$$P= =92.4 \mu W $$

However if, by the results of my equations I try that instant I get to a different answer (13.2 mW).

I already verified my equations three times and I don't find the mistake. Should I be calculating Joule heating in other components (besides the 2 resistors). What am I missing?
 

Answers and Replies

  • #2
NascentOxygen
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At t=12ms the capacitor voltage is ≈ 6 volts? So I calculate power dissipated at t=12ms to be 13.2μW
 
  • #3
158
7
Exactly, as I said I got the same... I don't why and how they got the 92.4 as an answer
 
  • #4
cnh1995
Homework Helper
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Should I be calculating Joule heating in other components (besides the 2 resistors).
In steady state, the voltage source is absorbing power. So it starts absorbing power at some instant before the steady state. Although it is not actually "heating", I think you should include it in your calculations.
 
  • #5
NascentOxygen
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In steady state, the voltage source is absorbing power. So it starts absorbing power at some instant before the steady state. Although it is not actually "heating", I think you should include it in your calculations.
It may be worth investigating whether the examiner did include power absorbed by the battery in his calculations, if only to be able to rule that out as the source of error.

From wikipedia: Joule heating, also known as ohmic heating and resistive heating, ....
 
  • #6
rude man
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In steady state, the voltage source is absorbing power. So it starts absorbing power at some instant before the steady state. Although it is not actually "heating", I think you should include it in your calculations.
That's not dissipated power. Disspiated power reaches a maximum.at some time 0 < t < infinity = MAX{iA2(t)RA + iB2(t)RB}, whatever that is.
The math is pretty messy, especially with the time delay for i2. I lost patience, but it's easy to make a mistake here somewhere. Looks like the current thru RA actually reverses at some point etc.
 
  • #7
cnh1995
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That's not dissipated power.
True, but I have seen it being treated as active power in ac circuits. Say you have two ac sources V1=100∠0° and V2=50∠0° connected to each other through a resistance R.
In this circuit, voltage across V2 and current through V2 are in phase and V2 is absorbing power. It is treated as real power along with the power dissipated in R.
 
  • #8
rude man
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True, but I have seen it being treated as active power in ac circuits. Say you have two ac sources V1=100∠0° and V2=50∠0° connected to each other through a resistance R.
In this circuit, voltage across V2 and current through V2 are in phase and V2 is absorbing power. It is treated as real power along with the power dissipated in R.
"Absorbed" is not "dissipated". Quite the reverse. Problem asks for dissipated power.
 

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