Consider the following problem:

Robby wants to put a light in the shed so he puts a cable between his house and the shed. In the cable there are 2 wires with a combined resistance of $0,92 \space \Omega$. Robby connects his transformator at his house on an electricity grid of $230 V$. He connects the secondary inductor($14.3V$) to the cable. A $12 \space V$ / $30 \space W$ lamp in the shed works perfectly.

There are a couple of questions about this piece of text, which have been bothering me:

  • Calculate the heat per second which is created in the cable.

My problem with questions like this is that I have an overwhelming amount of data and it is hard for me to distinguish which is needed and which is not. So my initial thought was that I'd calculate $I$ like this:

$ I = 230 : 0,92 = 250 \space A$, and then use the formula $P = I^2 \times R$ which would give me a ridiculous amount of $57500 \space J/s$. I knew I did something wrong when I saw this, so I looked at the correction model and found out that they found the $I$ of the lamp, and afterwards used that in he formula I gave in combination with $R= 0,92 \Omega$. Why is it done like this? I first thought that you could you $I_{lamp}$ because this is in series and $I$ is the same everywhere in series, but I have 2 problems with that idea:

  • How do we know it is in series?

  • Also, if we do $ I = 230 : 0,92 = 250 \space A$ this isn't equal to $I_{lamp}$, so I dismissed that idea and now I am back at square one..

Then we get the following passage which confused me even more:

Robby decides the put the transformator in the shed. The lamp functions perfectly when he puts the loop ratio to 958:50.

What?! Does this mean that he didn't use the transformator in the first passage? But if he didn't why is the secondary inductor mentioned in the first passage then?!

The following 2 questions are easier (if you made the first one correctly):

  • Calculate the primary voltage on the transformator

I just did $(958/50) \times 12 \approx 230V$

  • Explain without using calculations that he loses much less now

$I_{secondary}$ is much smaller now, and because of the formula $P=I^2 \times R$ this leads to less loss of energy.

So the bulk of my problem lies with 'analyzing' the first text. How can you conclude from that test that the transformator isn't used?

  • $\begingroup$ If anyone is interested in the copious amounts of homework questions I post on this site, look at my profile. If anyone has any suggestions on how to improve my homework questions I'd gladly hear them. I try to abide by the rules as much as possible because this site is my last resort basically $\endgroup$ – Ylyk Coitus Jan 22 '13 at 20:21

Look at the voltage and resistance in the secondary circuit.

The driving voltage is max 14.3 volts, on a sinusoidal curve. (We can assume the resistance in the secondary winding is small enough to ignore.) Since power is a function of $V^2$, the equivalent DC voltage is $14.3 \sqrt{2}/2 = 10.04V$.

The outgoing wire and returning wire each has resistance 0.46 ohms.

We need to know the resistance Rl of the lamp. It's power is $30W = IV = (V/Rl)V = V^2/Rl = 10.04^2/Rl$. So solve for $Rl$ and you get $Rl = 10.04^2 / 30$.

So the total resistance is $R_{tot} = Rl + 0.46\times 2$.

You take it from there :-)

| cite | improve this answer | |
  • $\begingroup$ How do you know for sure 14.3 is max? Wouldn't it be more logical if 14.3 is the rms? $\endgroup$ – Ylyk Coitus Jan 22 '13 at 22:01
  • 1
    $\begingroup$ @Ylyk: You'd have to read the placard on the transformer. I think it's typical to be rated in terms of peak voltage, not rms. $\endgroup$ – Mike Dunlavey Jan 22 '13 at 22:39
  • 1
    $\begingroup$ Oh, I didn't know that. But from the previous questions I made in this book I learned that they always use rms here. Also, I don't see how that would matter (it seems you are proposing an alternate way of doing it? $\endgroup$ – Ylyk Coitus Jan 22 '13 at 22:43
  • $\begingroup$ @Ylyk: Well, it's got to matter, because a factor of 1.4 in voltage is a factor of 2 in power, so you could burn things out if you misunderstood it. $\endgroup$ – Mike Dunlavey Jan 22 '13 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.