Homework Help: Draw circuit, % increase in the resistance

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1. Oct 7, 2016

moenste

1. The problem statement, all variables and given/known data
A 12 V, 24 W lamp and a resistor of fixed value are connected in some way inside a box with two external terminals. In order to discover the circuit arrangement inside the box a student connects a variable DC power source and an ammeter in series with the box and obtains the folliwing results:

Applied potential difference and current 1: 1.0 V and 1.17 A;
Applied potential difference and current 2: 12 V and 4.00 A.

(a) Draw a circuit diagram of the most likely arrangement inside the box, giving your reason.
(b) Use the 12 V, 4.00 A reading to deduce the value of the fixed resistor.
(c) What is the percentage increase in the resistance of the lamp as the applied potential difference changes from 1.0 V to 12 V?
(d) When the applied potential difference is increased to 24 V the current is again found to be 4.00 A. Explain this observation.

(You may neglect the internal resistance of the power supply and the resistance of the ammeter in all your calculations.)

Answers: (b) 6.0 Ω, (c) 500 %.

2. The attempt at a solution
(a) I would guess that they are connected in parallel in the box. But I am struggling to understand the logic of the whole circuit. If they are connected in parallel then they both have 12 V. How can they have 12 V (the lamp and the resistor) if the applied PD is 1 V?

Also if we calculate the current of the lamp: P = V I so ILamp = 24 / 12 = 2 A. How can current of the lamp be 2 A when it is 1.17 A in total?

But in general I would say that the circuit has a battery and the ammeter in series and the lamp and the resistor are in parallel in the box.

(b) RTotal = 12 / 4 = 3 Ω. ILamp = P / V = 24 / 12 = 2 A. RLamp = V / ILamp = 12 / 2 = 6 Ω. 1 / RTotal = 1 / RLamp + 1 / RResistor → 1 / 3 = 1 / 6 + 1 / RResistor → RResistor = 6 Ω.

(c) Shouldn't the resistance of the lamp stay the same? ILamp = P / V = 24 / 12 = 2 A. RLamp = V / ILamp = 12 / 2 = 6 Ω.

(d) Didn't start this one, since I'm not sure on the first three.

2. Oct 7, 2016

.Scott

First of all, both of your answers (b and c) are correct.
Your guess at "a" is also correct, they are connected in parallel.
The fact that the lamp is a 24-Watt, 12V lamp simply means that at 12V, the lamp will burn 24 Watts. It doesn't mean that there is always 12V across it.

The resistance of the lamp does not stay the same. It is refered to as a "lamp load" (appropriately enough). You see this with regular incandescent lamps (not LEDs ot Flourescents).
What happens with a lamp load is that as the lamps filament heats up, it becomes more resistive. This is important. It provides an automatic heat regulation. For example, if the opposite was true, becoming less resistive, as it heated up it would draw more current and heat up further - leading to thermal runaway.

3. Oct 7, 2016

moenste

Why the answer is 500 % in (c)?

4. Oct 7, 2016

.Scott

At 12V, the lamp is burning 24 watts. So it's current is 24W/12V = 2A. It's resistance will be 12V/2A = 6ohms.
Since the total current at 12V is 4A, the fixed resistor must also be 2A (2A+2A=4A). So the fixed resistor is 12V/2A = 6ohms.

At 1V, the fixed resistor will draw 1V/6ohms = 0.17A. That leaves 1.00A that the lamp must be drawing.
So the resistivity of the lamp at 1V it 1V/1A=1ohm. So the lamps resistance is increasing from 1ohm to 6ohms, a 500% increase.

Now you need to calculate the current being carried by the fix resistor at 24V to answer question d.

5. Oct 7, 2016

moenste

Clear.

I actually got these numbers but I didn't write them here. My doubt was -- isn't it a 600 % increase? I mean 1 * 600 % = 6 Ohm.

I = 24 / 6 = 4 A. But that's the current flowing through the resistor. Not through the whole circuit.

And any ideas on the (a) part please?

6. Oct 7, 2016

.Scott

Let's start at 1000. A 1% increase would by an increase of 10, yielding 1010. A 100% increase would be an increase of 1000, yielding 2000.
A six-fold increase means you are multiplying by 6. A 500% increase means that you are adding 500% of the original.

7. Oct 7, 2016

moenste

I also considered that definition. If we take 1 as 100 % and 6 as 1 + 5, then it's indeed a 500 % increase in terms of the original value 1.

What about (a) and (d) then?

8. Oct 7, 2016

.Scott

You've already said that you are applying the same 1, 12, or 24Volts to both the lamp and the fixed resistor. If you had deduced that you were applying the same current to them, you would be calling it a series circuit. So tell me what the answer to "a" is.

Tell me, at 24V, what portion of the 4A is going through the fixed resistor and what remaining portion of it is going through the lamp?

9. Oct 7, 2016

moenste

I think the lamp and the resistor should be in parallel. But I don't understand:
If the total current is 4 A and if we found that 4 A is the current of the resistor, then 0 A is going through the lamp.

10. Oct 7, 2016

.Scott

I'm not sure what part you don't understand. These two devices are in the box, the box has two wires coming out of it, those two devices are each connected to those wires and are connected in parallel. So whatever voltage you apply to one is also applied to the other. In the experiment, three voltages are applied. First 1 volt and then the total resistance is measured. Then 12 volts and then the total resistance is measured. Then 24 volts and the total resistance is measured.

That is correct. So when 24 volts is applied to the 12-volt-24-watt lamp, there is no current. Now tell my why there is no current. Think in realistic terms.

11. Oct 7, 2016

moenste

"A 12 V, 24 W lamp". So I assumed that the lamp ALWAYS has 12 V. Could you explain what the "12 V, 24 W" lamp means?

Maybe because we have a 12 V lamp? I mean if the V = 1 V or V = 12 V the lamp can work (somehow at V = 1 V, but still can work). And if we have 24 V then the lamp will just burn out, it can't handle so much energy through it? And so the resistor takes the energy and no current goes through the lamp.

12. Oct 7, 2016

.Scott

It's the rating of the lamp. If you go to the store you can by a 60W 120V light bulb. It doesn't mean that the light bulb always has 120volts on it. It means you are supposed to put 120 volts on it to use it.
Very good. That is correct. The lamp has burned out and is now an "open" circuit.

13. Oct 7, 2016

moenste

Alright, that was my confusion. If I got it correctly, then we can just rephrase the problem with "We have a lamp that should be used with 12 V and 24 W..." and the idea doesn't change. And then in the problem we give three different voltages through the circuit (1, 12 and 24 V) and discuss the results of the experiments.

14. Oct 7, 2016

.Scott

To be precise: It should be used with 12V, and when it is, it will draw 24W. (but not that bulb that was tested with 24V)