AS Physics questions - Resistivity

Stuck on a couple of questions, any help much appreciated.

1. The resistance of the tungsten filament of an electric lamp is found to be 27ohms at 273K. If the resistivity of tungsten increases by 0.0056 of its resistivity at 273K for each degree rise in temp, calculat the ratio

power of lamp when first switched on / power of lamp at working temp

2. A copper wire and an iron wire have the same length and diameter. The resistivity of iron is about 8 times that of copper. They are connected to a variable power supply a) in parallel b) in series
The p.d. provided by the power supply is gradually increased from zero. Explain which wire glows first in each situation.

Thanks!

Reply 1

Jonatan

13

dinkymints

Stuck on a couple of questions, any help much appreciated.

1. The resistance of the tungsten filament of an electric lamp is found to be 27ohms at 273K. If the resistivity of tungsten increases by 0.0056 of its resistivity at 273K for each degree rise in temp, calculat the ratio

power of lamp when first switched on / power of lamp at working temp

Not sufficient information. You need to know how quickly heat dissipates, or better yet, the working temperature of the lamp.

dinkymints

2. A copper wire and an iron wire have the same length and diameter. The resistivity of iron is about 8 times that of copper. They are connected to a variable power supply a) in parallel b) in series
The p.d. provided by the power supply is gradually increased from zero. Explain which wire glows first in each situation.

Thanks!

Treat them as resistors, assume one has resistivity 1 Ohm and teh other has 8 Ohm. Then calculate the current through, and voltage across, each wire. Power , P, equals current times voltage P = IV. The wire with the greatest power (due to the current through it) will heat up quicker and thus glow earlier since power is energy/time(actually you also need to take into consideration the density and specific heat capacity difference, but I assume that these factors are negliable as compared to the difference in resistivity.)

Reply 2

dinkymints

OP

12

Jonatan

Not sufficient information. You need to know how quickly heat dissipates, or better yet, the working temperature of the lamp.

Sorry, so stupid, the working temp is 2400K

Treat them as resistors, assume one has resistivity 1 Ohm and teh other has 8 Ohm. Then calculate the current through, and voltage across, each wire. Power , P, equals current times voltage P = IV. The wire with the greatest power (due to the current through it) will heat up quicker and thus glow earlier since power is energy/time(actually you also need to take into consideration the density and specific heat capacity difference, but I assume that these factors are negliable as compared to the difference in resistivity.)

Do you mean resistance 1Ohm and 8Ohm?
Could I just calculate resistances in parallel and series? Would the wire with the greatest resistance glow first or something? I have no voltage or current values. (This q has a * next to it so is meant to be hard...)

Thanks

Reply 3

Jonatan

13

dinkymints

Sorry, so stupid, the working temp is 2400K

Do you mean resistance 1Ohm and 8Ohm?
Could I just calculate resistances in parallel and series? Would the wire with the greatest resistance glow first or something? I have no voltage or current values. (This q has a * next to it so is meant to be hard...)

Thanks

Since you seem to find it difficult I can give you the solution (Just don't coppy this straight of teh internet. Your teachers will be able to tell that all my typoes are nto teh same as the ones you usually make, and more crucially they will notice that it is written in a different style. Furthermore, that woant help you on teh exam as you do not learn how to do it. I strongly recomend that you try to understand this, and if you don't you ask someone who is good at explaining stuff (not necessarily a teacher, you probably have one or two nerds in your class who are eager to help you (such as me))) Uff, many nested parantheses.

Heres the solution (algebraicly): the time it takes for a wire to light up depends on how quickly it gets hot. This in turn depends on two things. The rate at which heat is dissipated into the wire, and teh heat capacity of the wire (greater heat capacity means more heat must be suplied to raise the temperature. Hence it takes longer to boil 4 litres of water than 1 dl.) Im going to argue that the difference in heat capacity is small, and negliable (small in this case means that its small as compared to the difference between the rates at which heat is dissipated. Negliable means that it woant affect my conclusion).

So far there is no maths here, Well here it comes (In case you don't liek maths. Im sorry, but you have to get used to it. Maths is the language of physics and if you do not understand it Physics is going to be very difficult.)

The voltage V, relates to the current I and resistance R according to Ohms law:

V = IR.

Lets now considder the two wires in paralell. Because they are conected in paralell there is teh same voltage across them. Lets name the resistance of teh copper wire R, then the resistance of the iron wire is 8 * R. Since the voltage across teh two wires is the same (they are connected in paralell) the current through each wire will be:

Copper:
Ic = V / R

Ii = V / 8R

(Note that "Ic" is simply my way of writing "current in the copper wire" It does not mean that I should be multiplied with c or anything like that Just think of "Ic" as one symbol. Teh same goes for Ii and iron )

Now, the rate at which heat is dissipated in teh wires is given by the power.
P = IV (Do you see why? Hint: 'Energy is conserved' )

So the rate at which heat is dissipated in the copper, Pc, and teh rate at which energy is dissipated in iron , Pi, is simply:

Pc = Ic * V = (V / R) * V = V²/R
Pi = Ii * V = (V / 8R) * V = V²/8R

Now, V²/R is obviously larger than V²/8R Thus Pc > Pi and thus heat is dissipated in the copper at a faster rate than in teh iron, therefore the copper will light up quicker, assuming that the difference in heat capacity is small as compared to teh difference in resistance (Since the resistance in iron is 8 times as large as the resistance in copper, this is a fairly safe assumption to make. )

To work it out when the wires are serially connected you cannot use teh voltage, because here you do not know anything about the voltages across teh resistors. However, when two wires are serially connected, all current passing through the first wire will also pass through teh second one. Therefore, teh CURRENT , I , will be the same. Hence we can find the voltage in terms of teh resistance:

Vc = I * R
Vi = I * 8R

Once again teh power is given as teh product between voltage and current, and thus:

Pc = Vc * I = (I * R) * I = I² * R
Pi = Vi * I = (I * 8R) * I = I² * 8R

Clearly I² * 8R is larger than I² * R and hence Pi > Pc.

Therefore the IRON ill light up first when the wires are connected serially.

Thsi result can also be extended to lightbulbs. Where the lightbulb that shiones the brightest when two lightbulbs are conected in parallell will be the dimmest when the same light bulbs are connected in a serial circuit (Thus if you connect a 60W lightbulb in series with a 40W lightbulb, the 40W lightbulb will glow brighter. The Power values on lightbulbs are valid for parallell conection to teh net voltage only!!! )

Reply 4

dinkymints

OP

12

Thanks so much. I do actually understand this now, just wasn't too sure about how to go about it.

Still not sure about the first q I posted though, if anyone has any ideas?

Reply 5

Jonatan

13

dinkymints

Thanks so much. I do actually understand this now, just wasn't too sure about how to go about it.

Still not sure about the first q I posted though, if anyone has any ideas?

The only reason it is difficult is because the information is given in such a weird form. It sais that it increases with 0.0057 (or smth) times the resistivity at 237K. Well, then we simply have to find thsi ammount by taking the resistance at 237K and multiplying it with 0.0057. The number you get is then the number of Ohm the resistance increases with for every 1K you add.

Simply subtract 237K from the working temperature, multiply this with the original resistivity and multiply teh answer of that with 0.0057. Then you get the Increase in resistivity. Add this to the original resistivity and you have the resistivity at working temperature. Then you divide with teh original resistivity to get the ratio.

Very ankward, but rather straight forward. This is mainly about converting the information you are given from one form into another. Its like when you are asked to find the halflife of a radioactive compound, and rather than giving you the original activity and the activity after time t in propper Si units, they give it in something weird like the mass, molar weight and radiation intensity in Curie. I just hate those imperical questions... Long live dimensionless quantities like the radian!

AS Physics questions - Resistivity

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