Thread: Physics (electricity) Help

Hi Guys,

I've had a physics worksheet to do and there's been various questions that I couldn't answer so I was wondering if you guys out there could help me. I've got the answers as they are supplied but I'm just not sure on how they work and some of the formulas that comes with them. Hope you guys can explain it to me thanks. Ok here we go;

8. A cell of 2V and negligible resistance is connected to two resistors in series. The current through the cell is 0.2A. If one resistor consists of two resistance coils of 5Ohms and 10Ohms respectively arranged in parallel. Find the;

a) the other resistance.

Well I figured out this question. We use R=V/I, 2/0.2 = 10Ohms
Then we figuredout the total resistance in the parallel circuit and it worked out to be 10/3Ohmd. So we subtracted 10-(10/3) and we have the resistance of the resistor in the serious circuit not the parallel one.

b) The potential difference across the 5Ohm coil. <<<<<Didn;t understand.


9. A resistor of 10Ohms is connected in series with two other resistors. (6 ohms and 30ohms) that are in parallel, If a current of 3A flows from the battery find.

a) the total resistance

1/Rt=1/30+5/30=6/30=30/6=5.Ohms Since in series we just add to find the total resistance 5+10= 15 Ohms


b) the emf of the battery (could somone explain what emf is?)

3x15=45V <<<<<<wasn't quite sure how i did it I just guesses

c) the current through each resistor <<<didn't get this one


10. The resisivity of constantan is 4.9x10&-7Ohm Metres<(idono). What resistance of a long piece of constantan wire 1m long and a radius of 0.1mm.

Completley lost here.


13. A 100W, 250V lamp is used on a 200V supply.

a) What will be the current flowing through the lamp <<<didn't get


b) Explain why it is more economical to use a lamp labeled 250V but it is used on 200V mains. <<<also didn't get


Hopeu can help guys thanks =]

8. a) Find the total resistance of the resistors in parallel by

Then find the total resistance in the circuit using and subtract.

b) The PD across the coil means the voltage which acts on it. You'll need to find the current across it first, then calculate the PD from this by

9. emf is the electromotive force; the voltage applied by the battery basically.

Treating it as voltage, in this case you would calculate it by

The current through each resistor refers to the fact that, as they are in parallel, the current 'splits'. The voltage is constant , as they are in parallel, so you can calculate it by .

10. There is a formula for resistivity;

Rearranging this gives us

Where R is resistance, p is resistivity, A is cross-sectional area and l is length.

13.


Pick whichever.

Originally Posted by giga502

Hi Guys,

I've had a physics worksheet to do and there's been various questions that I couldn't answer so I was wondering if you guys out there could help me. I've got the answers as they are supplied but I'm just not sure on how they work and some of the formulas that comes with them. Hope you guys can explain it to me thanks. Ok here we go;

8. A cell of 2V and negligible resistance is connected to two resistors in series. The current through the cell is 0.2A. If one resistor consists of two resistance coils of 5Ohms and 10Ohms respectively arranged in parallel. Find the;

a) the other resistance.

Well I figured out this question. We use R=V/I, 2/0.2 = 10Ohms
Then we figuredout the total resistance in the parallel circuit and it worked out to be 10/3Ohmd. So we subtracted 10-(10/3) and we have the resistance of the resistor in the serious circuit not the parallel one.

Correct.

b) The potential difference across the 5Ohm coil. <<<<<Didn;t understand.

Potential difference is just another name for the voltage. The 5 ohm coil is part of the parallel combination which you already calculated to be 10/3 ohms and you calculated the total ohms in the circuit to be 10. Now you can use the voltage divider equation to calculate the voltage as 2volts*(10/3)/10=2/3 volt

9. A resistor of 10Ohms is connected in series with two other resistors. (6 ohms and 30ohms) that are in parallel, If a current of 3A flows from the battery find.

a) the total resistance

1/Rt=1/30+5/30=6/30=30/6=5.Ohms Since in series we just add to find the total resistance 5+10= 15 Ohms

Correct.

b) the emf of the battery (could somone explain what emf is?)

3x15=45V <<<<<<wasn't quite sure how i did it I just guesses

It's the voltage. You guessed right. 3 amps times 15 ohms equals 45 volts.

c) the current through each resistor <<<didn't get this one

Well, you know the total current, 3 amps, and you know each resistor 6 ohms and 30 ohms. The current will split through the resistors in inverse proportion to the resistance. So the 6 ohm resistor will get 5 times the current of the 30 ohm. You can take it from there.
Or, another way to do it would be, starting with the 45 volt battery voltage, the current through the 10 ohm resistor is 3 amps, so the voltage drop through the 10 ohm resistor is 30 leaving 45-30=15 volts across the other 2 resistors. Now calculate the current through each.

10. The resisivity of constantan is 4.9x10&-7Ohm Metres<(idono). What resistance of a long piece of constantan wire 1m long and a radius of 0.1mm.

Completley lost here.

First you have to calculate the cross-sectional area of the wire (remember pi*r-squared) which will give you a number in square millimeters and you will have to convert it to square meters to get everything into the same units of measure. The resistance will be equal to the length multiplied by the resistivity divided by the cross-sectional area.

13. A 100W, 250V lamp is used on a 200V supply.

a) What will be the current flowing through the lamp <<<didn't get

If you assume a constant resistance, which isn't quite true because the resistance will go up (edited) as the lamp gets hotter, then you could calculate the resistance of the lamp at 250 v, which gives you 100W. Power =V^2/R. R=V^2/power=250^2/100. Then solve for the new wattage based on 200 volts and the resistance you just calculated.

b) Explain why it is more economical to use a lamp labeled 250V but it is used on 200V mains. <<<also didn't get

Hopeu can help guys thanks =]

Because it will use less wattage at the lower voltage. It will also be dimmer, so you need more lamps to do the same job, so I don't necessarily agree it's more economical.

for his question 13, could it also be said that the lamp will hae a constant current draw independent of the voltage? and because of that the current will be a set value even if you decrease the voltage and would that way lower the wattage? or is it that an incorrect assumption?

Less energy will be lost as heat because less heat is produced, but less light is too.....

Originally Posted by Arcane_Mathamatition

for his question 13, could it also be said that the lamp will hae a constant current draw independent of the voltage? and because of that the current will be a set value even if you decrease the voltage and would that way lower the wattage? or is it that an incorrect assumption?

It would be an incorrect assumption. The lamp has a fixed resistance. And using Ohm's law it is easy to show that changing the voltage across a fixed resistance changes the current.

thanks guys. This has helped me get a better understanding of electricity and I appreciated all the help thanks

In electricity there are electrons