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Reading material on the basics of circuitry

F

foges

Senior member
My physics exams are coming up and circuits is part of it, i get some of it (like calculate the resistance of these resistors in parallel), but when wattage and parallel combined with series comes i get confused. So if anyone can recomend a website that would be awesome. 😀
 
well i was going over this past examination paper (page 8, see below) and i get the very first part (why the "students" explanation is incorrect), but the rest i dont understand. For example why would the brightness of bulb A and C not stay the same and the brightness in bulb B dim. I also calculated the wattage over bulb B to be 1.5, not 1.08. If anyone could walk me through that 😀

http://rapidshare.com/files/25680685/Physics_HL_-_Nov_2000_-_P2.pdf.html
 
8a: electrons don't matter; e-fields matter.
8b: 3 series equal resistors... they'll obviously all be equal
8c - this depends on the actual behavior of the bulbs. if they're acting like resistors, A and C will get brighter, because the resistance of all 4 bulbs together is lower than the resistance of the 3 in the initial setup. Since the resistance is lower, the current is higher, and all the current is flowing through A and C. The current won't double, and since its' split between B and D, B has to get dimmer. Now, if the bulbs are really rated for only 10V / 3W, A and C might burn out, but I don't think that's what they're looking for 😉
 
Thanx, that makes sense 🙂 how about question d (ii), thats the one i really dont get. Here is what the marscheme says:

One way is to look at resistor arrangement as a potential divider.
PD across B/D is one fifth of 30V since the parallel resistance is 1/5 of the
5 whole.
So PD across B is 30/5 = 6 V.
Original PD was 10 V.
Now P =V2/r i.e. P proportional to V2
So new power is
(6/10)2 * 3W = 0.36 * 3 =1.08 W

PS: the 2's after anything mean squared
 
Originally posted by: foges
Thanx, that makes sense 🙂 how about question d (ii), thats the one i really dont get. Here is what the marscheme says:

One way is to look at resistor arrangement as a potential divider.
PD across B/D is one fifth of 30V since the parallel resistance is 1 of the
5 whole.
So PD across B is 30/5 = 6 V.
Original PD was 10 V.
Now P =V2/r i.e. P proportional to V2
So new power is
(6/10)2 * 3W = 0.36 * 3 =1.08 W

PS: the 2's after anything mean squared
Click to expand...

Are you saying this is your guess or this is the answer?
 
This is the answer (from the marscheme), but i dont understand it (why is the resistance 1/5 of the whole?)
 
If the bulbs are 10V 3W, then they are 33.33 ohms. Total resistance 100 ohm. Add another 33.33 ohm resistor D across B and the restistance of B and D in parallel is 33.33/2 or 16.67 ohms. The total resistance of all 4 resistors is now 33.33 +16.67 + 33.33 = 83.33 ohms. Divide 83.33 by 16.67, and you have 5.

Rob Murphy.
 
Originally posted by: foges
This is the answer (from the marscheme), but i dont understand it (why is the resistance 1/5 of the whole?)
Click to expand...

The overall resistance is equal to the resistance of A, plus the resistance of B and D in parallel, plus the resistance of C.

The resistance of A (or C) is equal to twice the resistance of B and D in parallel. This is because putting two identical resistors in parallel is the same as halving the resistance.

So, if we name the parallel resistance (of B and D) R, the resistance of A will be 2R and the resistance of C will also be 2R.

So, the total resistance would be 2R + R + 2R = 5R.
So, the total resistance (5R) is 5 times the resistance of B and D in parallel (R). Or, the resistance of B and D in parallel is 1/5 of the total resistance.
 
Thank you. It makes sense now 😀😀, im still looking for some reading material though, can you recomend any books?
 
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