ALL PHYSICS DOUBTS - POST HERE !!

[holtadit]

So if the magnet is pushed into the coil the emf flows in the pink arrow's direction ?

And if the magnet is pulled out of the coil the emf flows in the red arrows's direction ?

[The Golden Girl =D]

yup

[The Golden Girl =D]

So if the magnet is pushed into the coil the emf flows in the pink arrow's direction ?

And if the magnet is pulled out of the coil the emf flows in the red arrows's direction ?

use the right hand rule of a wire and u'll get the whole thing , don't worry it's easy

[holtadit]

In diagram 1 its just a  normal circuit with a 12v supply and a voltmeter connected accross a normal wire. NO components.

In diagram 2 its another circuit with an OPEN switch, voltmeter and 12V supply. NO COMPONENTS.



First question: in Diagram one would the P.D. across the wire be 0 V ?  Because there should be no change in the PE of the individual charges

SECOND QUESTION: In diagram 2 would the P.D. across the OPEN switch be 12 V ?   

[nid404]

Diagrams correct....but voltage jst the opp

In the first one,  P.D across the wire would be 12V, since resistance of the wire is negligible and there is no other component....
Whenever you see a switch open, remember, no current flows, and hence, there is no P.D

[Amr Fouad]

Diagrams correct....but voltage jst the opp

In the first one,  P.D across the wire would be 12V, since resistance of the wire is negligible and there is no other component....
Whenever you see a switch open, remember, no current flows, and hence, there is no P.D

Hey, that is totally incorrect..

Yes it is true, in the 1st diagram, the voltage is zero. V=IR, and R=o, therefore V=0..

Again, in the second one, V=o, as I=0..

[nid404]

In the first diagram, R is not equal to zero, practically speaking, if you set up something like this in the lab, you will get a little less than 12V across the first circuit...im dead sure....cuz indirectly we r measuring the emf of the battery for the first one....

[haris94]

i agree with nid404
diagram 1: 12V
diagram 2: 0V

[Amr Fouad]

guys..this O-level..it is not A-L..

ask astar.pm him if u dont believe me...

V=RI...R=0...so V=0..period.

[nid404]

doesn't matter which level it is dear...

I am dead sure....anyway u want to ask him...go ahead...always better to clear ur doubts

[holtadit]

Okay never mind that.


Heres my next query. For a D.C. output how should I draw the waves ? The first picture (top) or second one (bottom) ?

[holtadit]

Earth TO STUDENT FORUMS !!!  COME IN !!! Do YOU READ ME !!?

[holtadit]

i guess not

[Adzel]

POSITIVE... 

ans: The first picture (top)

[Saladin]

yes, the first picture, because you are simply absorbing the negative current.

That is why, the rectified current simply looks the like original one with the bottom cut off.