Physics: Post your doubts here!

[ramith]

Please answer these questions about internal resistance.

1.Three cells, each of e.m.f. 1.5v, are connected in series to a 15 ohm light bulb. The current in the circuit is 0.27A. Calculate the internal resistance of each cell.

2. A cell of e.m.f 1.5v has an internal resistance of 0.50 ohm. Calculate the maximum current it can deliver. under what circuit does it deliver this maximum current. Calculate also the maximum power it can deliver to an external load. Under what circumstances does it deliver this maximum power?

 

[ad98ify]

please anyone???

 

[unique840]

hi guys. i m having sm conceptual problems in basic concepts in chapters like gravity and electric field like :-

• why is gravitational field strength maximum at poles rather than at equator when it should be vice versa as seen by the formula. g=GM/r^2 that is g is directly proportional to 1/r^2?
• in electric field, why do we say that inside a point charge, potential is constant that is an equipotential surface exists?

please please answer them

g is directly proportional to 1/r^2
at poles, radius is minimum so g will b maximum while at equator, radius is max to g will b minimum.

 

[unique840]

Please answer these questions about internal resistance.

1.Three cells, each of e.m.f. 1.5v, are connected in series to a 15 ohm light bulb. The current in the circuit is 0.27A. Calculate the internal resistance of each cell.

2. A cell of e.m.f 1.5v has an internal resistance of 0.50 ohm. Calculate the maximum current it can deliver. under what circuit does it deliver this maximum current. Calculate also the maximum power it can deliver to an external load. Under what circumstances does it deliver this maximum power?

1. v = ir
3 cells so v = 4.5 v
4.5 = 0.27 * r
r = 16.66 ohms
this is total resistance
r = 15 + internal resistance*3
1.66 = 3x
x = 0.55 ohms

2. v = ir
1.5 = 0.5 * i
i = 3 A
current divides in parallel n remains same in series so max current in series circuit.
p = vi
p = 3*1.5
p = 4.5 W

 

[ramith]

1. v = ir
3 cells so v = 4.5 v
4.5 = 0.27 * r
r = 16.66 ohms
this is total resistance
r = 15 + internal resistance*3
1.66 = 3x
x = 0.55 ohms

2. v = ir
1.5 = 0.5 * i
i = 3 A
current divides in parallel n remains same in series so max current in series circuit.
p = vi
p = 3*1.5
p = 4.5 W

Thanks alot!!!

 

[JD REBORN]

Can somebody plz explain to me how part c of question 2 is done?

 

[ad98ify]

g is directly proportional to 1/r^2
at poles, radius is minimum so g will b maximum while at equator, radius is max to g will b minimum.

in our syllabus we always take earth to be a perfect sphere with uniform density. so this concept cant be applied here =(

 

[code11]

hey guys , tomorow iz my physics practial exam of as level (school exam)
how do i prepare it ?
any relevant sites wud b helpful

 

[CaptainDanger]

hey guys , tomorow iz my physics practial exam of as level (school exam)
how do i prepare it ?
any relevant sites wud b helpful

Do past papers... As many as you can... Practice...

 

[abcde]

Hi can plz somebody explain to me how to answer the following two questions from physics 9702 AS level May/June 2010 Paper 23:

Question 2)Part C
Question 3)Part A

??????

Check the attachment for Q2(c).
For Q3(a), you need to find the area beneath the graph for which you'll have to employ a means of counting the boxes as the graph isn't a straight line and integration is not an option.

 

[unique840]

Thanks alot!!!

ur welcum

 

[ad98ify]

hi guys. i m having sm conceptual problems in basic concepts in chapters like gravity and electric field like :-

• why is gravitational field strength maximum at poles rather than at equator when it should be vice versa as seen by the formula. g=GM/r^2 that is g is directly proportional to 1/r^2?
• in electric field, why do we say that inside a point charge, potential is constant that is an equipotential surface exists?

please please answer them

 

[JD REBORN]

Check the attachment for Q2(c). View attachment 3231
For Q3(a), you need to find the area beneath the graph for which you'll have to employ a means of counting the boxes as the graph isn't a straight line and integration is not an option.

Thanks Very Much

 

[Unicorn]

Hi everyone, assalamoalaikum!!

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your AS/A2 PHYSICS doubts in this thread. InshaAllah other people here will help me and you all.

NOTE: If any doubts in the pastpapers, please post the link! You can find links here!

Any Physics related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites!
Thanks!
Jazak Allah Khair!

Physics Notes:

Some links & Notes - by 'destined007'

Physics Practical Tips - by arlery

Notes for A2 Direct Sensing (Applications) - shared by sweetiepie

The link isn't working :'(

 

[abcde]

The link isn't working :'(

They probably won't for the time being. Search for the topic headings in the search bar. You'll find 'em.

 

[Unicorn]

They probably won't for the time being. Search for the topic headings in the search bar. You'll find 'em.

ok but why?

 

[Unicorn]

Can someone please help me? My teacher is basically of no use and refuses to answer our questions. So I want to know what does Justify mean in physics? In question 1ciii

Thank you

 

[abcde]

Can someone please help me? My teacher is basically of no use and refuses to answer our questions. So I want to know what does Justify mean in physics? In question 1ciii

Thank you

If you're talking about Q1)(c), you have show whether the assumption is valid or not (justified or not).

 

[Unicorn]

If you're talking about Q1)(c), you have show whether the assumption is valid or not (justified or not).

when i look at the mark scheme the answer makes no sense | i got the two figures but can't relate them how

 

[abcde]

when i look at the mark scheme the answer makes no sense | i got the two figures but can't relate them how

Post the links for the paper and the mark scheme. I'll have a look.