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secondary 4 | A Maths
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Physics Qn! Please explain to me (: thank you!
Date Posted: 5 years ago
Views: 318
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done
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hope this helps.
if u hv any doubts, pls ask,
:)
if u hv any doubts, pls ask,
:)
Date Posted:
5 years ago
Does length not affect resistance?
Sorry, but I don’t understand how to get the red colour values :”)
not to worry, this is my style of teaching.
this is the 1st time u see this shortcut,
u not used to it. no worries :)
let's take a look at option A first:
r , A and R are the originals.
if radius is 0.5r, the new area would be 0.25A.
u wrote in the qn paper:
up r, down R right?
but u shld add:
up r, up A , down R
so conversely,
down r , down A, up R
since the new area is reduced by 4 times, the new resistance due to the reduced area is increases by 4 times.
hence, u see the 4R in red.
next, length:
u wrote in the qn paper:
up L, up R right?
so
down L, down R
for option A,
the new length is 0.5L, i,e. halved,
so the resistance is also halved.
hence, u the 0.5R in red
combine the 2 effects of area and length, u get the 2R in blue.
k? :)
this is the 1st time u see this shortcut,
u not used to it. no worries :)
let's take a look at option A first:
r , A and R are the originals.
if radius is 0.5r, the new area would be 0.25A.
u wrote in the qn paper:
up r, down R right?
but u shld add:
up r, up A , down R
so conversely,
down r , down A, up R
since the new area is reduced by 4 times, the new resistance due to the reduced area is increases by 4 times.
hence, u see the 4R in red.
next, length:
u wrote in the qn paper:
up L, up R right?
so
down L, down R
for option A,
the new length is 0.5L, i,e. halved,
so the resistance is also halved.
hence, u the 0.5R in red
combine the 2 effects of area and length, u get the 2R in blue.
k? :)
Oh ok! I think I got it. If I still have doubts, can I ask again?
give me some time, i do another write-up. :)
Ok! Thank u!
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hope this helps. :)
*aiyo, u dun hv to downvote my previous post mah,...
*aiyo, u dun hv to downvote my previous post mah,...
Date Posted:
5 years ago
Ok, I think I roughly got it. Thank you! (:
can u remove the downvote
and return the upvote? :)
#discouraged
and return the upvote? :)
#discouraged
and u dun sound convinced.
just rem:
magnitude of R depends on the magnitude of A , not r or diameter
just rem:
magnitude of R depends on the magnitude of A , not r or diameter
Ok! Thank you (:
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see this helps.
it is to see the effect of area change on resistance AND the effect of length change on resistance.
my guess is:
u r confused over radius change or the square effect.
the critical thing is not radius change, but Area change, and the effect that area change on resistance
it is to see the effect of area change on resistance AND the effect of length change on resistance.
my guess is:
u r confused over radius change or the square effect.
the critical thing is not radius change, but Area change, and the effect that area change on resistance
Date Posted:
5 years ago
if u still r confused over anything, dun worry, just feel free to ask.
u got the basics.
r doubles, area increases by 4 times,
so resistance decreases by 4 times
length doubles,
resistance doubles.
if u can see,
radius change is more complex
than length change.
u got the basics.
r doubles, area increases by 4 times,
so resistance decreases by 4 times
length doubles,
resistance doubles.
if u can see,
radius change is more complex
than length change.
Why need to consider area? Because question never say anything about area.
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