capacitance
so in the 3rd diagram our teacher said we can ignore the c5 when c1/c3=c2/c4
my question is how ? like i didnt understood how we can and if in more such future circuits how will i know tht i have to skip tht capacitor
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@Gyro Gearloose
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+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Ahhh a wheatstone bridge
a what
It's called a wheatstone bridge ifirc
oh
didnt knew tht
but like how can we ignore c5
BYJU'S NEET
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Wheatstone Bridge of Capacitor | Physics | NEET 2021/2022 | Concept...
Relevant for Class: 12th
Chapter: Capacitors
Prerequisites: Wheatstone bridge, Potential Difference.
A Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance/ capacitance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. Here you will learn about Wheatstone bridge of...
Idk wither maine nahi padha
But
Ok, so, look at it this way.
Some charge q flows from the cell to the beginning of the junction.
C1=C3 so the charge splits evenly, into q/2 and q/2 right?
Last year
ew byjus
yep
And then, when it needs to flow into C2 and C4, the ratio of the charges in each is going to be the same, because they're also equal
So there is no need for any charge to flow down C5 path at all
Because the same charge continues in each path
because flowing to c2 and c4 will be more favourable?
Not favourable per se. It's more that the amount of charge already present is the correct amount to fully fill the capacitor
There is no need for any exchange from upper path to lower path or vice-versa
didnot understood at all
didnt get the last part rest all yeah i got it
The ratio of charge in C2 and C4 is the same ratio as in C1 and C3 right?
yeah
And the charge flowing in the two paths are already in that ratio because they split according to C1 and C3
So the ratio doesn't need to change
It stays the same
Because of that, no charge moves through the middle path.
If any charge moved from upper path to lower path, the ratio would change, but that shouldn't happen
aahh okk
and like c5 is connected like | what if it was connected like --
then what will happen
will this still be followed?
No, then you have charge splitting into three paths in the beginning
It becomes a parallel capacitors situation
will it become normal parallel connection then?
ahh okk
thnkss
one more thing
can we say that charge doesnt want to divide and wants to go the path where it is least divided?
No, charge wants to go on the path where it is least resisted.
ah okkk
thnks
+solved @Opt
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