In the figure there is a four way key at the middle . if key is thrown from situation BD ton AD , then how much charge will flow through point O ????
1Q= CV
So charge on 4μF capacitor = 4X3 = 12 μ Coulombs.
When Switch is changed from BD to AD, both the plates of 4μF capacitor are short circuited. Hence charge on their plates will cancell out.
So 12 μ Coulombs charge will flow throught point O in A→D direction
1Edit : Thought the new switch was AC ..
But eventually the current through o must become 0 right ?
1Where the hell is point "O" its a four way switch right, so is "O" hanging in the air
1Assuming that Aman has made a mistake ...
If key is thrown from situation BD ton AC.
Then the net charge which flows through O is 0
@ Aman .. please check the variable names before you post here [1]
1I think o is in the middle of the 4-way key such that BD means BO - OD
and AD means AO - OD ...
And if it is AC , isn't total charge = 36*2 μC ?
1@ Ashish , please read the Q carefully , aman has written that the switch is changed form BD to AD , not AC.
1
O point is at the centre and not in the air..........
see the new image. the black spot at the centre is O.
and the ans is 72 μC.
1If the new switch is AC, then it is 72 μC ( 36 μC from both )... but AD, I am not sure...
Q= CV
So, charge on left capacitor = 4X3 = 12 μ Coulombs.
charge on right capacitor = 4X6 = 24 μ Coulombs.
Now for the left capacitor :-
When Switch is changed from BD to AD, both the plates of 4μF capacitor are short circuited.
Hence charge on their plates will cancell out.
So 12 μ Coulombs charge will flow throught point O in A→D direction
P.S: Ashish has solved upto this in very begining, but he missed to consider the change in charges on the plates of right capacitor [3]
For the right capacitor :-
When Switch is changed from BD to AD, net Voltage across its plates decreases from 6V to 3 V ( think why ...... [1] )
So, final charge on right capacitor = 4X3 = 12 μ Coulombs.
So 12 μ Coulombs charge will flow throught point O in A→D direction
So Total Charge Flowing throught point O in A→D direction = 24 μ Coulombs.
@ Varun
You are correct. If the new switch is AC, then net charge flowing through O in A→C direction will be:-
36μC + 36 μC = 72 μC
@ Aman
If you have found this question in any text book, then there is a misprint. It should be AC not AD. Answers for both case are discussed already above
Case AC : 72 μC
Case AD : 24 μC
