Think of this. Replace all three capacitors with a new capicator of some value that is equal to the total combined circuit capacitance of x,y,z .
Then the current vs time flowing through the new capacitor is exaclty the same that has flowed through X..which means the charge stored on the plate (A side) is the same as if x,y,z had been connnected. The current flow determines the charge accumulated.
Three uncharged capacitors, X, Y, Z, each with capacitance of 12uF each are connected
A---------[X]------[Y/Z]-----------B
*Y&Z is parallel to each other*
A potential difference of 9.0V is applied between points A and B.
Explain why when the potential difference of 9.0V is applied, the charge on one plate of capacitor X is 72uF.
Ans says:
>some discussion as to why all charge of one sign on one plate of X
>Q= 8.0x10^(-6) x 9.0 = 72uF
I'm not too sure of the explaination.
Please help? Thanks a lot!
1 answer
1 answer