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!
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.
To understand why the charge on one plate of capacitor X is 72uF when a potential difference of 9.0V is applied, let's break it down step by step.
1. When a potential difference is applied between points A and B, this creates an electric field between the two points. The potential difference of 9.0V means that point A is at a higher potential than point B.
2. Capacitor X is connected in series with a parallel combination of capacitors Y and Z. In a series connection, the same current flows through all the components, while in a parallel connection, the potential difference across each component is the same.
3. Since X is in series with Y and Z, the same current flows through X, Y, and Z. This means that the charge on one plate of X will be the same as the charge on one plate of Y and Z.
4. Capacitors Y and Z are in parallel, so they share the same potential difference. In this case, the potential difference across Y and Z is 9.0V. Since the potential difference across each component in parallel is the same, the charge on one plate of Y and Z will also be the same.
5. Now, we know that the capacitance of X, Y, and Z is 12uF each. The charge on one plate of a capacitor is given by the formula Q = C * V, where Q is the charge, C is the capacitance, and V is the potential difference.
6. Plugging in the values, the charge on one plate of Y and Z is Q = 12uF * 9.0V = 108uF.
7. Since the charge on one plate of X is the same as Y and Z, it will also be 108uF.
However, the given answer seems to have a typo. Instead of stating 72uF, it should be 108uF based on the explanation provided.