Picture of circuit: http://www.webassign.net/pse/p26-21alt.g…
Four capacitors are connected as shown in the figure below (C = 20.0 µF).
_______C______3uF__
| |
a_____________| |______20uF______b
| |
|_______6uF_________|
(a) Find the equivalent capacitance between points a and b.
I found this to be 6.02 uF and it was correct.
(b) Calculate the charge on each capacitor if ΔVab = 10.5 V.
20.0 µF capacitor _______________uC Found this to be 63.21 uC by 10.5V*6.02uF. I don't understand why this works so please explain.
6.00 µF capacitor _______________uC
3.00 µF capacitor ______________uC
capacitor C _______________uC
How do I calculate the charge on each individual capacitor? What is the 10.5V exactly... I thought the voltage stayed the same all through the circuit? Please explain this thoroughly. Thanks in advance for your help.
Four capacitors are connected as shown in the figure below (C = 20.0 µF).
_______C______3uF__
| |
a_____________| |______20uF______b
| |
|_______6uF_________|
(a) Find the equivalent capacitance between points a and b.
I found this to be 6.02 uF and it was correct.
(b) Calculate the charge on each capacitor if ΔVab = 10.5 V.
20.0 µF capacitor _______________uC Found this to be 63.21 uC by 10.5V*6.02uF. I don't understand why this works so please explain.
6.00 µF capacitor _______________uC
3.00 µF capacitor ______________uC
capacitor C _______________uC
How do I calculate the charge on each individual capacitor? What is the 10.5V exactly... I thought the voltage stayed the same all through the circuit? Please explain this thoroughly. Thanks in advance for your help.
-
For part b) the current flows from one end to the other. ALL the current must pass through the 20 uF capacitor before reaching the others.
As it acts like a 6.02 uF capacitor the circuit will gain a charge given by Q = CV
Which means that this current has flowed through the 20 uF capacitor too.
For the other capacitors, SOME of the current flows through the top branch and some flows through the bottom branch. The total of both must be the same as we calculated for part b above.
In working out part a) you would have worked out the equivalent capacitance of both the top and the bottom parts of the circuit. 6 uF on the bottom and just under 3 uF on the top.
And they will share the current in proportion to their capacitance.
i.e about 6/9 of the current ( charge) on the bottom capacitor and 3/9 of the current passing through BOTH of the capacitors in the upper branch. This is because they are both in series and must have had the same current through each one.
The thing that varies is the VOLTS across each element in the series branch.
As it acts like a 6.02 uF capacitor the circuit will gain a charge given by Q = CV
Which means that this current has flowed through the 20 uF capacitor too.
For the other capacitors, SOME of the current flows through the top branch and some flows through the bottom branch. The total of both must be the same as we calculated for part b above.
In working out part a) you would have worked out the equivalent capacitance of both the top and the bottom parts of the circuit. 6 uF on the bottom and just under 3 uF on the top.
And they will share the current in proportion to their capacitance.
i.e about 6/9 of the current ( charge) on the bottom capacitor and 3/9 of the current passing through BOTH of the capacitors in the upper branch. This is because they are both in series and must have had the same current through each one.
The thing that varies is the VOLTS across each element in the series branch.