Capacitors in Parallel and Series

In summary, to find the charge on each capacitor in this combination circuit, you can use the formula Q = CV and the voltage division rule for capacitors. First, calculate the equivalent capacitance, which is 1.34uF. Then, use this value to calculate the charge on each capacitor. Next, use voltage division to find the voltage across each capacitor. The sum of the voltages across the 2uF and 5uF capacitors will be the same as the voltage across the 1uF capacitor. From there, you can use the voltage division formula to calculate the voltages across the 2uF and 5uF capacitors.
  • #1
meadow
19
0
Hello. I have a brain fart and this combination circuit has me stumped. Here is a crude representation of the circuit. I had no problem finding the total capacitance (1.34uF), but the second part asks you to find the charge on each of the capacitors, assuming that the potential difference between b and a is 300 V.

2uF 5uF
----||------||----
a.---| |----- ----.b
3uF ------||----------
1uF
Ok, I know that the Q2=Q5 and Q3=Q7. The total charge is 402 uC, so Q3=Q7=402uF. Where do I go from here. I have tried all combinations...
 
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  • #2
Okay, I don't see that circuit of yours. Could you go maybe by circuit flow :P
 
  • #3
oops!

The circuit from point a goes to a capacitor of 3uF, then leads to 2 capacitors in series (2uF and 5uF) that are together parallel to a capacitor of 1uF. From the parallel circuit section you get to point b. Does that make sense? I tried to draw it out...but I guess it didn't work:) Thanks!
 
  • #4
So 3uF+( (2uF + 5uF)//(1uF ) ?
Your diagram just suggests otherwise. Anyway.
To calculate the charge on a capacitor in steady state, you use:

Q = CV

The only problem then would be finding V across each capacitor.
The voltage division rule applies to capacitors as V1 = C2/(C1+C2)
 
  • #5
  1. With equivalent capacitance, first calculate charge with Q=CV.
  2. Then calculate voltage across 3uF using V=Q(as obtained above)/3uF
  3. Voltage across the parallel combination of 1uF with series combination of 2uF and 5uF will Vab (300v)-V(across 3uF, as obtained above, in step 2). This will be voltage across 1UF.
  4. Using voltage division formula calculate voltages across 2uF and 5uF; sum of voltages across 2uF and 5Uf will be same as voltage across 1uF
 

What is the difference between capacitors connected in parallel and series?

When capacitors are connected in parallel, their equivalent capacitance increases and their voltage remains the same. In series, their equivalent capacitance decreases and their voltage adds up.

How do you calculate the equivalent capacitance of capacitors connected in parallel and series?

To calculate the equivalent capacitance of capacitors in parallel, simply add up all the individual capacitances. For capacitors in series, use the formula 1/Ceq = 1/C1 + 1/C2 + 1/C3 and so on, where Ceq is the equivalent capacitance and C1, C2, C3, etc. are the individual capacitances.

What impact does connecting capacitors in parallel have on their total capacitance?

Connecting capacitors in parallel increases their total capacitance, as the effective plate area and distance between the plates increases. This results in a larger capacitance value.

How does connecting capacitors in series affect their total capacitance?

Connecting capacitors in series decreases their total capacitance, as the effective plate area and distance between the plates decreases. This results in a smaller capacitance value.

What are some practical applications of capacitors connected in parallel and series?

Capacitors connected in parallel are commonly used in power supply circuits to increase capacitance and provide stable power output. In series, capacitors are used in circuits such as filters and timing circuits to achieve desired frequency responses and precise timing intervals.

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