Voltage across capacitors.

  • Thread starter rave7
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  • #1
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A 2.41microF and a 7.38microF capacitor are connected in series across a 30.0-V battery. A 10.3microF capacitor is then connected in parallel across the 2.41-microF capacitor. Determine the voltage across the 10.3-microF capacitor.

How do i approach this question. do i start by finding the total energy produced by the capacitors? since E=1/2 CV^2 . But how do i find voltage across the 7.38microF capacitor?
 

Answers and Replies

  • #2
LowlyPion
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A 2.41microF and a 7.38microF capacitor are connected in series across a 30.0-V battery. A 10.3microF capacitor is then connected in parallel across the 2.41-microF capacitor. Determine the voltage across the 10.3-microF capacitor.

How do i approach this question. do i start by finding the total energy produced by the capacitors? since E=1/2 CV^2 . But how do i find voltage across the 7.38microF capacitor?
Figure first the equivalent capacitance in the network including the new capacitor. Then you can determine the voltage straight away across the parallel capacitors. The relationship you need for voltage is V = Q/C.
 
  • #3
tiny-tim
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Welcome to PF!

Figure first the equivalent capacitance in the network including the new capacitor. Then you can determine the voltage straight away across the parallel capacitors. The relationship you need for voltage is V = Q/C.
Hi rave7! Welcome to PF! :smile:

(have a mu: µ and a squared: ² :smile:)

I'll just add this to what LowlyPion says …

never use energy of capacitors if you can avoid it! :cry:

and remember that Q will be the same on the parallel pair as on the single (7.8µF) one. :smile:
 
  • #4
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A 2.41microF and a 7.38microF capacitor are connected in series across a 30.0-V battery. A 10.3microF capacitor is then connected in parallel across the 2.41-microF capacitor. Determine the voltage across the 10.3-microF capacitor.

the equivalent capacitance are worked out as shown:
10.3 + 2.41 = 12.71
(1/12.71) + (1/7.38) = 1/(Ceqv)
my equivalent capacitance will be approximate 4.67 µF

Since Q = CV, i can find the total Q in this circuit.
Q = 4.67 * 30 = 140 µC

After that, how do i determine the voltage across the 10.3µF capacitor, the Q flowing into the parallel branch to the resistors are they the same or different. abit confused. can i use the method of current divider to determine the q flowing into the resistors in the parallel branch?
 
  • #5
tiny-tim
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the equivalent capacitance are worked out as shown:
10.3 + 2.41 = 12.71
(1/12.71) + (1/7.38) = 1/(Ceqv)
my equivalent capacitance will be approximate 4.67 µF

Since Q = CV, i can find the total Q in this circuit.
Q = 4.67 * 30 = 140 µC
Hi rave7! :smile:

Yes, that's fine. :smile:
the Q flowing into the parallel branch to the resistors are they the same or different.
(what resistors? i assume you mean capacitors? :confused:)

Remember, the voltage across a capacitor is the same as the voltage between any two points containing the capacitor and without anything else (in series).

So the voltage across either capacitor is the same as, for example, the voltage between the two points where their connecting wires meet … in other words, voltage across capacitors in parallel are always the same (just as charge across capacitors in series are always the same). :smile:
After that, how do i determine the voltage across the 10.3µF capacitor, …
abit confused. can i use the method of current divider to determine the q flowing into the resistors in the parallel branch?
The total voltage drop has to be 30.

Either treat the parallel capacitors as one, or caclulate the voltage drop across the 7.38 µF, and subtract from 30. :smile:
 
  • #6
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thanks for those who had replied to this thread. i have managed to get the solutions.

Best regards.
 

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