For the system of capacitors shown in the figure below, find the following. (Let

Question

For the system of capacitors shown in the figure below, find the following. (Let C1 = 8.00 µF and C2 = 2.00 µF.)
http://www.webassign.net/sercp9/16-p-037-alt.gif
(a) the equivalent capacitance of the system
________ µF

(b) the charge on each capacitor
on C1_______ µC
on C2_______µC
on the 6.00 µF capacitor______µC
on the 2.00 µF capacitor______µC

(c) the potential difference across each capacitor
across C1______V
across C2______V
across the 6.00 µF capacior_____V
across the 2.00 µF capacitor______V

Explanation / Answer

First C1 and the 6F capacitor are in series so the equivalent capacitance for them is found by...

1/CEQ = 1/6 + 1/8

CEQ of that part is = 3.43 F (Call this CEQA)

Next the 2F capacitor and C2 are in series so the equivalent capacitance for them is found by...

1/CEQ = 1/2 + 1/2

CEQ of that part is = 1F (Call this CEQB)

Those two equivalent capacitances are in parallel with each other so we can get total CEQ for the circuit

For parallel CEQ = CEQA + CEQB

Total CEQ = 3.43F + 1F = 4.43F

From here the total charge on the circuit can be found by Q = CV = (4.43 X 10-6)(90) = 3.99 X 10-4 C

or 399C

Now we can work backward. You need to remember that Voltage in parallel portions is equal but splits in series while charge in series is equal but splits in parallel

We then know that V for CEQA and for CEQB will be 90V but will be split by each of the two capacitors that make up those two CEQ's

Starting there we can find the charge in the upper portion

Q = CEQAV = (3.43 X 10-6)(90) = 3.09 X 10-4 C or 309 C

That means that the charge across C1 and the 6F is 309 C

Do the same for the lower portion

Q = CEQBV = (1 X 10-6)(90) = 9 X 10-5 C or 90 C

That means that the charge across C2 and and the 2F is 90F

Note that the two values of charge added up will total the entire circuit charge found previously

Now the separate voltages can be determines

V = Q/C

Since the Q is now known for each individual capacitor, solve for voltage in each one and note that the total in the top must be 90 as well as the total in the bottom

For the top first

V = (309 X 10-6)/(6 X 10-6) = 51.5 V for the 6F capacitor

V = (309 X 10-6)/(8 X 10-6) = 38.6 V for the 8F capacitor

For the bottom

V = (90 X 10-6)/(2 X 10-6) = 45 V and since they are both 2F capacitors there is no need to to the calculation twice, they each have 45 V across them.

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