12. ·09points I PenusAnswers Ser P11 16P009. Consider the following. (Letc.-12.2

Question

12. ·09points I PenusAnswers Ser P11 16P009. Consider the following. (Letc.-12.20 AF and C.-6.20 .) 6.00 F C, -C, 9.00 V (a) Find the equivalent capacitance of the capacitors in the figure. (b) Find the charge on each capacitor. on the right 12.20 F capacitor on the left 12.20 uF capacitor on the 6.20 F capacitor on the 6.00 F capacitor uC puC AC (e) Find the potential difference across each capacitor on the right 12.20 F capacitor on the left 12.20 F capacitor on the 6.20 pF capacitor on the 6.00 uF capacitor a Show My Work (optional)

Explanation / Answer

when capacitors are connected in series, net capacitance = C

1/C = 1/C1 + 1/C2

for parallel

C= C1 + C2

here 6 and 6.2 micro F are in parallel.

so net C = 12.2 micro F

now,

12.2 , 12.2 and 6.2 are in series.

Hence net capacitance = 3.07 micro Farad.

For series, Q is same and V is divided,

For parallel V is same and Q is divided.

Hence Q = 3.07 * 9 = 27.63 C

so Q across left 12.2 = 27.63 C

Q across right 12.2 = 27.63 C

but Q will split between 6 and 6.2

and Q is directly proportional to C

hence more C have more Q

so Q across 6 = 6*27.63/12.2 = 13.58C

hence across 6.2 = 27.63-13.58 = 14.05C

----------------------

we have 4 values of C and 4 values of Q

we can get using Q=CV

so V = Q/C

Q1=27.63 C1= 12.2 V1= 2.28V

Q2=27.63 C2= 12.2 V2= 2.28V

Q3=13.58 C3= 6 V3= 2.26

Q4=14.05 C4=6.2 V4 = 2.26

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