Question

So, I don't know what I'm missing here but I can't seem to be able to find the charge of capacitor C.

Four capacitors are connected as shown in the figure below. (C-12.0 μΕ) С 3.00 pLF 20.0 uF Ήτ (a) Find the equivalent capacitance between points a and b. 5.91 (b) Calculate the charge on each capacitor, taking AVab 12.0 V 20.0 HF capacitor 70.92 6.00μF capacitor 50.7 3.00 uF capacitor 20.22 capacitor C HC Enter a number. es you have calculated on other capacitors to find the charge on capacitor C. Hc

Answer 1

(a)

Equivalent capcitance of series= (C1*C2)/(C1+C2)

Equivalent capacitance of parallel = C1+C2

the capacitor C and 13 are in series which is in parallel with the micro farady capacitor.

This combination is in series with the 20 micro farady capacitor.

So Equivalent capacitance will be, of C, 3 and 6 will be,

C=(12*3/12+3)+6

C=8.4 micro farady.

Now this will be in series with 20, so

Ceq=(20*8.4)/(20+8.4)

Ceq= 5.915 micro farady.

(b)

Now we know that charge in series is same.

So, charge on Capacitor C will be same as of the charge 3 micro farady, that is

It will also be,20.22 micro coloumb.