In figure `C_1=C_5=8.4muF` and `C_2=C_3=C_4=4.2muF`.The applied potential is `V_(ab)=220V`

(a) What is the equivalent capacitance of the network between points `a` and `b`?
(b) Calculate the charge on each capacitor and the potential difference across each capacitor.

Answer

A

Explanation

a. Simple series and paralle grouping of capacitor.
b. `q_(n et)=C_(n et)V`
`=(2.5xx10^-6)(220)`
`=5.5xx10^-4C`
`C_1, C_5` and equivalent of other three capacitors are in series. Hence, charges across them are same

`q_("total")` will distribute between `C_2` and `C_34` in direct ratio of capacitance.
`:. q_2/q_34=4.2/2.1=2/1`
`:. q_2=2/3(5.5xx10^4)=3.7xx10^I-4C`
`q_32=1/3(5.5xx10^-4)=1.8xx10^-4C`
for finding PD across any capacitor use the equation
`C=q/V`