In the circuit shown in figure, if the point R is earthed and point P is given a potential of +1800V, then charges on `C_(2)andC_(3)` are respectively

Options

A. `2.4xx10^(-3)C,12xx10^(-3)C`
B. `1.6xx10^(-3)C,0.8xx10^(-3)C`
C. `32xx10^(-3)C,1.6xx10^(-3)C`
D. `4.8xx10^(-3)C,2.4xx10^(-3)C`

Answer

A

Explanation

`C_(eq)` of system = `(C_(2)" parallel "C_(3))" series "C_(1)`
= `1//(1/3+1/((4+2)))=2muF`
So, charge taken from source = `=q_(eq)=C_(eq)DeltaV=1800xx2xx10^(-6)C=3600muC`
Potential droop across
`C_(1)=(q_(c_(1)))/(C_(c_(1)))=(3600xx10^(-6))/(3xx10^(-6))=1200V`
So, potential drop across combination of `4muFand2muF` capacitors
= 1800 - 1200 = 600 V
Hence,
`q_(2)=C_(2)V_(QR)=4xx10^(-6)xx600=2.4xx10^(-3)C`
and `q_(3)=C_(3)V_(QR)=2xx10^(-6)xx600=1.2xx10^(-3)C`