AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Consider the circuit, The charge on 4µF capacitor is,
q4=C4V
q4=(4µF)(6)
q4=24µC
Given condition,

qx

q4

=

3

8

qx

24

=

3

8

qx=9µC
The potential difference on xµF capacitor is,
V1=

qx

C

V1=

9

x

Thus, the voltage across 2µF will also be

9

x

The voltage across 3µF will be (6−

9

x

) volt
The capacitors 3µF and (x+2)µF is in series, so there will be same charge on 3µF and (x+2)µF
So,
q3=q(x+2)
3(6−

9

x

)=(x+2)

9

x

6x−9=3x+6
x=5