Capacitor `C_3` in the circuit is a variable capacitor (its capacitance can be varied). `C_1 and C_2` are of fixed values. Graph is plotted between potential difference `V_1` (across capacitor `C_1)` versus `C_3`. Electric potential `V_1` approches on asymptote of 10 V as `C_3 prop oo`.
,
When `V_1 = 4V`m then `C_3` is equal to

Options

A. `5C_2//2`
B. `5C_1//3`
C. `5C_2//3`
D. none of these

Answer

D

Explanation

When `C_(3)toinfty` it means distance between the plates of `C_(3)` is zero or both plates will be at the same potential then `C_(2)` will be shorted entire pontential V will ber across `C_(1)` hence `V=V_(1)=10V`
From graphs when `C_(3)=0,V_(1)=2V. So C_(3)` will act like open switch `C_(1)` and `C_(2)` will be in series potential different across `C_(2)` is `V_(2)=10-V_(1)`
`=8V`
`q=C_(1)V_(1)=C_(2)V_(2)` or `C_(12)=C_(28)` or `C_(1)=4C_(2)`
`V_(1)=4V,V_(2)=10-4=6V`
`q=C_(1)V_(1)=(C_(2)+C_(3))V_(2)`
or `C_(14)=(C_(2)+C_(3))6` or `16C_(2)=6C_(2)+6C_(3)`
or `C_(3)=5C_(2//3)`