The capacitance of a variable capacitor cann be changed from `50muF` to `400muF` by turning the knob from `0^(@)` to `180^(@)` on a calibrated semicircular dial. When the knob is set at `180^(@)`, the capacitor is connected to a 100 V battery. When it is fully charged, the battery is disconnected and the knob is brought back to `0^(@)`. what is the P.D. across the capacitor when the knob is at `0^(@)`.

Options

A. 700 V
B. 750 V
C. 800 V
D. 850 V

Answer

C

Explanation

`C_(1)=50muF` at `0^(@) and C_(2)=400muF` at `180^(@)`
P.D. across `C_(2)=100V`
`therefore q=C_(2)V=400xx10^(-6)xx100=4xx10^(-2)C`
when it is brought to `0^(@)`, q remains constant and
`because q=C_(1)V_(1)`
`therefore V_(1)=(q)/(C_(1))=(4xx10^(-2))/(50xx10^(-6))=(4xx10^(3))/(5)=800V`