In the series `L-C-R` circuit shown in the figure, the rms voltage across the resistor and inductor are 400 V and 700 V respectively. If the applied voltage is `E=500sqrt(2)sin(omega t)`, then the peak voltage across the capacitor is

Options

A. 400 V.
B. 400`sqrt(2)`V.
C. 800 V
D. `800 sqrt(2)` V

Answer

B

Explanation

(b) : rms potential differences are related as follows :
`V^(2) =V_(R)^(2)+(V_(L)^(2)-V_(C)^(2))`
`implies 500^(2)=400^(2)+(700^(2)-V_(C)^(2))`
`implies V_(C)=400V `
Hence peak voltage can be written as follows :
`implies V_(C)^(0)= 400sqrt(2)V`
Hence option (b)is correct.