The voltage time (V-t) graoh for triangular wave having peak value `(V_0)` is as shown in fig.

Options

A. `(V_(0))/(3)`
B. `(V_(0))/(2)`
C. `(V_(0))/(sqrt(2))`
D. `(V_(0))/(sqrt(3))`

Answer

D

Explanation

`V (t) = ((4V_(0))/(T))t` for `0 le t le (T)/(4)` , `V^(2) = ((16 V_(0)^(2))/(T^(2))) t^(2)`
`:.lt V^(2) gt_(0-T//4) = (((16 V_(0)^(2))/(T^(2))) int_(0)^(T//4) t^(2) dt)/(int_(0)^(T//4) dt) = (V_(0)^(2))/(3)` ,
`:. V_("rms") = sqrt((V_(0)^(2))/(3)) = (V_(0))/(sqrt(3))`