The primary and secondary coils of a transmformer have `50` and `1500` turns respectively. If the magnetic flux `phi` linked with the primary coil is given by `phi=phi_(0)+4t`, where `phi` is in weber, `t` is time in second and `phi_(0)` is a constant, the output voltage across the secondary coil is

Options

A. `120 V`
B. `220 V`
C. `30 V`
D. ` 90 V`

Answer

A

Explanation

Here, `n_(P) = 50, n_(s) = 1500`
`phi_(P) = phi_(0) + 4 t`
`E_(P) = (d phi_(P))/(dt) = (d)/(dt) (phi_(0) + 4 t) = 4`
As `(E_(s))/(E_(P)) = (n_(s))/(n_(P))`
`:. E_(s) = (n_(s))/(n_(P)) xx E_(P) = (1500)/(50) xx 4 = 150 V`