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

Answer

B

Explanation

The magnetic flux linked with the primary coil is given by
`phi = phi_(0) + 4t`
So, voltage across primary
`V_(p) = (d phi)/(dt) = (d)/(dt) (phi_(0) + 4t)`
`= 4V` (as `phi_(0) =` constant)
Also, we have
`N_(p) = 50` and `N_(s) = 1500`
From relation,
`(V_(s))/(V_(p)) = (N_(s))/(N_(p))` or `V_(s) = V_(p) (N_(s))/(N_(p)) = 4 ((1500)/(50)) = 120 V`