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)=(dphi)/(dt)=d/(dt)(phi_(0)+4t)=4` volt (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)=120V`