The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $ϕ$ linked with the primary coil is given by $ϕ = ϕ_{o} + 4 t$ where $ϕ$ is weber, t is time in second and $ϕ_{o}$ is a constant, the output vltage across the secondary coil is

90 v

120 V

220 V

30 V

Solution

120 V

The magnetic flux linked with the primary coil is given by

$ϕ = ϕ_{o} + 4 t$

So voltage across primary

$V_{P} = \frac{dϕ}{dt} = \frac{d}{dt} (ϕ_{o} + 4 t) V_{P} = 4 volt (as ϕ_{o} = constant)$

Also we have

N_P = 50 and N_S = 1500

From relation

$\frac{V_{S}}{V_{P}} = \frac{N_{S}}{N_{P}} \Rightarrow V_{S} = V_{P} \frac{N_{S}}{N_{P}} \Rightarrow V_{S} = 4 (\frac{1500}{50}) \Rightarrow V_{S} = 120 V$

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