The primary and secondary coils of a transformerhave 50 and 1500 turns respectively- If themagnetic flux linked with the primary coil isgiven by - - 0-4t- where - is in webers- t istime in seconds and -0 is a constant- the outputvoltage across the secondary coil is-

The correct option is C 120 voltsInduced emf in primary coil E_p=dϕ/dt=d/dt ( ϕ _0+4t )=4Volt Induced emf in secondary coil E_s/E_p=N_s/N_P⇒E_s/4=1500/ ...

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Byju's Answer

Standard XII

Physics

Flux as the Flow

The primary a...

Question

The primary and secondary coils of a transformer
have 50 and 1500 turns respectively. If the
magnetic flux linked with the primary coil is
given by $ϕ = ϕ_{0} + 4 t,$ where $ϕ$ is in webers, t is
time in seconds and $ϕ_{0}$ is a constant, the output
voltage across the secondary coil is:

30 volts

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90 volts

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120 volts

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220 volts

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Solution

The correct option is C 120 volts
Induced emf in primary coil
$E_{p} = \frac{d ϕ}{d t} = \frac{d}{d t} (ϕ_{0} + 4 t) = 4 V o l t$
Induced emf in secondary coil
$\frac{E_{s}}{E_{p}} = \frac{N_{s}}{N_{P}} \Rightarrow \frac{E_{s}}{4} = \frac{1500}{50} \Rightarrow E_{s} = 120 v o l t$

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The primary and secondary coils of a transformerhave 50 and 1500 turns respectively. If themagnetic flux linked with the primary coil isgiven by ϕ=ϕ0+4t, where ϕ is in webers, t istime in seconds and ϕ0 is a constant, the outputvoltage across the secondary coil is:

The correct option is C 120 voltsInduced emf in primary coilEp=dϕdt=ddt(ϕ0+4t)=4VoltInduced emf in secondary coilEsEp=NsNP⇒Es4=150050⇒Es=120volt

The primary and secondary coils of a transformer
have 50 and 1500 turns respectively. If the
magnetic flux linked with the primary coil is
given by $ϕ = ϕ_{0} + 4 t,$ where $ϕ$ is in webers, t is
time in seconds and $ϕ_{0}$ is a constant, the output
voltage across the secondary coil is:

The correct option is C 120 volts
Induced emf in primary coil
$E_{p} = \frac{d ϕ}{d t} = \frac{d}{d t} (ϕ_{0} + 4 t) = 4 V o l t$
Induced emf in secondary coil
$\frac{E_{s}}{E_{p}} = \frac{N_{s}}{N_{P}} \Rightarrow \frac{E_{s}}{4} = \frac{1500}{50} \Rightarrow E_{s} = 120 v o l t$