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(A) Given: Number of turns in the primary coil $N_p = 50$. Number of turns in the secondary coil $N_s = 1500$. The magnetic flux linked with the prima

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$120$

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(A) Given: Number of turns in the primary coil $N_p = 50$. Number of turns in the secondary coil $N_s = 1500$. The magnetic flux linked with the primary coil is $\phi = \phi_0 + 4t$.According to Faraday's law of electromagnetic induction,the induced voltage $(EMF)$ is given by $V = \frac{d\phi}{dt}$.Therefore,the voltage across the primary coil is $V_p = \frac{d}{dt}(\phi_0 + 4t) = 4\, V$.Using the transformer ratio formula,$\frac{V_s}{V_p} = \frac{N_s}{N_p}$.Substituting the values,$V_s = V_p 	imes \frac{N_s}{N_p} = 4 	imes \frac{1500}{50} = 4 	imes 30 = 120\, V$.

The primary and secondary coils of a transformer have $50$ and $1500$ turns respectively. If the magnetic flux $\phi = \phi_0 + 4t$,where $\phi$ is in webers,$t$ is time in seconds and $\phi_0$ is a constant,the output voltage across the secondary coil is......$V$

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