Question

The primary and secondary coil of a transformer have $50$ and $1500$ turns respectively. If the magnetic flux $\varphi$ linked with primary coil is given by $\varphi ={\varphi }_0+4t$, where is in webers, $t$ is time in seconds and $\varphi$ is a constant, the output voltage across the secondary coil is:

• A. 150 V
• B. 90 V
• C. 120 V
• D. 180 V

Answer 1

The correct option is C 120 V

Magnetic flux linked with the primary coil $\varphi ={\varphi }_o+4t$
So, induced emf in the primary coil $|V_p|=\frac{d\varphi }{dt}=\frac{d}{dt}({\varphi }_o+4t)=4V$
Number of turns in primary coil $N_p=50$
Number of turns in secondary coil $N_s=1500$
So, output voltage in secondary coil $V_s=\frac{N_s}{N_p}{V}_p$
$⟹V_s=\frac{1500}{50}\times 4=120V$