Question

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

• A. 90 V
• B. 120 V
• C. 220 V
• D. 30 V

Answer 1

Answer: (B)

The magnetic flux linked with the primary coil is given by
$\varphi ={\varphi }_0+4t$
So, voltage across primary
$V_p=\frac{d\varphi }{dt}=\frac{d}{dt}\left({\varphi }_0+4t\right)$
$=4$ volt (as ${\varphi }_0=$ constant)
Also, we have
$N_p=50$ and $N_s=1500$
From relation,
$\frac{V_s}{V_p}=\frac{N_s}{N_p}$
or $V_s=V_p\frac{N_s}{N_p}$
$=4\left(\frac{1500}{50}\right)$
$=120V$