Question

The primary and secondary coils of a transformer have $50$ and $1500$ turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi=\phi_{0}+4 t$, where $\phi$ is in weber, $t$ is time in second and $\phi_{0}$ 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
$\phi=\phi_{0}+4\, t$
So, voltage across primary
$V_{p} =\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_{0}+4 t\right)$
$=4$ volt (as $\phi_{0}=$ constant)
Also, we have
$N_{p}=50$ and $N_{s}=1500$
From relation,
$\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}$
or $V_{s} =V_{p} \frac{N_{s}}{N_{p}}$
$=4\left(\frac{1500}{50}\right)$
$ = 120 \,V$