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, then the output voltage across the secondary coil is:

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

Answer 1

Answer: (B)

Voltage across primary coil
$V_p=\frac{d\varphi }{dt}=\frac{d}{dt}\left({\varphi }_0+4t\right)=4V$
(as ${\varphi }_0=$ constant)
Also, we have
$N_p=50$ and $N_s=1500$
From relation,
$\frac{V_s}{V_p}=\frac{N_s}{N_p}$
$\Rightarrow V_s=V_p\frac{N_s}{N_p}=4\left(\frac{1500}{50}\right)$
$=120V$