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

Answer 1

Answer: (B)

Given :
Number o f turns in a primary coil $N_p=50$
Number o f turns in a secondary coil $N_s=1500$
Magnetic flux linked with primary
$\varphi ={\varphi }_0+4t$
Voltage across the primary coil
$V_p=\frac{d\varphi }{dt}$
$=\frac{d}{dt}({\varphi }_0+4t)=4V$
Also, for an ideal transformer
$\frac{V_S}{V_P}=\frac{N_S}{N_P}$
$\therefore V_S=\left(\frac{N_S}{N_P}\right)\times V_P$
$V_S=\left(\frac{1500}{50}\right)\times 4$
$=120V$