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)

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
$\phi=\phi_{0}+4t$
Voltage across the primary coil
$V_{p}=\frac{d \phi}{dt}$
$=\frac{d}{dt}(\phi_{0}+4t)=4V$
Also, for an ideal transformer
$\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}}$
$\therefore V_{S}=\left(\frac{N_{S}}{N_{P}}\right)\times V_{P}$
$V_{S}=\left(\frac{1500}{50}\right)\times4$
$=120\,V$