Q. 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 $\phi = \phi_{0} + 4t,$ 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) 30V
Explanation
(b) The magnetic flux linked with the primary coil is given by
$\phi = \phi_{0} + 4t$
So, voltage across primary
$\ \ \ \ \ \ \ \ \ \ V_{P} = \frac{d\phi}{dt} = \frac{d}{dt}(\phi + 4t)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4\ V(as\ \phi_{0} = constant)$
Also, we have
$\ \ \ \ \ \ \ \ \ N_{P} = 50\ and\ N_{S} = 1500$
From relation,
$\ \ \ \ \ \ \ \ \frac{V_{S}}{V_{P}} = \frac{N_{S}}{N_{P}}$
Or $\ \ \ V_{S} = V_{P}\frac{N_{S}}{N_{P}} = 4\left( \frac{1500}{50} \right) = 120V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.