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 $\phi=\phi_{0}+4t$, where $\phi$ is in weber, $\phi$ 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)

120 V

Answer 2

The magnetic flux linked with the primary coil is given by $\phi=\phi_{0}+4 t$ So, voltage across primary $V_{p} =\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_{0}+4 t\right)$ $=4 V \quad\left(\text { as } \phi_{0}=\text { constant }\right)$ Also, we have $N_{p}=50$ and $N_{s}=1500$ From relation, $\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}$ or $V_{s}=V_{p} \frac{N_{s}}{N_{p}}=4\left(\frac{1500}{50}\right)$ $=120\, V$