Alternating Current: Practice Problem & Solution
47. The primary and secondary coil 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 Webers, t is time in seconds and $\phi_0$ is a constant, the output voltage across the secondary coil is: (2007)
Solution Explained:
To solve this problem, we apply the core principles of Alternating Current. Understanding the underlying formula is key to arriving at the correct answer below:
Voltage across primary is $V_p = \left| \frac{d\phi}{dt} \right| = \left| \frac{d}{dt}(\phi_0 - 4t) \right| = 4 V$.
Using the transformer ratio $V_s = V_p \left(\frac{N_s}{N_p}\right)$.
Secondary voltage $V_s = 4 \times \left(\frac{1500}{50}\right) = 4 \times 30 = 120 V$.
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