The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $\p

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, $t$ is time in seconds, and $\phi_0$ is a constant, the output voltage across the secondary coil is:

Accepted Answer

1. **Induced EMF in Primary ($V_p$):** According to Faraday's law of induction, the induced electromotive force (emf) is the rate of change of magnetic flux. Given the flux linked with the primary coil is $\phi = \phi_0 + 4t$, the induced voltage in the primary is:
   $$ V_p = \left| \frac{d\phi}{dt} ight| = \frac{d}{dt}(\phi_0 + 4t) = 4 	ext{ V} $$

2. **Transformer Equation:** For an ideal transformer, the ratio of voltages across the primary and secondary coils is equal to the ratio of their number of turns:
   $$ \frac{V_s}{V_p} = \frac{N_s}{N_p} $$
   Where:
   *   $V_s$ is the secondary voltage (output).
   *   $V_p = 4 	ext{ V}$ (primary voltage).
   *   $N_s = 1500$ (secondary turns).
   *   $N_p = 50$ (primary turns).

3. **Calculation:**
   $$ V_s = V_p 	imes \frac{N_s}{N_p} $$
   $$ V_s = 4 	imes \frac{1500}{50} $$
   $$ V_s = 4 	imes 30 $$
   $$ V_s = 120 	ext{ V} $$

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