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 webers, `t` is time in second and `phi_(0)` is a constant, the output voltage across the secondary coil is

Options

A. 90 V
B. 120 V
C. 220 V
D. 30V

Answer

To solve the problem, we need to find the output voltage across the secondary coil of the transformer given the number of turns in the primary and secondary coils and the magnetic flux linked with the primary coil.

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns in the primary coil, \( N_P = 50 \)
- Number of turns in the secondary coil, \( N_S = 1500 \)
- Magnetic flux linked with the primary coil, \( \Phi = \Phi_0 + 4t \)

2. **Calculate the induced EMF in the primary coil:**
- The induced EMF (\( E_P \)) in the primary coil can be calculated using Faraday's law of electromagnetic induction, which states:
\[ E_P = -\frac{d\Phi}{dt} \]
- Differentiate the magnetic flux with respect to time:
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(\Phi_0 + 4t) = 0 + 4 = 4 \text{ volts} \]
- Therefore, the induced EMF in the primary coil is:
\[ E_P = 4 \text{ volts} \]

3. **Use the transformer equation to find the output voltage:**
- The relationship between the voltages and the number of turns in the primary and secondary coils is given by:
\[ \frac{E_S}{E_P} = \frac{N_S}{N_P} \]
- Rearranging this equation to find \( E_S \):
\[ E_S = E_P \cdot \frac{N_S}{N_P} \]
- Substitute the known values:
\[ E_S = 4 \cdot \frac{1500}{50} \]

4. **Calculate the output voltage:**
- Simplifying the equation:
\[ E_S = 4 \cdot 30 = 120 \text{ volts} \]

5. **Conclusion:**
- The output voltage across the secondary coil is:
\[ E_S = 120 \text{ volts} \]