The primary and secondary coils of a transmformer 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

Options

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

Answer

To find the output voltage across the secondary coil of the transformer, we can follow these steps:

### Step 1: Determine the EMF induced in the primary coil
The magnetic flux linked with the primary coil is given by:
\[ \phi = \phi_0 + 4t \]
To find the EMF (Electromotive Force) induced in the primary coil, we use Faraday's law of electromagnetic induction, which states:
\[ \text{EMF} = -\frac{d\phi}{dt} \]
Differentiating the magnetic flux with respect to time \( t \):
\[ \frac{d\phi}{dt} = \frac{d}{dt}(\phi_0 + 4t) = 0 + 4 = 4 \text{ V} \]
Thus, the EMF induced in the primary coil \( E_P \) is:
\[ E_P = 4 \text{ V} \]

### Step 2: Use the turns ratio to find the EMF in the secondary coil
The relationship between the EMF in the primary coil and the EMF in the secondary coil is given by the turns ratio:
\[ \frac{E_S}{E_P} = \frac{N_S}{N_P} \]
Where:
- \( E_S \) is the EMF in the secondary coil
- \( E_P \) is the EMF in the primary coil
- \( N_S \) is the number of turns in the secondary coil (1500 turns)
- \( N_P \) is the number of turns in the primary coil (50 turns)

Substituting the values we have:
\[ \frac{E_S}{4} = \frac{1500}{50} \]
Calculating the right side:
\[ \frac{1500}{50} = 30 \]
Thus, we have:
\[ \frac{E_S}{4} = 30 \]

### Step 3: Solve for \( E_S \)
Multiplying both sides by 4 to find \( E_S \):
\[ E_S = 30 \times 4 = 120 \text{ V} \]

### Final Answer
The output voltage across the secondary coil is:
\[ \boxed{120 \text{ V}} \]

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