In a transformer the number of primary turns is four times that of the secondary turns. Its primary is connected to an AC source of voltage V. Then

Options

A. current through its secondary is about four times that of the current through its primary.
B. voltage across its secondary is about four times that of the voltage across its primary.
C. voltage across its secondary is about two times that of the voltage across the primary.
D. voltage across its secondary is about `(1)/(2sqrt2)` times of the voltage across its primary.

Answer

To solve the problem regarding the transformer, we will follow these steps:

### Step 1: Understand the relationship between primary and secondary turns
Given that the number of primary turns \( n_p \) is four times that of the secondary turns \( n_s \):
\[ n_p = 4n_s \]

### Step 2: Use the transformer voltage equation
The voltage relationship in a transformer is given by:
\[ \frac{V_p}{V_s} = \frac{n_p}{n_s} \]
Substituting \( n_p = 4n_s \) into the equation:
\[ \frac{V_p}{V_s} = \frac{4n_s}{n_s} = 4 \]
This implies:
\[ V_p = 4V_s \]

### Step 3: Determine the current relationship using power conservation
For an ideal transformer, the power input to the primary coil is equal to the power output from the secondary coil:
\[ P_p = P_s \]
This can be expressed as:
\[ V_p \cdot I_p = V_s \cdot I_s \]
Substituting \( V_p = 4V_s \) into the equation:
\[ 4V_s \cdot I_p = V_s \cdot I_s \]
Dividing both sides by \( V_s \) (assuming \( V_s \neq 0 \)):
\[ 4I_p = I_s \]
Thus, we can express the current in the secondary coil in terms of the primary coil:
\[ I_s = 4I_p \]

### Step 4: Conclusion
From the above calculations, we conclude that the current in the secondary coil is four times that of the current in the primary coil:
\[ I_s = 4I_p \]

### Final Answer
The current through the secondary is 4 times that of the current through the primary.

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