A transformer is having 2100 turns in primary and 4200 turns is secondary. An ac source of 120 V, 10 A is connected to its primary. The secondary voltage and current are

Options

A. 240 V, 5 A
B. 120 V, 10 A
C. 240 V, 10 A
D. 120 V, 20 V

Answer

To solve the problem step by step, we will use the transformer equations that relate the primary and secondary voltages and currents to the number of turns in the primary and secondary coils.

### Step 1: Identify the given values
- Number of turns in the primary coil, \( N_p = 2100 \)
- Number of turns in the secondary coil, \( N_s = 4200 \)
- Voltage in the primary coil, \( V_p = 120 \, \text{V} \)
- Current in the primary coil, \( I_p = 10 \, \text{A} \)

### Step 2: Use the transformer voltage equation
The relationship between the primary and secondary voltages and the number of turns is given by the formula:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
Where:
- \( V_s \) is the secondary voltage
- \( V_p \) is the primary voltage

Substituting the known values into the equation:
\[ \frac{V_s}{120} = \frac{4200}{2100} \]

### Step 3: Simplify the ratio of turns
Calculating the ratio of turns:
\[ \frac{4200}{2100} = 2 \]
Thus, the equation becomes:
\[ \frac{V_s}{120} = 2 \]

### Step 4: Solve for the secondary voltage
To find \( V_s \):
\[ V_s = 2 \times 120 = 240 \, \text{V} \]

### Step 5: Use the transformer current equation
The relationship between the primary and secondary currents and the number of turns is given by the formula:
\[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \]
Where:
- \( I_s \) is the secondary current
- \( I_p \) is the primary current

Substituting the known values:
\[ \frac{I_s}{10} = \frac{2100}{4200} \]

### Step 6: Simplify the ratio of turns for current
Calculating the ratio of turns:
\[ \frac{2100}{4200} = \frac{1}{2} \]
Thus, the equation becomes:
\[ \frac{I_s}{10} = \frac{1}{2} \]

### Step 7: Solve for the secondary current
To find \( I_s \):
\[ I_s = \frac{1}{2} \times 10 = 5 \, \text{A} \]

### Final Results
- Secondary Voltage, \( V_s = 240 \, \text{V} \)
- Secondary Current, \( I_s = 5 \, \text{A} \)

### Summary
The secondary voltage is 240 V and the secondary current is 5 A.

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