A transformer has 500 primary turns and 10 secondary turns If `V_(p)` is 120 V (rms) and a resistive load of `15 Omega` is connected across the secondary , the currents in the primary and secondary coil of transformer are -

Options

A. 3.2 mA, 0.16 A
B. 0.16 A, 3.2 mA
C. 3.2 mA, 3.2 mA
D. 0.16A, 0.16A

Answer

To solve the problem step by step, we will use the transformer equations and Ohm's law.

### Step 1: Calculate the Secondary Voltage (Vs)
The formula for the secondary voltage in a transformer is given by:

\[ V_s = V_p \times \frac{N_s}{N_p} \]

Where:
- \( V_p = 120 \, \text{V (rms)} \) (primary voltage)
- \( N_p = 500 \) (number of primary turns)
- \( N_s = 10 \) (number of secondary turns)

Substituting the values:

\[ V_s = 120 \times \frac{10}{500} \]

Calculating this:

\[ V_s = 120 \times 0.02 = 2.4 \, \text{V (rms)} \]

### Step 2: Calculate the Secondary Current (Is)
Using Ohm's law, the current in the secondary coil can be calculated as:

\[ I_s = \frac{V_s}{R_s} \]

Where:
- \( R_s = 15 \, \Omega \) (resistive load)

Substituting the values:

\[ I_s = \frac{2.4}{15} \]

Calculating this:

\[ I_s = 0.16 \, \text{A} \]

### Step 3: Calculate the Primary Current (Ip)
The relationship between the primary and secondary currents in a transformer is given by:

\[ I_p = I_s \times \frac{N_s}{N_p} \]

Substituting the values:

\[ I_p = 0.16 \times \frac{10}{500} \]

Calculating this:

\[ I_p = 0.16 \times 0.02 = 0.0032 \, \text{A} = 3.2 \, \text{mA} \]

### Final Results
- The current in the primary coil \( I_p \) is \( 3.2 \, \text{mA} \).
- The current in the secondary coil \( I_s \) is \( 0.16 \, \text{A} \).

### Summary
- \( I_p = 3.2 \, \text{mA} \)
- \( I_s = 0.16 \, \text{A} \)

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