A transformer with turns ratio `(N_(1))/(N_(2))=(50)/(1)` is connected to a 120 volt AC supply. If primary and secondary circuit resistance are `1.5kOmega and 1Omega` respectively then find out power out put.

Options

A. 5.76 W
B. 11.4 W
C. 2.89 W
D. 7.56 W

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Turns ratio: \( \frac{N_1}{N_2} = \frac{50}{1} \)
- Primary voltage (\( V_1 \)): 120 V
- Primary resistance (\( R_1 \)): 1.5 kΩ = 1500 Ω
- Secondary resistance (\( R_2 \)): 1 Ω

### Step 2: Calculate the secondary voltage (\( V_2 \))
Using the turns ratio, we can relate the primary and secondary voltages:
\[ \frac{N_1}{N_2} = \frac{V_1}{V_2} \]
Substituting the known values:
\[ \frac{50}{1} = \frac{120}{V_2} \]
Cross-multiplying gives:
\[ 50 \cdot V_2 = 120 \]
\[ V_2 = \frac{120}{50} = 2.4 \text{ V} \]

### Step 3: Calculate the power output (\( P \))
The power output can be calculated using the formula:
\[ P = \frac{V_2^2}{R_2} \]
Substituting the values we found:
\[ P = \frac{(2.4)^2}{1} \]
Calculating \( (2.4)^2 \):
\[ (2.4)^2 = 5.76 \]
Thus, the power output is:
\[ P = 5.76 \text{ W} \]

### Final Answer
The power output of the transformer is **5.76 W**.

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