The overall efficiency of a transformer is 90%. The transformer is rated for an output of 9000 watt. The primary voltage is 1000 vot. The ratio of turns in the primary to the secondary coil is 5: 1. The iron losses at full ioad are 700 watt. The primary coil has a resistance of 1 ohm
The voltage in secondary coil is:

Options

A. 100volt
B. 5000volt
C. 200volt
D. zero volt

Answer

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

### Step 1: Understand the given data
- Overall efficiency of the transformer, η = 90% = 0.9
- Output power (P_out) = 9000 W
- Primary voltage (V1) = 1000 V
- Turns ratio (N1:N2) = 5:1
- Iron losses = 700 W
- Resistance of primary coil (R1) = 1 Ω

### Step 2: Calculate the input power (P_in)
The efficiency of the transformer is given by the formula:
\[ \eta = \frac{P_{out}}{P_{in}} \]
Rearranging this gives:
\[ P_{in} = \frac{P_{out}}{\eta} \]
Substituting the values:
\[ P_{in} = \frac{9000 \, \text{W}}{0.9} = 10000 \, \text{W} \]

### Step 3: Calculate the total losses
The total losses in the transformer can be calculated as:
\[ \text{Total losses} = P_{in} - P_{out} \]
Substituting the values:
\[ \text{Total losses} = 10000 \, \text{W} - 9000 \, \text{W} = 1000 \, \text{W} \]

### Step 4: Calculate the copper losses
The copper losses can be calculated by subtracting the iron losses from the total losses:
\[ \text{Copper losses} = \text{Total losses} - \text{Iron losses} \]
Substituting the values:
\[ \text{Copper losses} = 1000 \, \text{W} - 700 \, \text{W} = 300 \, \text{W} \]

### Step 5: Calculate the current in the primary coil (I1)
Using the formula for power:
\[ P_{in} = V_1 \times I_1 + \text{Copper losses} \]
Rearranging gives:
\[ I_1 = \frac{P_{in} - \text{Copper losses}}{V_1} \]
Substituting the values:
\[ I_1 = \frac{10000 \, \text{W} - 300 \, \text{W}}{1000 \, \text{V}} = \frac{9700 \, \text{W}}{1000 \, \text{V}} = 9.7 \, \text{A} \]

### Step 6: Calculate the current in the secondary coil (I2)
Using the turns ratio:
\[ \frac{N_1}{N_2} = \frac{I_2}{I_1} \]
Given that \( \frac{N_1}{N_2} = 5 \), we can rearrange this to find \( I_2 \):
\[ I_2 = \frac{I_1}{5} \]
Substituting the value of \( I_1 \):
\[ I_2 = \frac{9.7 \, \text{A}}{5} = 1.94 \, \text{A} \]

### Step 7: Calculate the voltage in the secondary coil (V2)
Using the power formula for the secondary side:
\[ P_{out} = V_2 \times I_2 \]
Rearranging gives:
\[ V_2 = \frac{P_{out}}{I_2} \]
Substituting the values:
\[ V_2 = \frac{9000 \, \text{W}}{1.94 \, \text{A}} \approx 4645.4 \, \text{V} \]

### Step 8: Use the turns ratio to find V2 directly
Using the relationship between primary and secondary voltages:
\[ \frac{V_1}{V_2} = \frac{N_1}{N_2} \]
Substituting the known values:
\[ \frac{1000 \, \text{V}}{V_2} = 5 \implies V_2 = \frac{1000 \, \text{V}}{5} = 200 \, \text{V} \]

Thus, the voltage in the secondary coil is **200 V**.