A transformer having efficiency of `90%` is working on `200 V` and `3 kW` power supply. If the current in the secondary coil is `6A`, the voltage across the secondary coil and current in the primary coil respectively are

Options

A. `300 V,15 A`
B. `450 V,15 A`
C. `450 V,13.5 A`
D. `600 V,15 A`

Explanation

To solve the problem step by step, we need to find the voltage across the secondary coil (Vs) and the current in the primary coil (Ip) of the transformer.

### Step 1: Understand the efficiency of the transformer
The efficiency (η) of a transformer is given by the formula:
\[ \eta = \frac{P_{out}}{P_{in}} \times 100 \]
where \(P_{out}\) is the output power and \(P_{in}\) is the input power.

Given that the efficiency is 90%, we can express this as:
\[ 0.90 = \frac{P_{out}}{P_{in}} \]

### Step 2: Calculate the input power (P_in)
The input power is given as 3 kW, which is equal to:
\[ P_{in} = 3000 \, W \]

### Step 3: Calculate the output power (P_out)
Using the efficiency formula, we can find the output power:
\[ P_{out} = \eta \times P_{in} = 0.90 \times 3000 = 2700 \, W \]

### Step 4: Relate output power to secondary voltage and current
The output power can also be expressed in terms of the secondary voltage (Vs) and current (Is):
\[ P_{out} = V_s \times I_s \]
Given that the secondary current (Is) is 6 A, we can substitute this into the equation:
\[ 2700 = V_s \times 6 \]

### Step 5: Solve for secondary voltage (Vs)
Now, we can solve for Vs:
\[ V_s = \frac{2700}{6} = 450 \, V \]

### Step 6: Calculate the primary current (Ip)
Now we need to find the primary current (Ip). The input power can also be expressed as:
\[ P_{in} = V_p \times I_p \]
where \(V_p\) is the primary voltage, given as 200 V. We can rearrange this to find Ip:
\[ I_p = \frac{P_{in}}{V_p} = \frac{3000}{200} = 15 \, A \]

### Final Results
Thus, the voltage across the secondary coil (Vs) is 450 V, and the current in the primary coil (Ip) is 15 A.

### Summary of Results
- Voltage across the secondary coil (Vs) = 450 V
- Current in the primary coil (Ip) = 15 A

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