The number of turns of the primary and the secondary coils of a transformer are 10 and 100 respectively. The primary voltage and the current are given as 2 V and 1 A. Assuming the efficiency of the transformer as 90%, the secondary voltage and the current respectively are

Options

A. 20 V and 0.1 A
B. 0.2 V and 1 A
C. 20 V and 0.09 A
D. 0.2 V and 0.9 A

Explanation

To solve the problem, we will use the principles of transformers and the given data. Here are the steps to find the secondary voltage and current:

### Step 1: Understand the given data
- Number of turns in the primary coil (N1) = 10
- Number of turns in the secondary coil (N2) = 100
- Primary voltage (V1) = 2 V
- Primary current (I1) = 1 A
- Efficiency (η) = 90% = 0.9

### Step 2: Use the efficiency formula
The efficiency of a transformer is given by the formula:
\[ \eta = \frac{P_{out}}{P_{in}} = \frac{V_s \cdot I_s}{V_p \cdot I_p} \]
Where:
- \(P_{out}\) = output power
- \(P_{in}\) = input power
- \(V_s\) = secondary voltage
- \(I_s\) = secondary current
- \(V_p\) = primary voltage
- \(I_p\) = primary current

### Step 3: Rearranging the efficiency equation
From the efficiency equation, we can express the output power in terms of the input power:
\[ 0.9 = \frac{V_s \cdot I_s}{2 \cdot 1} \]
This simplifies to:
\[ V_s \cdot I_s = 1.8 \quad \text{(1)} \]

### Step 4: Use the transformer voltage ratio
The voltage ratio of a transformer is given by:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
Substituting the known values:
\[ \frac{V_s}{2} = \frac{100}{10} = 10 \]
Thus:
\[ V_s = 2 \cdot 10 = 20 \, \text{V} \quad \text{(2)} \]

### Step 5: Substitute \(V_s\) into equation (1)
Now we can substitute \(V_s\) from equation (2) into equation (1):
\[ 20 \cdot I_s = 1.8 \]
Solving for \(I_s\):
\[ I_s = \frac{1.8}{20} = 0.09 \, \text{A} \quad \text{(3)} \]

### Final Results
From our calculations:
- The secondary voltage \(V_s = 20 \, \text{V}\)
- The secondary current \(I_s = 0.09 \, \text{A}\)

### Summary
Thus, the secondary voltage and current are:
- **Secondary Voltage (V_s)**: 20 V
- **Secondary Current (I_s)**: 0.09 A

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