Calculate current drawn by primary coil of a transformer, Which steps down 200 V to 20 V to operate a device of 20 ohm resistance. Assume efficiency of transformer 80 %.

Options

A. `0.125 A`
B. `0.225 A`
C. `0.325 A`
D. `0.425 A`

Explanation

To solve the problem of calculating the current drawn by the primary coil of a transformer that steps down 200 V to 20 V, we can follow these steps:

### Step 1: Write down the given quantities
- Primary Voltage (\(V_P\)) = 200 V
- Secondary Voltage (\(V_S\)) = 20 V
- Load Resistance (\(R_L\)) = 20 Ω
- Efficiency (\(\eta\)) = 80% = 0.8

### Step 2: Use the formula for efficiency
The efficiency of a transformer is given by the formula:
\[ \eta = \frac{P_{out}}{P_{in}} \times 100 \]
Where:
- \(P_{out} = V_S \times I_S\) (Output Power)
- \(P_{in} = V_P \times I_P\) (Input Power)

Rearranging the efficiency formula gives:
\[ \eta = \frac{V_S \times I_S}{V_P \times I_P} \]
Substituting the known values:
\[ 0.8 = \frac{V_S \times I_S}{V_P \times I_P} \]

### Step 3: Substitute the known values into the equation
Substituting \(V_S = 20 V\) and \(V_P = 200 V\):
\[ 0.8 = \frac{20 \times I_S}{200 \times I_P} \]

### Step 4: Simplify the equation
Cancelling out the zeros:
\[ 0.8 = \frac{I_S}{10 \times I_P} \]
Rearranging gives:
\[ I_P = \frac{I_S}{8} \]
This is our first equation.

### Step 5: Calculate the secondary current \(I_S\)
Using Ohm's Law, we can find the secondary current:
\[ I_S = \frac{V_S}{R_L} \]
Substituting the known values:
\[ I_S = \frac{20 V}{20 \Omega} = 1 A \]

### Step 6: Substitute \(I_S\) back into the equation for \(I_P\)
Now substituting \(I_S = 1 A\) into the equation we derived:
\[ I_P = \frac{1 A}{8} = 0.125 A \]

### Final Answer
The current drawn by the primary coil of the transformer is:
\[ I_P = 0.125 A \]