A loss free transformer has `500` turns on its primary winding and `2500` in secondary. The meters of the secondary indicate `200` volts at `8` amperes under these condition. The voltage and current in the primary is

Options

A. `100 V,16A`
B. `40 V,40A`
C. `160 V,10A`
D. `80 V,20A`

Explanation

To solve the problem, we will use the principles of a transformer, which relate the number of turns in the primary and secondary coils to the voltages and currents in those coils.

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns in the primary winding, \( N_p = 500 \)
- Number of turns in the secondary winding, \( N_s = 2500 \)
- Voltage in the secondary winding, \( V_s = 200 \) volts
- Current in the secondary winding, \( I_s = 8 \) amperes

2. **Use the transformer voltage ratio:**
The relationship between the primary and secondary voltages and the number of turns is given by:
\[ \frac{N_s}{N_p} = \frac{V_s}{V_p} \]
Substituting the known values:
\[ \frac{2500}{500} = \frac{200}{V_p} \]
Simplifying the left side:
\[ 5 = \frac{200}{V_p} \]

3. **Solve for \( V_p \):**
Rearranging the equation gives:
\[ V_p = \frac{200}{5} = 40 \text{ volts} \]

4. **Use the transformer current ratio:**
The relationship between the primary and secondary currents is given by:
\[ \frac{N_s}{N_p} = \frac{I_p}{I_s} \]
Rearranging gives:
\[ I_p = I_s \cdot \frac{N_p}{N_s} \]
Substituting the known values:
\[ I_p = 8 \cdot \frac{500}{2500} \]
Simplifying the fraction:
\[ I_p = 8 \cdot \frac{1}{5} = \frac{8}{5} = 1.6 \text{ amperes} \]

5. **Final results:**
- Voltage in the primary winding, \( V_p = 40 \) volts
- Current in the primary winding, \( I_p = 1.6 \) amperes

### Summary:
The voltage in the primary winding is \( 40 \) volts and the current in the primary winding is \( 1.6 \) amperes.