Question

A transformer of 100% efficiency has 200 turns in the primary and 40000 turns in the secondary. It is connected to a 220 V main supply and secondary feeds to a 100 K$\Omega$ resistance. The potential difference per turn is

• A. 11 V
• B. 18 V
• C. 25 V
• D. 1.1 V

Hint:

The transformer equation relates the number of turns and the voltage in the primary and secondary coils. Make sure to calculate the potential difference per turn by dividing the total potential difference by the number of turns in the secondary.

Answer 1

The Correct Option is D

The potential difference per turn in a transformer is given by the formula: \[ V_s = \frac{V_p \cdot N_s}{N_p} \] Where: 
- \( V_s \) is the potential difference in the secondary coil, 
- \( V_p \) is the potential difference in the primary coil (220 V), 
- \( N_s \) is the number of turns in the secondary coil (40000 turns), 
- \( N_p \) is the number of turns in the primary coil (200 turns). 
We know that the transformer is 100% efficient, meaning the input power equals the output power. So: \[ P = V_s \cdot I_s = V_p \cdot I_p \] Given that the secondary coil is connected to a \( 100 \, \text{K}\Omega \) resistance, we can calculate the current: \[ I_s = \frac{V_s}{R} = \frac{V_s}{100 \times 10^3} \] Now, we can find the potential difference per turn: \[ V_s = \frac{V_p \cdot N_s}{N_p} = \frac{220 \times 40000}{200} = 440 V \] The potential difference per turn is: \[ \frac{V_s}{N_s} = \frac{440}{40000} = 1.1 \, \text{V} \] 
Thus, the potential difference per turn is \( 1.1 \, \text{V} \).