Question

What is the working principle of a transformer? The ratio of the number of turns in the primary and secondary coils in an ideal step-down transformer is 20 : 1. When input voltage of 250 V is applied, then the output current is 8 A. Calculate:
i) Current in the primary coil
ii) Output power

Hint:

In an ideal transformer, the power input to the primary coil is equal to the power output from the secondary coil. The voltage and current ratios depend on the turns ratio of the coils.

Answer 1

Working Principle of a Transformer:
A transformer works on the principle of mutual induction. It consists of two coils: the primary coil and the secondary coil, wound on a common iron core. When an alternating current is passed through the primary coil, it creates a changing magnetic field, which induces a voltage in the secondary coil. The voltage ratio between the primary and secondary coils is proportional to the ratio of the number of turns in each coil. For an ideal transformer, the input power is equal to the output power. The voltage and current are related by the following equations:
\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \] \[ \frac{I_p}{I_s} = \frac{N_s}{N_p} \] Where:
- \( V_p \) and \( V_s \) are the voltages in the primary and secondary coils,
- \( N_p \) and \( N_s \) are the number of turns in the primary and secondary coils,
- \( I_p \) and \( I_s \) are the currents in the primary and secondary coils.
Given: - Number of turns ratio \( \frac{N_p}{N_s} = 20:1 \),
- Input voltage \( V_p = 250 \, \text{V} \),
- Output current \( I_s = 8 \, \text{A} \).
Step 1: Calculate the current in the primary coil.
Using the current ratio equation:
\[ \frac{I_p}{I_s} = \frac{N_s}{N_p} = \frac{1}{20} \] Thus, the current in the primary coil is:
\[ I_p = \frac{I_s}{20} = \frac{8}{20} = 0.4 \, \text{A} \] So, the current in the primary coil is \( 0.4 \, \text{A} \).
Step 2: Calculate the output power.
The output power \( P_{\text{out}} \) is given by:
\[ P_{\text{out}} = V_s \times I_s \] From the voltage ratio, we can find the secondary voltage:
\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} = 20 ⇒ V_s = \frac{V_p}{20} = \frac{250}{20} = 12.5 \, \text{V} \] Thus, the output power is:
\[ P_{\text{out}} = 12.5 \, \text{V} \times 8 \, \text{A} = 100 \, \text{W} \] Thus, the output power is \( 100 \, \text{W} \).