Question

The current drawn by the primary coil of an ideal transformer, which steps up 22 V into 220 V, to operate a device having a load resistance of 110 \(\Omega\) is:

• A. 20 A
• B. 1 A
• C. 2 A
• D. 10 A

Hint:

For an ideal transformer, the current ratio is inversely proportional to the voltage ratio. Use the formula \( \frac{V_p}{V_s} = \frac{I_s}{I_p} \) to solve for the unknown current in either coil.

Answer 1

The Correct Option is A

We are given a transformer that steps up the voltage from 22 V to 220 V, and the load resistance is 110 \(\Omega\). We need to find the current drawn by the primary coil of the transformer. Step 1: Use the Transformer Equation The voltage and current relationship for an ideal transformer is given by: \[ \frac{V_p}{V_s} = \frac{I_s}{I_p} \] Where: - \( V_p \) is the primary voltage, - \( V_s \) is the secondary voltage, - \( I_p \) is the current in the primary coil, - \( I_s \) is the current in the secondary coil. Given: - \( V_p = 22 \, \text{V} \), - \( V_s = 220 \, \text{V} \). We need to find the current in the primary coil \( I_p \). Step 2: Find the Secondary Current First, we calculate the secondary current using Ohm’s law: \[ I_s = \frac{V_s}{R} \] Where \( R \) is the load resistance. Given: - \( R = 110 \, \Omega \), - \( V_s = 220 \, \text{V} \). Substitute the values: \[ I_s = \frac{220 \, \text{V}}{110 \, \Omega} = 2 \, \text{A} \] Step 3: Use the Transformer Equation to Find the Primary Current Now, using the transformer equation: \[ \frac{V_p}{V_s} = \frac{I_s}{I_p} \] Substitute the values: \[ \frac{22 \, \text{V}}{220 \, \text{V}} = \frac{2 \, \text{A}}{I_p} \] Solving for \( I_p \): \[ I_p = \frac{2 \, \text{A} \times 220 \, \text{V}}{22 \, \text{V}} = 20 \, \text{A} \] Thus, the current drawn by the primary coil is 20 (A) Therefore, the correct answer is Option (A): 20 (A)