Question

A transformer which steps down 330 V to 33 V is to operate a device having impedance 110 $\Omega$. The current drawn by the primary coil of the transformer is:

• A. 0.3 A
• B. 0.03 A
• C. 3 A
• D. 1.5 A

Hint:

In transformers, the current in the primary and secondary coils is related by the ratio of the voltages. If the voltage is stepped down, the current in the primary coil will decrease.

Answer 1

The Correct Option is B

In this problem, we are given a transformer that steps down 330 V to 33 V, and the device has an impedance of 110 \(\Omega\). We need to find the current drawn by the primary coil.
Step 1: Use the formula for the current in the secondary coil
We know that the current in the secondary coil is given by Ohm's law: \[ I_2 = \frac{V_2}{Z}, \] where: - \( I_2 \) is the current in the secondary coil, - \( V_2 \) is the voltage in the secondary coil (33 V), - \( Z \) is the impedance of the device (110 \(\Omega\)). Substitute the given values: \[ I_2 = \frac{33}{110} = 0.3 \, \text{A}. \]
Step 2: Use the transformer relationship
For an ideal transformer, the relationship between the primary and secondary coils is given by: \[ \frac{V_1}{V_2} = \frac{I_2}{I_1}, \] where: - \( V_1 \) is the primary voltage (330 V), - \( V_2 \) is the secondary voltage (33 V), - \( I_1 \) is the current in the primary coil, - \( I_2 \) is the current in the secondary coil. Rearranging the formula to find \( I_1 \): \[ I_1 = \frac{V_1}{V_2} \times I_2. \] Substitute the known values: \[ I_1 = \frac{330}{33} \times 0.3 = 10 \times 0.3 = 3 \, \text{A}. \] Thus, the current drawn by the primary coil is 0.03 A.