Question

The ratio of the number of turns of the primary to the secondary coils in an ideal transformer is 20:1. If 240 V AC is applied from a source to the primary coil of the transformer and a 6.0 \( \Omega \) resistor is connected across the output terminals, then the current drawn by the transformer from the source will be:

• A. 4.0 A
• B. 3.8 A
• C. 0.97 A
• D. 0.10 A

Hint:

In transformers, the ratio of the primary to secondary voltage is equal to the ratio of the number of turns in the coils.

Answer 1

The Correct Option is D

To solve this problem, we will use the transformer equations and Ohm's Law. We know that the ratio of turns in the primary coil (\(N_p\)) to the secondary coil (\(N_s\)) is 20:1, and an ideal transformer maintains the ratios of voltage and current according to these turns. Therefore, the voltage across the secondary coil (\(V_s\)) can be obtained using the formula: 

\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \] 

Given \(V_p = 240 \text{V}\), \(N_p:N_s = 20:1\), we find:

\[ V_s = \frac{V_p \times N_s}{N_p} = \frac{240 \times 1}{20} = 12 \text{V} \]

Now, using Ohm's Law (\(V = IR\)), where \( R \) is the resistance connected to the secondary coil (\(R = 6.0 \, \Omega\)), the current through the secondary coil (\(I_s\)) is:

\[ I_s = \frac{V_s}{R} = \frac{12}{6} = 2 \text{A} \]

An ideal transformer conserves power, so power in the primary coil (\(P_p\)) equals power in the secondary coil (\(P_s\)). Therefore:

\[ P_p = V_p \times I_p = P_s = V_s \times I_s \]

Solving for the primary current (\(I_p\)):

\[ I_p = \frac{V_s \times I_s}{V_p} = \frac{12 \times 2}{240} = 0.10 \text{A} \]

Thus, the current drawn by the transformer from the source is \(0.10 \, \text{A}\).

Correct Answer: 0.10 A