Question

A particular step-up transformer has 100 turns on the primary coil and 4,000 turns on the secondary coil. If the current in the secondary circuit is 3 A, what is the current in the primary coil?
Group of answer choices

30 A

3,200 A

120 A

40 A

Answer 1

To find the current in the primary coil of a step-up transformer, you can use the law of conservation of energy, which states that in an ideal transformer, the power in the primary coil equals the power in the secondary coil. This can be expressed as:

\[
V_p I_p = V_s I_s
\]

Where:
- \(V_p\) and \(I_p\) are the voltage and current in the primary coil, respectively.
- \(V_s\) and \(I_s\) are the voltage and current in the secondary coil, respectively.

Also, the voltage and turns ratio can be related by the following equation:

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

Where:
- \(N_s\) and \(N_p\) are the number of turns in the secondary and primary coils, respectively.

Given:
- \(N_p = 100\) turns (primary)
- \(N_s = 4000\) turns (secondary)
- \(I_s = 3\) A (current in the secondary)

First, let's determine the turns ratio:

\[
\frac{N_s}{N_p} = \frac{4000}{100} = 40
\]

This means the secondary voltage is 40 times the primary voltage:

\[
\frac{V_s}{V_p} = 40
\]

Because it's a step-up transformer, we can also say that:

\[
\frac{I_s}{I_p} = \frac{N_p}{N_s}
\]

Substituting the known values:

\[
\frac{I_s}{I_p} = \frac{100}{4000} = \frac{1}{40}
\]

Now we can solve for the primary current \(I_p\):

\[
I_p = I_s \times 40
\]

Substituting \(I_s = 3\) A:

\[
I_p = 3 \, \text{A} \times 40 = 120 \, \text{A}
\]

Therefore, the current in the primary coil is **120 A**.

The correct answer is **120 A**.