Question

The number of windings on the primary coil of a transformer

is 1.5 times greater than on the secondary coil. The primary
coil has a current of 3.0 A and a voltage of 12.0 V. Determine
the voltage and current on the secondary coil.

Answer 1

I. Vs = 8.0 Volts.

II. Is = 4.5 Amperes.

Step-by-step explanation:

Given the following data;

Np = 1.5Ns =
`\frac {N_(P)}{N_(S)} = 1.5` ..... equation 1

Ip = 3.0 A

Vp = 12 V

To find the voltage and current on the secondary coil;

I. For the voltage in the secondary coil (Vs), we would use the following formula;

`\frac {V_(P)}{V_(S)} = \frac {N_(P)}{N_(S)}` ...... equation 2.

Substituting eqn 1 into eqn 2, we have;

`\frac {V_(P)}{V_(S)} = 1.5`

`\frac {12}{V_(S)} = 1.5`

Cross-multiplying, we have;

`V_(S) * 1.5 = 12`

`V_(S) = \frac {12}{1.5}`

Vs = 8.0 V

II. For the current in the secondary coil (Is), we would use the following formula;

`\frac {I_(S)}{I_(P)} = \frac {N_(P)}{N_(S)}` .... equation 3

Substituting eqn 1 into eqn 3, we have;

`\frac {I_(S)}{I_(P)} = 1.5`

`\frac {I_(S)}{3.0} = 1.5`

Cross-multiplying, we have;

`I_(S) = 1.5 * 3.0`

Is = 4.5 A