If the primary winding of a transformer has an applied voltage of 120V and a current of 1 amp, what is the maximum current available on the 30V secondary side of the transformer?

To understand why the maximum current available on the 30V secondary side of the transformer is 4 amps, it's important to apply the principle of conservation of energy and the equations for transformers.

Transformers operate on the principle of electromagnetic induction and are designed to convert voltage from one level to another while preserving power. The relationship between the primary and secondary voltages and currents can be described by the formula for transformers:

(Vp / Vs) = (Ip / Is)

Where:

• Vp is the primary voltage (120V in this case)

• Vs is the secondary voltage (30V in this case)

• Ip is the primary current (1 amp in this case)

• Is is the secondary current, which we are trying to find.

First, we calculate the turns ratio based on the voltages:

Turns ratio = Vp / Vs = 120V / 30V = 4.

This means that the voltage is stepped down by a factor of 4, leading to a corresponding increase in current on the secondary side. To find the secondary current, rearranging the transformer equation gives us:

Is = Ip * (Vp / Vs)

Substituting in the known values:

Is = 1 amp * (120V /