Question

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

Answer 1

The maximum current available on the 30V secondary side of the transformer is 4 amps.

Step-by-step explanation:

To determine the maximum current available on the secondary side of a transformer, we need to use the concept of power in power out. The power on the primary side is given by the product of voltage and current, which is 120V x 1A = 120W. Since transformers are assumed to be 100% efficient in this case, the power on the secondary side will be the same, so we can set up an equation to solve for the current:

120W = 30V x I

Where I is the maximum current on the secondary side. Solving for I, we get:

I = 120W / 30V = 4A

Therefore, the maximum current available on the 30V secondary side of the transformer is 4 amps.