A step-down transformer has 79 turns in its primary coil and 15 turns in its secondary coil. The primary coil is connected to standard household voltage (e m frms = 120.0 V, 60.0 Hz). Suppose the secondary coil is connected to a 12.0-Ω resistor (a) What is the maximum current in the resistor?

I = 1.9A

Explanation:

The transformer has the relation given by the equation Np/Ns = Vp/Vs. Solving the equation with Np = 79 turns, Ns = 15 turns, Vp = 120V amd Vs = ?:

79 turns/15 turns = 120V/Vs

Vs = (120V) (15 turns) / 79 turns

Vs = 22.8V

To calculate the maximum current in the resistor we use the equation I = Vs/R:

I = 22.8V/12.0Ω

I = 1.9A