Question 15.10: Consider the source, transformer, and load shown in Figure 1...

Because of the applied source, the primary voltage is V_{1rms} = 110 V. The primary and secondary voltages are related by Equation 15.51:

V_{2rms} = \frac{N_2}{ N_1} V_{1rms}      (15.51)
V_{2rms} = \frac{N_2} {N_1} V_{1rms} = \frac{1} {5} × 110 = 22 V
Rearranging Equation 15.56, we have

I_{2rms} =\frac{ N_1} {N_2} I_{1rms}          (15.56)
I_{1rms} =\frac{ N_2} {N_1} I_{2rms}
a. With the switch open, the secondary current is zero. Therefore, the primary current I_{1rms} is also zero, and no power is taken from the source.
b. With the switch closed, the secondary current is
I_{2rms} = \frac{V_{2rms}}{ R_L} = \frac{22} {10} = 2.2 A

Then, the primary current is

I_{1rms} =\frac{ N_2} {N_1} I_{2rms} = \frac{1}{5} \times 2.2 = 0.44 A

Let us consider the sequence of events in this example. When the source voltage is applied to the primary winding, a very small primary current (ideally zero) flows, setting up the flux in the core. The flux induces the voltage in the secondary winding. Before the switch is closed, no current flows in the secondary. After the switch is closed, current flows in the secondary opposing the flux in the core. However, because of the voltage applied to the primary, the flux must be maintained in the core. (Otherwise, Kirchhoff’s voltage law would not be satisfied in the primary circuit.) Thus, current must begin to flow into the primary to offset the magnetomotive force of the secondary winding.