Question 15.10: Analysis of a Circuit Containing an Ideal Transformer Consid...

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_{2\mathrm{rms}}=\frac{N_2}{N_1}V_{1\mathrm{rms}}=\frac{1}{5}\times 110=22 \mathrm{~V}

Rearranging Equation 15.56, we have

I_{2\mathrm{rms}}=\frac{N_1}{N_2}I_{1\mathrm{rms}}           (15.56)

I_{1\mathrm{rms}}=\frac{N_2}{N_1}I_{2\mathrm{rms}}

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_{2\mathrm{rms}}=\frac{V_{2\mathrm{rms}}}{R_L}=\frac{22}{10}=2.2\mathrm{~A}

Then, the primary current is

I_{1\mathrm{rms}}=\frac{N_2 }{N_1}I_{2\mathrm{rms}}=\frac{1}{5}\times 2.2=0.44\mathrm{~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 ow into the primary to offset the magnetomotive force of the secondary winding.