Practice Problems (KEY)

Problem 1

\[V_{S,\text{RMS}} = \frac{V_{S,\text{pk}}}{\sqrt{2}} = \frac{120}{\sqrt{2}} = \boxed{84.85\,\text{V}_{\text{RMS}}}\]

\[a = \frac{V_1}{V_2} = \frac{120}{8} = \boxed{15}\]

\[P_{\text{payload}} = IV = (6)(20) = 120\,\text{W}\]

\[P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{120\,\text{W}}{0.9} = \boxed{133\,\text{W}}\]

Problem 2

\[a = \frac{V_1}{V_2} = \frac{120}{24} = \boxed{5}\]

\[P_{\text{load}} = IV = (6)(15) = 90\,\text{W}\]

\[P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{90\,\text{W}}{0.75} = 120\,\text{W}\]

\[I_S = \frac{P_{\text{in}}}{V_S} = \frac{120\,\text{W}}{120\,\text{V}} = \boxed{1\,\text{A}}\]

Problem 3

Use a transformer. This is an AC circuit, and transformers are designed for AC. Voltage adapters (resistive voltage dividers) work in both AC and DC circuits but dissipate power as heat — they cannot be 100% efficient. Transformers are treated as ideal (100% efficient) in this course, making them the correct and more efficient choice for stepping down AC voltage.

Problem 4

a. Diode

Problem 5

c. Smooths the output