Practice Problems (KEY)

Practice Problems (KEY)#

  1. Plot the following equations:

a)
\(v(t) = -2 + 2\cos(360^\circ \cdot 100\ \mathrm{Hz} \cdot t)\ \mathrm{mV}\)

b)
\(v(t) = 2\cos(360^\circ \cdot 10\ \mathrm{Hz} \cdot t)\ \mathrm{V}\)

c)
\(v(t) = 4\cos(360^\circ \cdot 1\ \mathrm{kHz} \cdot t)\ \mathrm{mV}\)

d)

What equation is associated with the graph?

\[ v(t) = 2\ \mathrm{V} + \cos(360^\circ \cdot 10\ \mathrm{kHz} \cdot t)\ \mathrm{V} \]

e) What equation is associated with the graph?

\[ v(t) = \cos(360^\circ \cdot 400\ \mathrm{kHz} \cdot t)\ \mathrm{mV} \]

f) What equation is associated with the graph (note: the period is 0.01667 ms)?

\[ i(t) = 1 + \cos(360^\circ \cdot 60\ \mathrm{kHz} \cdot t)\ \mathrm{mA} \]

  1. If
    \(v_{S}(t) = 4\ \mathrm{V} + 8\cos(360^\circ \cdot 10\ \mathrm{kHz} \cdot t)\ \mathrm{V}\),
    graph \(i_{X}(t)\) for the signal below.

\[ i_{X}(t) = \frac{v_{S}(t)}{R_{EQ}} \]
\[ i_{X}(t) = \frac{4 + 8\cos(360^\circ \cdot 10\ \mathrm{kHz} \cdot t)}{400\ \Omega} \]
\[ i_{X}(t) = 0.01 + 0.02\cos(360^\circ \cdot 10\ \mathrm{kHz} \cdot t)\ \mathrm{A} \]
\[ i_{X}(t) = 10 + 20\cos(360^\circ \cdot 10\ \mathrm{kHz} \cdot t)\ \mathrm{mA} \]

  1. Find the power consumed by the circuit below given \(V_{S} = 1.5\ \mathrm{V}\).

\[ R_{eq} = (R_{3} \parallel R_{2}) + R_{1} = \frac{1}{\frac{1}{500\ \Omega} + \frac{1}{500\ \Omega}} + 50\ \Omega = 300\ \Omega \]
\[ I_{S} = \frac{V_{S}}{R_{eq}} = \frac{1.5\ \mathrm{V}}{300\ \Omega} = 5\ \mathrm{mA} \]
\[ P_{tot} = I_{S} V_{S} = (0.005\ \mathrm{A})(1.5\ \mathrm{V}) = 7.5\ \mathrm{mW} \]

  1. Find the average power consumed by the circuit below given:

\(v_{S}(t) = 2.12\cos(360^\circ \cdot 100\ \mathrm{kHz} \cdot t)\ \mathrm{V}\)

\[ i_{S}(t) = \frac{v_{S}(t)}{R_{EQ}} = \frac{2.12\cos(360^\circ \cdot 100\ \mathrm{kHz} \cdot t)}{300\ \Omega} = 7.07\cos(360^\circ \cdot 100\ \mathrm{kHz} \cdot t)\ \mathrm{mA} \]
\[ I_{RMS} = \frac{7.07}{\sqrt{2}} = 5\ \mathrm{mA} \]
\[ V_{RMS} = \frac{2.12}{\sqrt{2}} = 1.5\ \mathrm{V} \]
\[ P_{AVG} = V_{RMS} I_{RMS} = (1.5\ \mathrm{V})(5\ \mathrm{mA}) = 7.5\ \mathrm{mW} \]

Compare the power consumed by the two circuits above.
The power consumed is the same.

  1. What is \(\frac{1}{\mathrm{Hz}}\) equivalent to?

a. s
b. 1/s
c. f
d. None of the above

  1. A 9 V battery is connected to a resistor that consumes 7.22 mW of power. Which AC source would cause the same resistor to consume 7.22 mW of average power?

a. \(7.22\cos(360^\circ \cdot 2\ \mathrm{kHz} \cdot t)\ \mathrm{mV}\)
b. \(9\cos(360^\circ \cdot 2\ \mathrm{kHz} \cdot t)\ \mathrm{V}\)
c. \(7.22\ \mathrm{mV}_{RMS}\)
d. \(9\ \mathrm{V}_{RMS}\)

  1. A B-52 generator produces the signal
    \(v(t) = 290\cos(360^\circ \cdot 400\ \mathrm{Hz} \cdot t)\ \mathrm{V}\).

a) Graph the signal as a function of time.

b) What is the RMS voltage for the generator?

  1. The fuse for a 2000-lb general-purpose bomb includes a spinner producing
    \(v(t) = 15\cos(360^\circ \cdot 2\ \mathrm{kHz} \cdot t)\ \mathrm{mV}\).
    The arming circuit is modeled as three resistors. Graph the current signal \(I_{S}(t)\).

  1. Which of the two sources below produces more average power?

  1. The circuit below has a current source providing
    \(i(t) = 11.75\cos(360^\circ \cdot 50\ \mathrm{Hz} \cdot t)\ \mathrm{A}\).

Find \(v_{1}(t)\).

  1. An AC-powered electric fan, modeled as a 150 \(\Omega\) resistor, is plugged into a \(120\ \mathrm{V}_{RMS}\) outlet. The fan requires \(90\ \mathrm{V}_{RMS}\) to operate, so a resistor is added to form a voltage divider.

Given:

  • \(V_{S} = 120\ \mathrm{V}_{RMS}\)

  • \(R_{fan} = 150\ \Omega\)

Find the resistor value \(R\) to provide the fan with \(90\ \mathrm{V}_{RMS}\).