Practice Problems (KEY)

1. Plot the following equations:

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

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

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

d)

What equation is associated with the graph?

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

e) What equation is associated with the graph?

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

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

\[i(t) = 1 + \cos(360{^\circ}*60\ kHz*t)\ MA\]

2. If \(v_{s}(t) = 4V + 8\cos(360{^\circ}\ 10kt)\)V, graph i~x~(t) for the signal below

\[i_{X}(t) = \frac{v_{S}(t)}{R_{EQ}}\]

\[i_{X}(t) = \frac{4 + 8\cos(360{^\circ}*10kHz*t)V}{400\mathrm{\Omega}}\]

\[i_{X}(t) = 0.01 + 0.02\cos(360{^\circ}*10\ kHz*t)A\]

\[\mathbf{i}_{\mathbf{X}}\left( \mathbf{t} \right)\mathbf{= 10 + 20}\mathbf{\cos}\left( \mathbf{360{^\circ}*10}\mathbf{kHz*t} \right)\mathbf{mA}\]

* *

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

R~eq~ =\(\ \left( R_{3}ǁR_{2} \right) + R_{1} = \frac{1}{\frac{1}{500\ \Omega} + \frac{1}{500\ \Omega}} + 50\ \Omega = 300\ \mathrm{\Omega}\)

I~S~ = \(\frac{V_{S}}{R_{EQ}} = \frac{1.5\ V}{300\ \Omega} = 5\ mA\)

P~tot~ = \(I_{s}*V_{s} = .005\ A*1.5\ V = 7.5\ mW\)

4. Find the average power consumed by the circuit below given

v~s~(t) = 2.12 cos(360° * 100 kHz * t) V

V~S~(t) =2.12 cos(360° * 100 kHz * t) V

\[\mathbf{I}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{V}_{\mathbf{S}}}{\mathbf{R}_{\mathbf{EQ}}}\mathbf{=}\frac{\text{2.12~cos(360}\text{°}\text{ }\text{* }\text{100}\text{ kHz }\text{* }\text{t)}\text{ V}}{\mathbf{300}\mathbf{\Omega}}\mathbf{= 7.07\ }\text{cos(360}\text{°}\text{ }\text{*}\text{ }\text{100}\text{ kHz }\text{*}\text{ }\text{t)}\text{ }\mathbf{mA}\]

\[\mathbf{I}_{\mathbf{RMS}}\mathbf{=}\frac{\mathbf{7.07}}{\sqrt{\mathbf{2}}}\mathbf{= 5\ mA}\]

\[\mathbf{V}_{\mathbf{RMS}}\mathbf{=}\frac{\mathbf{2.12}}{\sqrt{\mathbf{2}}}\mathbf{= 1.5\ V}\]

\[\mathbf{P}_{\mathbf{AVG}}\mathbf{=}\mathbf{V}_{\mathbf{RMS}}\mathbf{*}\mathbf{I}_{\mathbf{RMS}}\mathbf{= 1.5}\mathbf{V*5\ mA = 7.5\ mW}\]

Compare the power consumed by the two circuits above. What can you say about the power consumed?

The power consumed is the same

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

   a. s

   b. 1/s

   c. f

   d. None of the above

6. A 9-V battery is connected to a resistor which consumes 7.22 mW of power. Which of the following AC sources would cause the same resistor to consume 7.22 mW of average power?

(a) 7.22 cos(360° * 2 kHz * t) mV

(b) 9 cos(360° * 2 kHz * t) V

(c) 7.22 mV~RMS~

(d) 9 V~RMS~

7. A B-52 generator produces a signal of v(t) = 290 cos(360° * 400 Hz * t) V.

a) Graph this signal as a function of time.

b) What is the RMS voltage for the generator?

8. The fuse for a 2000-pound general purpose bomb includes a spinner which produces an AC signal of v(t) = 15 cos(360° * 2 kHz t) mV. The arming circuit is modeled as three resistors, as shown below. Graph the current signal coming out of the spinner, I~S~(t).

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

10. The circuit below has a current source providing

\(i(t) = 11.75\cos(360{^\circ} \times 50\ Hz*t)A\).

Find v~1~(t).

11. An AC-powered electric fan, modeled as a 150-Ω resistor, is to be plugged into a standard 120 VRMS wall outlet. Since the fan requires 90 VRMS to operate, a resistor is added to form a voltage divider.

Radapter

Vs=120VRMS

Rfan=150Ω

Find the resistor value, R, to provide the fan 90 V~RMS~.