Practice Problems (KEY)

1. Answer the following questions about the circuit below

a. Find I~s~:

\[\mathbf{R}_{\mathbf{3,4,5}}\mathbf{= 3}\mathbf{\Omega}\mathbf{+ 4}\mathbf{\Omega}\mathbf{+ 5}\mathbf{\Omega}\mathbf{= 12}\mathbf{\Omega}\]

\[\mathbf{R}_{\mathbf{2,3,4,5}}\mathbf{=}\frac{\mathbf{R}_{\mathbf{2}}\mathbf{*}\mathbf{R}_{\mathbf{3,4,5}}}{\mathbf{R}_{\mathbf{2}}\mathbf{+}\mathbf{R}_{\mathbf{3,4,5}}}\mathbf{=}\frac{\mathbf{2}\mathbf{\Omega}\mathbf{*12}\mathbf{\Omega}}{\mathbf{2}\mathbf{\Omega}\mathbf{+ 12}\mathbf{\Omega}}\mathbf{= 1.71}\mathbf{\Omega}\]

\[\mathbf{R}_{\mathbf{eq}}\mathbf{=}\mathbf{R}_{\mathbf{1}}\mathbf{+}\mathbf{R}_{\mathbf{2,3,4,5}}\mathbf{+}\mathbf{R}_{\mathbf{6}}\mathbf{= 1}\mathbf{\Omega}\mathbf{+ 1.71}\mathbf{\Omega}\mathbf{+ 6}\mathbf{\Omega}\mathbf{= 8.71}\mathbf{\Omega}\]

\[\mathbf{I}_{\mathbf{S}}\mathbf{=}\frac{\mathbf{V}_{\mathbf{S}}}{\mathbf{R}_{\mathbf{eq}}}\mathbf{=}\frac{\mathbf{5}\mathbf{V}}{\mathbf{8.71}\mathbf{\Omega}}\mathbf{= 574}\mathbf{mA}\]

b. Find the power consumed by R~2~

Current Divider, where \(\mathbf{R}_{\mathbf{eq}}\) will be the equivalent resistance given by the portion of the circuit that is in parallel (\(\mathbf{R}_{\mathbf{2,3,4,5}}\)):

\[\mathbf{I}_{\mathbf{2}}\mathbf{=}\mathbf{I}_{\mathbf{S}}\frac{\mathbf{R}_{\mathbf{eq}}}{\mathbf{R}_{\mathbf{2}}}\mathbf{=}\mathbf{I}_{\mathbf{S}}\frac{\mathbf{R}_{\mathbf{2,3,4,5}}}{\mathbf{R}_{\mathbf{2}}}\mathbf{= (574}\mathbf{mA)*}\frac{\mathbf{1.71}\mathbf{\Omega}}{\mathbf{2}\mathbf{\Omega}}\mathbf{= 491}\mathbf{mA}\]

\[\mathbf{P}_{\mathbf{2}}\mathbf{=}\mathbf{I}_{\mathbf{2}}^{\mathbf{2}}\mathbf{R}_{\mathbf{2}}\mathbf{=}{\mathbf{(0.491}\mathbf{A)}}^{\mathbf{2}}\mathbf{*2}\mathbf{\Omega}\mathbf{= 482}\mathbf{mW}\]

Or with voltage divider:

\[\mathbf{V}_{\mathbf{2}}\mathbf{=}\mathbf{V}_{\mathbf{S}}\mathbf{*}\frac{\mathbf{R}_{\mathbf{2,3,4,5}}}{\mathbf{R}_{\mathbf{2,3,4,5}}\mathbf{+}\mathbf{R}_{\mathbf{1}}\mathbf{+}\mathbf{R}_{\mathbf{2}}}\mathbf{= 5}\mathbf{V*}\frac{\mathbf{1.71}\mathbf{\Omega}}{\mathbf{8.71}\mathbf{\Omega}}\mathbf{= 982}\mathbf{mV}\]

\[\mathbf{P}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{V}_{\mathbf{2}}^{\mathbf{2}}}{\mathbf{R}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{0.982}\mathbf{V}}^{\mathbf{2}}}{\mathbf{2}\mathbf{\Omega}}\mathbf{= 482}\mathbf{mW}\]

2. 

Find \(R_{EQ}\), \(I_{S}\), \(P_{2}\), and \(P_{Total}\) for the following circuit

\[R_{eq} = R_{1} + R_{2,3} = 50\Omega + \left( \frac{1}{500\Omega} + \frac{1}{500\Omega} \right)^{- 1} = 50\Omega + 250\Omega = 300\mathrm{\Omega}\]

\[I_{S} = \frac{V_{S}}{R_{eq}} = \frac{1.5V}{300\Omega} = 5mA\]

\[P_{TOTAL} = I_{S}V_{S} = 1.5V*5mA = 7.5mW\]

\[I_{2} = \frac{R_{eq}}{R_{2}}\left( I_{S} \right) = \frac{R_{2,3}}{R_{2}}\left( I_{S} \right) = \frac{250\Omega}{500\Omega}*(5mA) = 2.5mA\]

\[P_{2} = I_{2}^{2}*R_{2} = (2.5mA)^{2}*500\Omega = 3.125mW\]

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3. 

Find the equivalent resistance of the following circuit and determine if the fuse rating is high enough

\[R_{eq} = \left( \frac{1}{50\Omega + 50\Omega} + \frac{1}{50\Omega} \right)^{- 1} = 33.3\mathrm{\Omega}\]

\[I_{S} = \frac{V_{S}}{R_{eq}} = \frac{15V}{33.3\Omega} = 450mA\]

\[No,\ fuse\ rating\ is\ not\ high\ enough!\]

4. 

   Determine the appropriate sized resistor so the voltage across the load is 45V.

\[R_{Adapter} = \frac{V_{Adapter}}{I_{s}}\]

\[V_{Adapter} = V_{s} - V_{Lamp} = 115V - 45V = 70V\]

\[R_{Adapter} = \frac{70V}{2A} = 35\mathrm{\Omega}\]

5. For resistors in series, a bigger resistor drops more voltage than a smaller resistor.

   a. True (b) False

6. For resistors in series, more current flows through the bigger resistor.

(a) True (b) False

7. In a current divider, the greater current always flows through the smaller resistor.

(a) True (b) False

8. For resistors in parallel, the equivalent resistance is always less than the smallest resistor.

(a) True (b) False

9. What is the purpose of a fuse? Choose all that apply.

(a) To keep a circuit from working under normal conditions.

(b) To allow a circuit to work under normal conditions.

(c) To protect the wiring of a circuit.

(d) To break the circuit during excessive currents.

10. Solve for the unknown parameter(s) in each of the following circuits.

a. Circuit #1

b. Circuit #2

11. Four 36 Ω light bulbs are connected in parallel to a 9-volt battery, as shown. How much power does the battery produce?

12. 

    A SATCOM control module is modeled as the circuit below. Which of the following is true? Select all that apply.

    a. I~1~ = I~2~

    b. V~2~ = V~3~

    c. I~2~ > I~3~

    d. V~1~ + V~3~ = V~S~

13. The proposed arming circuitry of an experimental munition is modeled as the following resistive circuit. Answer the following questions:

R~1~=600Ω R~2~=300Ω

Is

Vs R~4~=500Ω

a) What is Req as seen by the source, Vs?

b) What is the source voltage, V~S~?

c) How much power is being dissipated at R2?

d) What is the voltage across R4?

14. The circuit below shows a variable power supply capable of providing up to 50 V. Assume each bulb can be modeled by a 3 Ω resistor.

a) Calculate the maximum voltage setting on the power supply before the fuse blows.

b) Calculate the voltage for each light bulb assuming maximum current flow (i.e. just before the fuse blows).

15. The circuit below shows a TV modeled by a 5kΩ load and needing 50V to operate, connected up to a 120V source.

a) Given only 2kΩ and 3kΩ resistors (use as many

of each as you need), how could you create a voltage adapter to make this circuit work?

b) Given only 2kΩ resistors, how could

you create a voltage adapter to make this circuit work?