EE215 Class Problems, Week 2 Solutions All questions are initially listed in black. After class, those that were covered in class using Top Hat (as class activities or shortly after homework) will be highlighted in blue. Questions remaining in black (after class) may be used for additional review. Solutions are noted in italics. Question #1 What is the minimum power rating required of either of the 4 ohm resistors in the
following circuit?
a.
b.
c.
1/4 Watt
1/2 Watt
1 Watt
Solution: By Voltage Divider: vx = k1/k2 * 10V
k1 = 4 || 4 = 2 ohms
k2 = 2 + (4 || 4) + (10 || 10) + 1 = 2 + 2 + 5 + 1 = 10
Therefore, vx = 2V
2
And the power dissipated by either 4 ohm resistor is: 2 /4 = 1W
Therefore, the resistor must have a power rating of at least 1W.
Question #2 You would like to heat your dorm room using a couple of electric radiators. You are not sure how you should wire the two radiators with the power supplied to the dorm. Which configuration do you think will keep you warmest? a.
b.
The circuit on the left The circuit on the right Solution: In the circuit to the left, each radiator dissipates 2402/Req (or 57,600/Req) where Req is the equivalent resistance of the radiator. In the circuit to the right, each radiator dissipates 1202/Req (or 14,400/Req) where Req is the equivalent resistance of the radiator. Clearly, the configuration on the left delivers more power (and heat). Question #3 It’s late at night and you and your lab partner in electronic circuits lab have one last connection to make on your circuit before you can go home for the night. Alas, you realize you have no jumper wires left in your lab kit. Eager to go home, you grab a rather long length of 24 AWG (THW insulated) wire that someone left on the table, neatly coiled (into 15 tidy, 2” diameter loops) and tied with a rubberband. You connect the wire in series with the 1mH inductor in your circuit. What is the effective inductance of the wire in microHenries? Solution: 24 AWG (THW insulated) wire has about 0.059” diameter: http://www.panduit.com/heiler/SelectionGuides/WW-­‐
WASG03%20Electrical%20Wire%20Sizes-­‐WEB%207-­‐7-­‐11.pdf If the wire is neatly coiled, we can assume no overlap and the length of the coil is simply: 0.059” diameter * 15 coils = 0.885” The number of coils = 15 and The diameter of the coil = 2” Using an inductance calculator (for a single layer coil; formula given on the calculator page): http://www.66pacific.com/calculators/coil_calc.aspx The equivalent inductance is about 12.6 microHenries. Question #4 In the following circuit, your goal is to deliver at least 2.0V to the load resistor RL for at least three hours using Duracell Ultra Power AA Batteries (google ‘Duracell Ultra Power AA Batteries datasheet’ to find specifics). Part A: What is the minimum value of the battery voltages (assuming all four AA batteries degrade at the same rate) before the load voltage drops below 2.0V? Solution: The voltage across RL, by voltage divider is equal to: Vsource*(400||400)/[(400||400) + 100 + 100] = Vsource * (200/400) = 0.5 Vsource Vsource has to be at least 4V for the voltage across RL to be 2V 4V/4 batteries = 1V per battery Part B: What is the maximum current that this circuit will draw from the batteries? Solution: When the batteries are fully charged and fresh, Vsource =6V And the equivalent resistance of the four resistors in the circuit is 400 ohms Therefore maximum current = 6/400 ohms = 15mA Part C: Approximately how long will the batteries last (= power the circuit as designed) at this maximum current level? Solution: The datasheet provides loading curves for 10mA and 25mA. Service Hours before the batteries drop below 1V = between 200 and 300 hours Part D: How many components can you eliminate from the circuit and still achieve at least 2V across the load resistance for between 200 and 300 hours? Solution: If we reduce the circuit to 2 batteries and eliminate all the peripheral resistors, we will have two batteries driving a 400 ohm resistor. The maximum current will be the “fresh” values of the two batteries in series (3V) divided by the 400 ohm resistance = 7.5mA. Looking at the datasheet, the batteries can supply at least 1V each for at least 300 hours at this current. Therefore, we can reduce the circuit from 4 batteries and 4 resistors to two batteries and 1 resistor. Part E: Suppose you make 1 million units a year and the average consumer depletes the batteries every six months. How many batteries did you rescue from a landfill every year? Solution: 1,000,000 * 4 batteries (2 batteries saved used up twice a year) = 4,000,000 batteries. Part F: Did you make any difference to the world with this effort? In the United States, we dispose of about 3 billion batteries a year. 4,000,000 batteries would make up 0.13% of that waste stream. Not bad for one company. Question #5 Is the following circuit valid? Solution: no, two voltage sources of unequal value cannot be in parallel. Question #6 Is the following circuit valid? Solution: yes, as shown the 2 9V sources are in parallel, ideal, and of the exact same value. Question #7 Is the following circuit valid? Assume the current sources are ideal. Solution: yes, current sources in series must be equal. Question #8 Is the following circuit valid? Solution: no, current sources in series must be equal. Question #9 What one value of resistance (in Ohms) can all the R1 resistors in the following circuit be replaced with, to most closely approximate the equivalent value of these resistors so that the load resistor “sees” a similar resistance? a. 2R1 b. 5/8 R1 c. 8/3 R1 d. None of the above Solution: b. 5/8 R1 Question #10 Which of the following inductance values could most closely “mimic” the effect of the three inductors in the following circuit. Assume the voltage source V has not yet reached steady state (and is time varying). a.
b.
c.
d.
1mH 2mH 3mH 4mH Solution: the equivalent inductance is 1.67mH, so the closest answer is b 2mH Question #11 What one value of resistance (in Ohms) can all the 1000 Ohm resistors in the following circuit be replaced with, so that the current i remains the same? Solution: [(1000 + 1000) || 1000 || 1000] + 1000 + 1000 = 2400 ohms Question #12 What is the minimum value of R (in ohms) such that the current through the LED can never exceed the 20mA maximum current rating of the LED? Solution: in the worst case scenario, the equivalent resistance of the LED is zero. Maximum tolerable current = 20mA = 3/(6 + R) – giving a minimum R of 144 ohms Question #13 What one value of resistance (in Ohms) can all the 1000 Ohm resistors in the following circuit be replaced with, so that the current i remains the same? Solution: The two 1,000 ohm resistors on the far right are in series = 2,000 Ohms The two 1,000 ohm resistors (one diagonal, one up/down) are also in parallel = 500 Ohms. This 500 ohm resistor is then in series with the 1000 ohm resistor connected to the + terminal of the dependent source = 1500 ohms. Then the 1,500 ohms is in parallel with 2,000 ohms = 860 ohms Finally, the 860 ohms is then in series with the 1,000 ohm resistor on the bottom of the circuit = 1860 ohms