Homework - Basic Circuit Theory | ENEE 204, Study notes of Electrical and Electronics Engineering

Download Homework - Basic Circuit Theory | ENEE 204 and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! HW #1 ENEE 204 Prof. Lawson (020X) Due February 8th by 9:30AM Problem #1 - John is an electrical engineering student and Jasmine is a chemistry student. John doesn’t think anything important happens the first day of classes, so he skips his Basic Electric Circuit Theory class to go visit Jasmine. She says that a 25 W light bulb in her house is burned out and asks John if he has a spare. He says that he only has a 25 W bulb for a light in his car, but that he is certain it will work in her apartment since it has the same power rating. She says that she doesn’t think that sounds right, and so they make a bet. The loser has to clean the other person’s apartment. Who wins the bet and why? Note: To get full credit, you need to give a quantitative explanation based on electrical properties, INTEGRATED with a conceptual explanation; don’t JUST use equations. I think John would be cleaning! The power rating of the bulb is designed for a particular voltage in mind. Take any one of the light bulbs. Say, first I connect it to a 10V battery, and then to a 20 V battery. The bulb would be brighter for the 20V battery – it must be drawing more power! Just because there is a power rating on the bulb does not mean that it draws that same power in all situations. Mathematically (assuming the same resistance for the bulb) Ohm’s law would argue that the current through the bulb is doubled when it is connected to a 20V battery as opposed to a 10V one (V = IR; If V doubles, I doubles (for a constant R)). And since power consumed is given by P = VI – the bulb would consume 4 times as much power when connected to a 20V battery rather than a 10V one. So then, what does the power rating on the bulb mean? A 25W bulb is supposed to consume 25W of power for a specific value of the voltage. This specific voltage value is also specified on the bulb (typically as 25W, 110V or 25W, 220V etc.). So if the Voltage rating is different for two bulbs with the same power rating, they would draw different amounts of power when plugged into the same supply. Specifically in the case of John and Jasmine, the apartment probably has a supply voltage of 110V while the car battery for John would have a much lower voltage about 12-14V. By my first argument, since the car bulb draws 25W of power from the 12-14 V battery, if connected to the apartment outlet it would lead to a much larger power consumption than 25W, causing the car bulb to burn out. John gets to clean! Problem #2 -Hours later, Jasmine goes to study in the library while John is still cleaning. Soon he gets bored, and decides to take apart her 1200 W hairdryer, to see how long the heater wire is. John discovers that the wire is made of Nichrome AWG-20 wire. Nichrome has a conductivity of 106 Mhos/m (a) If two pieces of Nichrome wire are equally long but one piece is thicker (greater in diameter) than the other, which wire would you expect to have greater resistance? Give a conceptual explanation; don’t just quote an equation. I would expect the thicker wire to have a lower resistance. Resistance, is like how difficult is it for the free electrons in the wire to flow through it when there is an electric potential difference between the two ends of the wire. The more difficult it is for the electrons to flow, the higher the resistance. What drives the electrons to flow? The potential difference across the ends of the wire HW #1 ENEE 204 Prof. Lawson (020X) sets up a field inside the wire, the electrons respond to it, setting up a flow of electrons all through the wire. If the wire is thicker, there is more material and thus there are just more electrons that can flow (without changing anything else, such as the distance that the electrons have to flow etc.). In other words, for the same potential difference, I get a much more bang for the buck – in terms of the total charge that flows through the wire in one second. Interpreting resistance as ‘resistance to the flow of this charge’, an increased flow would imply lower resistance. (b) Now consider two pieces of Nichrome wire that are equally thick, but one is longer than the other. Which has higher resistance, i.e., which allows less current to flow when both wires are connected to the same voltage source? Again, give a conceptual explanation. Hint: In which wire is the electric field greater? Here, the longer wire will have a higher resistance. For a longer wire, the same potential difference would produce a smaller electric field inside the wire. Recall that mathematically, E = - (derivative of V with respect to distance). In other words, field is determined by how fast the potential is changing along a given direction with the sign in that equation giving the direction of the field. For a longer wire with the same potential difference, the average field value of the |E| = V/d is lower! So, the electrons in the longer wire feel a smaller push, leading to decreased flow (as if the wire were more clogged when longer) – which implies that the longer wire ahs more resistance. (c) The Nichrome wire in the hair dryer has a diameter of 0.032 in. When Jasmine returns home, how many feet of Nichrome wire are lying on her apartment floor? Assuming that V=110 V for her apartment, the power rating of 1200W gives a resistance of about 10 ohms for the dryer – Using P = VI and I = V/R we get R = V2/P. Now using the formula R =  L/A where  is the resistivity, L is the length and A is the cross- sectional area of the wire, we get L = .... [ Here  = 1/(conductivity) = 10-6 ohms/m; A = D2/4, where D = 0.032 in; and R 10 ohms] (d) Did the equations you used in part (c) “agree” with your qualitative explanations in part (a) and (b)? Explain. Yes the equation I used did agree with my qualitative reasoning in earlier parts. From part (a) I see that thicker the wire, lower the resistance. In other words, the resistance is inversely proportional to the cross-sectional area of the wire R  1/A From (b) I see that longer the wire, higher the resistance. In other words, resistance is directly proportional to the length R  L Both of these dependencies are also encoded in the equation that I used in part (c) R =  L/A. Problem #3 – John goes to Radio Hut and buys a 10 F capacitor. Jasmine goes too but she buys a 10 mH inductor. Each plans on connecting the part they bought to a mysterious voltage source that John’s Uncle Henry gave him, that has the following time dependence, once it is plugged in: HW #1 ENEE 204 Prof. Lawson (020X) (c) What is the MINIMUM power rating of the resistor necessary to avoid “overheating”? (Note, the answer could be more or less than one watt, depending on your answer to the previous part.) 72W (d) What is the MINIMUM power rating of the resistor necessary to achieve no heat increase whatsoever, while the resistor is plugged into the outlet. Explain your answer conceptually. Infinite – there will always be some heat generated – that’s what resistors do! (e) Even if the 1k resistor is way above the minimum required power capability, could anything else possibly go wrong when John plugs the resistor into the wall that could really get his mom smoking-mad? Explain your answer conceptually. Problem #5 – What is the energy stored in each of the components that John connected to the outlet in the previous problem (as a function of time)? Once John removes the inductor from the outlet, he flips it by 180 degrees and connects it again to the outlet (he has reversed the connections to the terminals). Now what is the time expression for the energy stored in the inductor? The stored energy is zero for the resistor. For the capacitor it is E = 0.5Cv(t)2 = 36(1+cos(754t)) mJ For the inductor it is E = 0.5Li(t)2 = 176(1-cos(754t)) J - this is unchanged after the inductor is flipped. Problem #6 – Ashanti’s Mother has an old current supply lying around the attic that supplies a constant current of 2 mA. While it’s very reliable, it can only supply a maximum power of 60W. If someone tries to draw more than 60W, the output current is reduced to zero and stays there until someone presses a “reset” button on the supply. When John comes to visit Ashanti, he brings his favorite 10 F capacitor, as usual. When he sees the old supply, he immediately takes out his capacitor, touches the two terminal leads together, then separates the leads and connects it to the current supply. He also connects the output terminals to an oscilloscope (also found lying around the attic), and then the turns the current source on. Plot the voltage that is displayed on the oscilloscope starting from when the supply is turned on. )( 150000,30 1500200 )(15030150*200)( tan0,;sec150 4.060)( )(4.0)(*)()( )(200002.010)( 1 )0()( 0 5 0 V st stt tvstforkVtv tconsaisvandithatafterondsT TTp Wttitvtp Vtddi C vtv tt             HW #1 ENEE 204 Prof. Lawson (020X)