University:
Boston University
Course:
CAS PY 242 | Principles of General Physics 2
Academic year:

2022
Views:

440

Pages:

5
Author:

arrolo17w4

Complex Circuits, Effective Resistance, and Electrical Power: Physics 2 This lesson is to teach how to add resistors in series and parallel in order to determine an effective resistance. Why this resistance is important is then discussed. Finally, a demonstration was shown to show about power in series vs. parallel circuits. - You’ve learned Ohm’s Law: V = IR, now what do you do when you have more than 1 resistor in a circuit? - You’ve also learned that resistors in series add as Rtot = R1+R2+… - In addition resistors in parallel add as 1/Rtot = 1/R1+1/R2+… - But what do you do if you have a complex circuit with many different groups of resistors in both series and parallel? - You want to find the effective resistance o Because that will tell you the current coming out of the battery - How do you do this? o Start with a small simple chunk that you know how to work with that’s either all series or all parallel. o Then keep redrawing circuit at each step as it simplifies. o This is easy o Do following examples: (all values in examples are in Ohms) Practice Problem: - Find the effective Resistance of the following circuit. Then find what the current is coming out of the battery. Electical Power: Series vs. Parallel - Show demo of two lightbulbs: one 100 W and other 40 W. At first have them wired in parallel. This is how the students are used to seeing them wired in their house. They know that the 100W bulb is brighter than the 40 W bulb. - Now tell them that you’re going to wire them in series. Ask them what they think is going to happen. They’ll usually say that the 100 will still be brighter. - Turn them on: the 40W bulb is ridiculously brighter than the 100W bulb. - Note: works best with clear bulbs, otherwise you can’t see the 100W bulb light up at all. - Ask them why that happened? They’ll say they have no idea - Tell them they already have all they need to know to figure it out except for a couple equations - Electrical Power in a circuits is given by P = IV, or in the case of resistors such as lightbulbs, P = I2R. - Now, they can figure out what happened using the Power equations and the effective resistance that they just learned. Here are what the two circuits look like: Parallel Solution: Known: P = IV 100W bulb: P100 = I100V100 I100 = 100W/120V I100 = .833 A V100 = I100R100 R100 = V100/I100 R100 = 120V/.833A R100 = 144 Ω P = I2R V = IR 40W bulb: P40 = I40V40 I40 = 40W/120V I40 = .33 A V40 = I40R40 R40 = V40/I40 R40 = 120V/.33A R40 = 360 Ω Now, find Reff : 1/144 + 1/360 = 1/Reff Reff = 102.9Ω which means that I coming out of the battery is I = 1.16 A I = V/Reff = 120V/102.9Ω Notice that current doesn’t split evenly between bulbs. Now Series Solution: V = IR Again, Known: P = IV P = I2R Now, we also know the resistance in each bulb. So first find the Reff = R100 + R40 Reff = 144Ω + 360Ω = 504Ω Therefore, you can find the current going through the circuit: V = IR I = V/R Itot = 120V/504Ω Itot = .24 A Now, find the power in each light bulb: P40 = I2R40 P40 = .242*360Ω = 20.7W P100 = I2R100 P100 = .242*144Ω = 8.3W WOW, the power in the 100W bulb is much less than the 40W bulb, that’s why it’s not as bright! That’s pretty cool, huh, that you were able to solve this whole problem knowing what you know, which you didn’t think was much at first, but is actually very powerful!