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Learn how to calculate potential changes around a closed loop, incorporate emf, terminal voltage, internal resistance in circuit calculations, and calculate electric power dissipated in circuit components. Understand the concept of electric power and its relation to work and energy.
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Today’s agenda: Potential Changes Around a Circuit. You must be able to calculate potential changes around a closed loop. Emf, Terminal Voltage, and Internal Resistance. You must be able to incorporate all of the above quantities in your circuit calculations. Electric Power. You must be able to calculate the electric power dissipated in circuit components, and incorporate electric power in work-energy problems. Examples.
Electric Power Last semester you defined power in terms of the work done by a force. We’d better use the same definition this semester! So we will. We focus here on the interpretation that power is energy transformed per time, instead of work by a force per time. The above equation doesn’t appear on your equation sheet, but it should appear in your brain.
However, we begin with the work aspect. We know the work done by the electric force in moving a charge q through a potential difference: Using the above, the work done by the electric force in moving an infinitesimal charge dq through a potential difference is: The instantaneous power, which is the work per time done by the electric force, is
Let’s get lazy and drop the in front of the V, but keep in the back of our heads the understanding that we are talking about potential difference. Then But wait! We defined I = dQ/dt. So And one more thing… the negative sign means energy is being “lost.” So everybody writes and understands that P<0 means energy out, and P>0 means energy in.
Also, using Ohm’s “law” V=IR, we can write P = I2R = V2/R. I can’t believe it, but I got soft and put P = I2R = V2/R on your starting equation sheet. Truth in Advertising I. The V in P=IV is a potential difference, or voltage drop. It is really a V. Truth in Advertising II. Your power company doesn’t sell you power. It sells energy. Energy is power times time, so a kilowatt-hour (what you buy from your energy company) is an amount of energy.
Example: an electric heater draws 15.0 A on a 120 V line. How much power does it use and how much does it cost per 30 day month if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh. For simplicity assume the current flows steadily in one direction. What’s the meaning of this assumption about the current direction? The current in your household wiring doesn’t flow in one direction, but because we haven’t talked about current other than a steady flow of charge, we’ll make the assumption. Our calculation will be a reasonable approximation to reality.
An electric heater draws 15.0 A on a 120 V line. How much power does it use. How much does it cost per 30 day month if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh.
How much energy is a kilowatt hour (kWh)? So a kWh is a “funny” unit of energy. K (kilo) and h (hours) are lowercase, and W (James Watt) is uppercase.
That’s a ton of joules! Good bargain for $17. That’s about 34,000,000 joules per dollar (or 0.0000029¢/joule). OK, “used” is not an SI unit, but I stuck it in there to help me understand. And joules don’t come by the ton. One last quibble. You know from energy conservation that you don’t “use up” energy. You just transform it from one form to another.
