ΔV of battery. = ε - Ir or εmf = I(R+r) (for this particular series circuit) March 04, Emf and internal resistance. Emf and internal resistance

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1 Emf and internal resistance Emf and internal resistance ΔV of battery = ε - Ir or εmf = I(R+r) (for this particular series circuit) As the current in the circuit increases the voltage, supplied to the external circuit, by the battery goes down. Sources of electromotive force? Application temporary dimming of lights with large current draw

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5 I total = 3.14 A March 04, 2013

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7 I A Which Bulb is Brighter - Concepts I B What determines brightness? If all bulbs have the same resistance, order the bulbs in order of brightest to least bright? Can resistance of a bulb change? explain I D What happens to order of brightness if bulb B is removed? I C Will bulb D glow? why or why not? explain page 598

8 Adjustable or Variable Resistor in a circuit adjustable or variable resistor (dimmer switch) ΔV battery = 120 volts Variable resistor = R v Resistance of Light Bulb = 220 Ω Light Bulb - that emits light of varying brightness, due to the variable resistor (that limits current flow) What current would be required in the circuit to give the light bulb an output of 60 watts? What resistance would the variable resistor have to be? If the variable resistance doubled, what would happen to the current flow through the circuit? and the power output of the light bulb? (brightness) If I want to add a stereo to this circuit that required 120 volts all the time (even if I adjusted the variable resistor), how would I do it? explain? Adjustable or Variable Resistor in a circuit adjustable or variable resistor (dimmer switch) ΔV battery = 120 volts Variable resistor = R v Resistance of Light Bulb = 220 Ω Light Bulb - that emits light of varying brightness, due to the variable resistor (that limits current flow) What current would be required in the circuit to give the light bulb an output of 60 watts? p = I 2 R ( of light only) What resistance would the variable resistor have to be? find I V = IR eq, find R eq for full circuit, it is larger than 220, If the variable resistance doubled, what would happen to the current flow through the circuit? and the power output of the light bulb? (brightness) If I want to add a stereo to this circuit that required 120 volts all the time (even if I adjusted the variable resistor), how would I do it? explain?

9 Electrical Power and Mechanical Power P = W/Δt Rate at which work is done or energy is used, measured in watts (J/s) Power = Work/time Power can also be: P = F*v Work = F*d P = W/t = F*d/t or F*v Work is done in raising an elevator upward. The power consumption of the motor will be related to the rate at which work is done or, in this case, the rate at which the elevator is lifted. P = N*m/s or J/s or w (watt) Simple Resistor and Capacitor in Combination Circuits switch If ΔV = 10 volts, R= 5 ohms and C = 4 Farad What is charged stored on capacitor? (while the switch is closed) V R C What is the current through the resistor when the circuit is closed? What is the power of the bulb? (resistor) When the switch is opened R How long would the light bulb "run" with the energy stored in the capacitor at designed power rating of the light bulb? C What about this orientation? It is different than your initial logical thoughts.

10 Power Loss in high voltage lines March 04, (modified) A high-voltage transmission line with a resistance of 0.31Ω/km carries a current of 1000 A. The line is at a potential of 700kV at the power station and carries the current to a city located 160 km from the station. a. What is the power loss due to the resistance of the line? b. What total power is transmitted? c. What is the fraction of the transmitted power does this loss represent? fractional loss = power loss/ Power in d. What is the actual potential at the city (adjusting for loss)? e. If the distribution cost (cost related to this energy loss above) is dollars/ kwhr, how much would it cost to run this line for a month, (just for energy loss)? f. How much energy is lost (distribution costs) in transporting the energy each month in kwhr and Joules? (with additional) Power Loss in high voltage lines A high-voltage transmission line with a resistance of 0.31Ω/km carries a current of 1000 A. The line is at a potential of 700kV at the power station and carries the current to a city located 160 km from the station. a. What is the power loss due to the resistance of the line? P = I 2 R = 49,500,000 watts b. What total power is transmitted? P = IV = 1000A * 700,000 Volts = 700,000,000 watts c. What is the fraction of the transmitted power does this loss represent? fractional loss = power loss/ Power in 50 MW/ 700 MW = 7.1 % (loss) d. What is the actual potential at the city (adjusting for loss)? current remains the same, energy per charge goes down P = IV, 650,000 W = 1000A *V, V = 650 kvolts e. If the distribution cost (cost related to this energy loss above) is dollars/ kwhr, how much would it cost to run this line for a month, (just for energy loss)? dollars/kwhr * 50,000 kw*24hr*30 days/month =.95 million dollars/month (lost) f. How much energy is lost (distribution costs) in transporting the energy each month in kwhr and Joules?

11 Kirchoff's Laws Junction rule demonstrates conservation of? The loop rule demonstrates conservation of?

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