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Electricity Part 2: Electric Current A conductor/dielectric is 1. A material in which charges are free of move about 2. A material in which charges are NOT free to move about 3. A material that shields a region from electric fields 0% 1 0% 2 0% 3 When a charged particle with charge, Q, is moved through a distance, d, in a uniform electric field E, the work done is given by 1. QE/d 2. Qd/E 3. QEd 0% 1 0% 2 0% 3 The Unit for electric fields is 1. 2. 3. 4. 5. 6. 7. Volt Newton/Coulomb Volt/meter All of the above 1 and 2 2 and 3 None of the above 0% 0% 0% 0% 0% 0% 0% 1 2 3 4 5 6 7 Electric current is the amount of charge moving past a point Maxwell Demon counting the charges passing a given point Definition of current I = current Q = amount of charge that passes point. t = time for charge to pass by. Q I t Units of Current Q Coulomb I t sec 1 Ampere (Amp) = 1 Coulomb/second Electric currents only flow in wires. 1. True 2. False 0% 1 0% 2 Examples of Currents Solar wind interacting with the earths magnetic field 1) Coronal Mass Ejection 2) Aurora from space 3) Aurora from ground CME animation Aurora from space Aurora effects 3 electron beams in a color TV Tokamak Fusion Experiments JET discharge We usually think of currents in wires Opening switch Simple circuit When a charged particle passes through the battery, it gains energy. When the particle passes through the light bulb it gives up the energy as heat. Ohm’s Law V=IR V= Voltage of the Battery. I=current in circuit. R=Resistance in the bulb/resistor. (Depends on materials and geometry.) Units of Resistance R=V/I (volts/amps) By definition, 1Ohm = 1 volt/amp, or 1=1V/A. Log Ride Analogy Water circuit analogy Example problem How many amps of current would flow in a light bulb that has a resistance of 60 if it is connected to a 12 V battery. I V/R 12V I 0.2 A 60 Power in a circuit When Charge Q passes through the battery it gains an amount of energy E=(Q)V (This is the amount of work the battery does on the charge.) If the charge takes an amount of time t to pass through the battery, the battery supplies a power of (does work at a rate of) E Q P V t t But Q I t Thus P IV The power supplied by the battery must be dissipated in the resistor. We also know the V=IR. Power dissipated in resistor P IV I ( IR) P I R 2 Example Calculation What is the resistance and how much current flows through a 100 W bulb? Note: The wattage on a bulb is its power output and assumes that you will use it in the US where the voltage in 110 V. I P /V I 100W 0 . 91 A 110V RV/I R 110V 0.91 A 121 Redo the calculations for a 60 W bulb I P /V I 60W 110V 0.55 A RV/I R 110V 0.55 A 200 If we want to increase the power output of a light bulb we should 1. Increase the resistance of the filament 2. Decrease the resistance of the filament 3. None of the above. 0% 1 0% 2 0% 3 If we were to use a bulb made for the US (where V=110V) and use it in Europe (V=220V) what would happen to the current in the bulb relative to68% the US value. 1. The current would double 2. The current would be cut it half 28% 3. The current would remain the same. 4% 1 2 3 What would happen to the power dissipated by the bulb? 1. It would quadruple 2. It would double 3. It would not change 4. It would be cut in half 5. It would be cut by a factor of 4. 44% 33% 15% 4% 1 2 3 4% 4 5