Chapter 23 Electric Current
Exercises
5. Increase current the bulb glows brighter.
9. #5
11. Appliance uses energy. Converts electrical energy to another for – heat energy for example.
18. 95% of the energy is converted to heat. Electrical energy becomes Thermal Energy.
19. Thick wires have less resistance, will carry more current with less heat.
24. I=V/R Remember: R is constant.
With the same resistance the 220V will have more current than 110V.
26. Refer to Table 22.1 page 403. 0.1
 | Amp will cause electrocution is the path is hand across the heart to the other hand. |
| Hand to elbow does not cross the heart and is safer. |
29. First use Power = current * voltage and solve for current, then determine R=V/I
| 40W | 50W | |
| I=P/V | 40W/12V=3.33A | 50W/12V=4.17A |
| R=V/I | 12V/3.33A=3.6 Ω | 12V/4.17A=2.9 Ω |
Less resistance in the high power lamp due to a larger filament; More Power because there is more current.
31. Series. Req is the sum of the resistors, so Req > R1
32. Parallel. Req will always be less than the least amount, so Req < R1
40. Electrons in a circuit move not by collisions, but by interacting with an electric field (wave) set up in the conductor. The electric field moves at the speed of light.
46. C will have full voltage and more current. C will be brighter because A&B share voltage.
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If A is removed then A&B will not burn, but C will stay the same. |
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If C is removed then A&B will stay the same as before. |
47. All become dimmer if added while connected in series. Stay same brightness if added while connected in parallel. Use battery power faster in parallel because more current.
48. More devices in series: the circuits total current (line current) will decrease. More devices in parallel: the circuits total current will increase.
60. In parallel: I=P/V both have the same voltage. Note the Resistance is constant, whether in series or parallel.
| Parallel | 60W | 100W |
| I=P/V | 60W/120V = 0.5A | 100W/120V = 0.83A |
| R=V/I | 120V/0.5A = 240 Ω | 120V/0.83A = 144.6 Ω |
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60W is dimmer - less power and less Current. |
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100W is brighter - more power and more Current. |
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Both have the same Voltage drop (120V). |
In series: V=P/I both have the same current I=V/R = 120V/384.6 Ω = 0.31A The voltage will be split and the power rating will changed based upon the new voltage.
| Series | 60W | 100W |
| V=I * R | 0.312A * 240 Ω = 74.88V | 0.312A * 144.6 Ω = 45.12V |
| P=I * V | 0.312A * 74.88V = 23W | 0.312A * 45.12V = 14W |
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60W has more resistance and more Voltage drop. |
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100W has less resistance and less Voltage drop. |
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60W has a higher power rating due than the 100W due to the new higher voltage, so 60W will be brighter than the 100W bulb in series. |
Problems
1. I=P/V=60W/120V=0.5A
2. R=V/I=120V/20A=6 Ω
3. I=P/V=1200W/120V=10A
4. 60 Amp-hr battery
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(3A/headlight)*(2headlights)=6A |
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60Amp-hr/6A=10 hrs |
5. 100W = 0.1kW @ 15 cents/kW-hr=$0.15/kW-hr for 1 week.
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($0.15/kW-hr)*(0.1kW)*(24hrs/day)*(7days/week)=$2.52/week or $131.40/yr |
6. 4W in 120V @ 15 cents/kW-hr for 1 year.
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Low Current: I=P/V=4W/120V=0.033A |
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High Resistance: R=V/I=120V/0.033A=3,636.4 Ω |
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(.004 kW)*(24hrs/day)*(365days/yr)=35.04 kW-hr/yr |
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($0.15/kW-hr)*(.004 kW)*(24hrs/day)*(365days/yr)=$5.26/yr |
7. 1 Watt = 1 Joule/sec
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P=V*I=110V*9A=990W |
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990W=990J/s |
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(990J/s)*(60s/minute)=59,400J/min |
8. 1 Amp = 1 Coulomb/sec
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9A=9C/s |
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(9C/s)*(60s/min)=540C/min |
9. Find I=P/V, then check which V gives 95 Ω using R=V/I
| 120V | 220V | |
| I=P/V | 150W/120V = 1.25A | 150W/220V = 0.68A |
| R=V/I | 120V/1.25A = 96 Ω | 220V/0.68A = 324 Ω |
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In 120V a bulb with 95 Ω resistance would be rated at 150 Watts of power. |
| At 220V for a bulb to be rated 150 Watts of power it would need 324 Ω of resistance. |
10. Find the R of the toaster because it will be the same for both voltages.
| 120V | 108V | |
| R=V/I | 120V/10A = 12 Ω | 12 Ω |
| I=V/R | 120V/12 Ω = 10A | 108V/12 Ω = 9A |
| P=V * I | 120V * 10A = 1200W | 108V * 9A = 972W |
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Voltage decreased by 10% (120V to 108V) |
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Current decreased by 10% (10A to 9A) |
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Power decreased by 19% (1200W to 972W). You and the power company saved 19% power!! |