Current Electricity

Documents you may need:

Ohm's Law Lab (pdf or google doc) If you were absent complete this alternate (pdf or google doc)

Basic Circuit Explanations pdf

Genecon & bulbs Challenge (pdf or google doc) If you were absent complete this alternate (pdf or google doc)

Series Lab (pdf or google doc) If you were absent complete this alternate (pdf or google doc)

Parallel Lab (pdf or google doc) If you were absent complete this alternate (pdf or google doc)

Ch. 22 Review worksheet (pdf or google doc)

Ch. 23 Review worksheet (pdf or google doc)

Additional Review Materials:

Ohm's Law video

Series vs Parallel circuits - how they are different (pdf)

Series-Parallel combination circuits study guide (pdf)

Series-Parallel combination circuits talking tutorial #1 (pdf) and #2 (pdf) This is a Livescribe tutorial, you must download the file to play it with the audio.

Ch. 20-23 Review Guide (pdf or google doc)

Ch. 20-23 Review Guide answers (pdf) This is a Livescribe tutorial, you must download the file to play it with the audio. Look at it after you try the worksheet on your own.

Homework Help:

Ch. 23 Review & pg. 605 #28, pg. 612 #82, 86, 87, 89

11. b. Sum the resistances by adding them together. (60) c. Use this resistance (R) and voltage (V) to solve for the current (I) using V=IR. (2) d. Since it’s a series circuit the current through each resistor is the same but the voltage changes across each. Use the current (I) and the resistance (R) to solve the voltage (V) using V=IR. (40)

12. b. Sum the reciprocals of the resistors to get the reciprocal of the total equivalent resistance. (>5) c. Given this resistance (R) and voltage (V) solve for the current (I) using V=IR. (6)

13. b. First find the equivalent resistance. (20) Just like today’s warm-up. Using this resistance (R) and voltage (V) solve for the current (I) using V=IR. (6) c. Since the two 10 ohm resistors are in parallel they have the same amount of voltage going through them. Use their equivalent resistance (R) and the current you found in (b) to solve for the voltage across the two parallel resistors. (30) Now that you know the voltage across the parallel part, you can use V=IR with the resistance (R) to solve for the current (3). Since the two resistors in parallel are the same, it makes sense that the current splits equal between them. d. With that resistance (R) and current (I) solve for the voltage (V) using V=IR. (90)

14. Since its hooked up to the wall the voltage is 120V. Sum the reciprocals of the resistors to get the reciprocal of the total equivalent resistance.(>4) With that resistance (R) and voltage (V) to solve or the current (I) using V=IR. (<30)

15. Just like the warm-up but more complex. Solve for the equivalent resistance of the four resistors in parallel (10 ohm, 10 ohm, 5 ohm and 15 ohm). (>2) Next, solve for the equivalent resistance of the two resistors in parallel along the bottom (10 ohm and 30 ohm). (7.5) Now you have a series circuit with 20, >2, 10, and 7.5 ohms. Sum the resistors to get the total resistance. (<40)

28. Use the current and the voltage to find the power, then use the power in kW and the time in hours to find the energy in kWh. The total cost divided by the amount of energy is the cost per kWh. (>30)

82. Use the current and the voltage to find the power, then use the power in kW and the time in hours to find the energy in kWh. The total cost divided by the amount of energy is the cost per kWh. (>500)

86. Use the voltage and the resistance to find the power (P=V^2/R which is a combination of V=IR and P=IV) and then use the power and the time to find the Energy in kWh. Make sure that you convert 30days to hours and remember that the heater is on only 1/4 the time. Multiply the total amount of energy in kWh by the price per kWh to find the total cost. (>200)

87. First, find out how many kWh $50 will buy if every kWh costs you $0.09. Set this amount of energy equal to E=Pt; you know the time (30 days/2) and solve for power. Convert the power to Watts and use P=IV to solve for the current. (>12)

89. First, find out how many kWh $5 will buy if every kWh costs you $0.15. Set this amount of energy equal to E=Pt; you know the power in kW and solve for time.( ~7days)

pg. 638 #82, 83

82. a. Find the equivalent resistance of four 240 ohm lightbulbs in parallel. The equivalent resistance should be less than 240 ohms. With that equivalent resistance and voltage, use Ohm's Law to solve for the current. (<3) b. Find the equivalent resistance of six 240 ohm lightbulbs in parallel. The equivalent resistance should be less than 240 ohms and even smaller than when four resistors were on. With that equivalent resistance and voltage, use Ohm's Law to solve for the current. (3) c. Find the equivalent resistance of six240 ohm lightbulbs and a heater in parallel. The equivalent resistance should be less than 240 ohms and even smaller than when six resistors were on. With that equivalent resistance and voltage, use Ohm's Law to solve for the current. (<10)

83. Refer to the section about fuses.