2. Circuits and Ohm’s Law SPH4UW

3. Electric Terminology Current: Moving Charges Symbol: I Unit: Amp  Coulomb/second Count number of charges which pass point/sec Power: Energy/Time Symbol: P Unit: Watt  Joule/second = Volt Coulomb/sec P = IV

4. Physical Resistor Resistance: Resistance: Traveling through a resistor, electrons bump into things which slows them down. R =  L /A  : L: A: Ohms Law I = V/R Double potential difference  double current A L Resistivity (E/J or Ωm ) Length Area

5. Understanding Two cylindrical resistors are made from the same material. They are of equal length but one has twice the diameter of the other. 1. 1. R 1 > R 2 2. 2. R 1 = R 2 3. 3. R 1 < R 2 21

6. Simple Circuit Practice… Practice… Calculate I (current) when  = 24 Volts and R = 8  R  I I Ohm’s Law: V =IR

7. Resistors in Series One wire: Effectively adding lengths: R eq =  (L 1 +L 2 )/A Since R  L, add resistance: RR = 2R R eq = R 1 + R 2

8. Resistors in Series Resistors connected end-to-end: If current goes through one resistor, it must go through other. I 1 = I 2 = I eq Both have voltage drops: V 1 + V 2 = V eq R1R1 R2R2 R eq R eq = V eq = V 1 + V 2 = R 1 + R 2 I eq I eq

9. Understanding Compare I 1 the current through R 1, with I 10 the current through R 10. 1. 1. I 1 < I 10 2. 2. I 1 = I 10 3. 3. I 1 > I 10 R 1 =1  00 R 10 =10  Compare V 1 the voltage across R 1, with V 10 the voltage across R 10. 1. V 1 >V 10 2. V 1 =V 10 3. V 1 <V 10 ACT: Series Circuit Note: I is the same everywhere in this circuit! V 1 = I 1 R 1 = 1 x I V 10 = I 10 R 10 = 10 x I V 1 = I 1 R 1 = 1 x I V 10 = I 10 R 10 = 10 x I

10. Practice: Resistors in Series = 22 volts. Calculate the voltage across each resistor if the battery has potential V 0 = 22 volts. R 12 = V 12 = I 12 = R 12 00 Expand: V 1 = V 2 = R 1 =1  00 R 2 =10  Check: V 1 + V 2 = V 12 ? Simplify (R 1 and R 2 in series): R 1 =1  00 R 2 =10  R 1 + R 2 = 11  V 1 + V 2 = 22 V I 1 =I 2 =V 12 /R 12 = 2 Amps I 1 R 1 = 2 x 1 = 2 Volts I 2 R 2 = 2 x 10 = 20 Volts Yes

11. Resistors in Parallel Two wires: Effectively adding the Area Since R  1/A add 1/R: R R = R/2

12. Resistors in Parallel Both ends of resistor are connected: Both ends of resistor are connected: Current is split between two wires: I 1 + I 2 = I eq Voltage is same across each: V 1 = V 2 = V eq R eq R2R2 R1R1

13. Understanding What happens to the current through R 2 when the switch is closed? 1) 1) Increases 2) 2) Remains Same 3) 3) Decreases What happens to the current through the battery? (1)Increases (2)Remains Same (3)Decreases ACT: Parallel Circuit V 2 = ε = I 2 R 2 I battery = I 2 + I 3

14. Practice: Resistors in Parallel Determine the current through the battery. Let E = 60 Volts, R 2 = 20  and R 3 =30 . R2R2 R3R3  R 23  Simplify: R 2 and R 3 are in parallel

15. Johnny “Danger” Powers uses one power strip to plug in his microwave, coffee pot, space heater, toaster, and guitar amplifier all into one outlet. 1. The resistance of the kitchen circuit is too high. 2. The voltage across the kitchen circuit is too high. 3. The current in the kitchen circuit is too high. Toaster Coffee PotMicrowave 10 A5 A10 A 25 A This is dangerous because… (By the way, power strips are wired in parallel.) Understanding

16. Resistors –PhysicalR =  L/A –Series R eq = R 1 + R 2 –Parallel1/R eq = 1/R 1 + 1/R 2 –Power P = IV Summary

17. Voltage Current Resistance Series Parallel Summary Different for each resistor. V total = V 1 + V 2 Increases R eq = R 1 + R 2 Same for each resistor I total = I 1 = I 2 Same for each resistor. V total = V 1 = V 2 Decreases 1/R eq = 1/R 1 + 1/R 2 Wiring Each resistor on the same wire. Each resistor on a different wire. Different for each resistor I total = I 1 + I 2 R1R1 R2R2 R1R1 R2R2

18. Understanding Which configuration has the smallest resistance? 1 2 3 123 Which configuration has the largest resistance? 1 2 3 R 2R R/2

19. Parallel + Series Tests Resistors R 1 and R 2 are in series if and only if every loop that contains R 1 also contains R 2 Resistors R 1 and R 2 are in parallel if and only if you can make a loop that has ONLY R 1 and R 2

20. Understanding Determine the voltage and current in each resistor R 3 =17   0 =10V R 4 =5  R 2 =25  R 1 =15  First we notice the voltage drop through these two resistor groupings total 10V. Let’s combine to find R EQ Let’s now add R 4 to this R EQ

21. Understanding R 3 =17   0 =10V R 4 =5  R 2 =25  R 1 =15   0 =10V R T =11.04  R4:R4: R4:R4: R1:R1: R1:R1: R2:R2: R2:R2: R3:R3: R3:R3: Since voltage drop has to total 10V, V EQ =10V-4.5V=5.5V Let’s determine the current, I T R 1, R 2, R 3, experience the same voltage This current will flow through each of the resistor groupings

22. 5Ω5Ω 7Ω7Ω 12V 10 Ω The V-I-R Chart VIR R1 R2 R3 Total Determine the Voltage, Current, and Resistance Step 1: Fill out the table with known resistors and the Total Voltage for circuit Step 1: Fill out the table with known resistors and the Total Voltage for circuit VIR R15 R27 R310 Total12

23. 5Ω5Ω 7Ω7Ω 12V 10 Ω The V-I-R Chart VIR R15 R27 R310 Total12 Determine the Voltage, Current, and Resistance Step 2: Using resistor laws, determine total resistance of circuit. Step 2: Using resistor laws, determine total resistance of circuit. VIR R15 R27 R310 Total129.117

24. 5Ω5Ω 7Ω7Ω 12V 10 Ω The V-I-R Chart Determine the Voltage, Current, and Resistance Step 3: Using Ohm’s law, determine the Current of circuit. Step 3: Using Ohm’s law, determine the Current of circuit. VIR R15 R27 R310 Total129.117 VIR R15 R27 R310 Total121.329.117 Step 4: Since initial current amount will also pass through resistor R1, we can determine its voltage drop. Step 4: Since initial current amount will also pass through resistor R1, we can determine its voltage drop. VIR R16.61.325 R27 R310 Total121.329.117

25. 5Ω5Ω 7Ω7Ω 12V 10 Ω The V-I-R Chart Determine the Voltage, Current, and Resistance Step 5: Since R2 and R3 have the same Voltage drop, we then must have 12V-6.6V=5.4V. Step 5: Since R2 and R3 have the same Voltage drop, we then must have 12V-6.6V=5.4V. VIR R16.61.325 R27 R310 Total121.329.117 VIR R16.61.325 R25.47 R35.410 Total121.329.117

26. 5Ω5Ω 7Ω7Ω 12V 10 Ω The V-I-R Chart Determine the Voltage, Current, and Resistance Step 6: Use ohm’s law to find currents Step 6: Use ohm’s law to find currents VIR R16.61.325 R25.47 R35.410 Total121.329.117 VIR R16.61.325 R25.40.777 R35.40.5410 Total121.329.117

27. Understanding R 3 =17   0 =10V R 4 =5  R 2 =25  R 1 =15  Let’s follow a conventional current path through R 1 You can pick any path through the circuit and the total voltage increases and decrease will balance You can reverse the direction of the current and thus the signs, (batteries increase the voltage, resistors drop the voltage) and obtain the same results.

28. Let us calculate the Current and the Power (used/generated) by the elements of the following circuit. What happens to the Power delivery and consumption if another identical bulb is place in parallel or in series with the first?

29. Let us calculate the Current and the Power (used/generated) by the elements of the following circuit when bulbs are in parallel. 2A 4A Because the bulbs (resistors) are in parallel, we use the parallel law to determine total resistance of the circuit.

30. Let us calculate the Current and the Power (used/generated) by the elements of the following circuit when bulbs are in series. 1A1A Because the bulbs (resistors) are in series, we use the series law to determine total resistance of the circuit.

31. Understanding v A a b J A voltmeter is a device that’s used to measure the voltage between two points in a circuit. An ammeter is used to measure current. Determine the readings on the voltmeter and the ammeter 2400V

32. Understanding v A a b J Let’s first combine the 2 parallel resistors Now we have 3 resistors in series Thus we can determine the current supplied by the battery 2400V

33. Understanding v A a b J At junction J this 6A current splits. Since the top resistor is twice the bottom resistor, half as much current will flow through it. Therefore the current through the top resistor is 2A and the bottom resistor is 4A 6A 2A 4A Thus the reading of the ammeter is 4A 2400V

34. Understanding v A a b J 6A 2A 4A The voltmeter will read a voltage drop of : Between points a and b Therefore the potential at point b is 900V lower than at point a 2400V