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Electrical Current (5)

Mr. Klapholz Shaker Heights High School. Electrical Current (5). Problem Solving. Problem 1. A current of 0.24 A is in a wire for 2.0 s . How much charge went by a single point?. Solution 1. I = Q ÷ T Q = I × T Q = (0.24 A) × (2.0 s ) Q = 0.48 C. Problem 2.

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Electrical Current (5)

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  1. Mr. Klapholz Shaker Heights High School Electrical Current (5) Problem Solving

  2. Problem 1 A current of 0.24 A is in a wire for 2.0 s. How much charge went by a single point?

  3. Solution 1 I = Q ÷ T Q = I × T Q = (0.24 A) × (2.0 s) Q = 0.48 C

  4. Problem 2 If 1.5 V (a ‘D-cell’) is attached to a bulb with a resistance of 30.0 W, then how much current will be in the circuit?

  5. Solution 2 I = V / R I = (1.5 V) / (30.0 W) I = 0.05 A

  6. Problem 3 A 9.0 V battery is connected to a bulb. How much energy gets turned to heat when 0.50 C moves through the circuit?

  7. Solution 3 Voltage = Energy / Charge Energy = V × Q Energy = (9.0 V) ×(0.50 C) Energy = 4.5 J

  8. Problem 4 A copper wire is 1.0 m long, and has a diameter of 4.0 mm. What is its resistance? The resistivity of copper is 1.72 x 10-8Wm.

  9. Solution 4 The radius of the wire is 2 mm. In meters, the radius is 2.0 x 10-3m R = rL / A R = rL / [pr2] R = (1.72 x 10-8Wm)(1.0m) / [p(2.0 x 10-3m)2] Please do the calculation. R = 1.4 x 10-3W

  10. Problem 5 Compare to problem 2. Again find the current when 1.5 V is attached to a resistance of 30.0 W. This time include the 1.0 W internal resistance of the battery. Take the 1.5 V to be the “Electromotor Force”, the Emf of the cell.

  11. Solution 5 (1 of 2) The internal resistance acts as a small resistor in series with the other components of the circuit. Resistors in series add up to make the equivalent resistance. The “Electromotive Force” is not a force; it is a voltage. REQ = R1 + R2

  12. Solution 5 (2 of 2) I = Voltage / Resistance I = E / (R + r) I = (1.5 V) / (30.0 W + 1.0 W) I = 0.048 A Notice that internal resistance of the battery decreases the current.

  13. Problem 6 Compare to problem 5. A battery with an emf of 1.5 V (and an internal resistance of 1.0 W) is attached to a bulb with a resistance of 30.0 W. What potential differences are created across the bulb and the battery?

  14. Solution 6 (1 of 2) We already know that the current in every part of the circuit will be 0.048 A. For any resistor, if you know the current and the resistance, then you can calculate the voltage drop across the resistor: V = I × R V = (0.048 A) × (30.0 W) V = 1.45 V Since the voltage drop across the bulb is completely due to the voltage rise across the battery, there is a 1.45 V rise across the battery.

  15. Solution 6 (2 of 2) Also, V =E - Ir) V = (1.5 V) – (0.048 A )( 1.0 W) V = 1.5 V – 0.048 V V = 1.45 V (This agrees with our other method.)

  16. Problem 7 A 1.5 V battery produces a current of 0.050 A in a 30.0 W bulb. How much power comes out of the bulb, and is most of it in the form of light (electromagnetic radiation) or as heat? (Ignore the internal resistance of the battery.)

  17. Solution 7 P = V × I P = (1.5 V) × (0.050 A) P = 0.075 W Or… P = I2 R = (0.050 A)2 (30.0 W) = 0.075 W Or… P = V2 / R = (1.5)2 / (30.0 W) = 0.075 W Almost all of this comes out as heat… So, it’s not so bad to leave a light on in the winter, but it’s short-sighted to do so on a summer day.

  18. Problem 8 A current of 3 A is made by a potential difference of 12 V, for 100 seconds. How much will this change the temperature of a solid (1 kg) that has a specific heat of 24 J kg-1 ˚C-1?

  19. Solution 8 Power = Voltage × Current P = (12 V) (3 A) = 36 W Power = Energy ÷ Time E = P t = (36 W) (100 s) = 3600 J Also, Q = mcDT DT = Q / mc = 3600 J / { (1 kg) × (24 J Kg-1 ˚C-1) } DT = 150 ˚C

  20. Problem 9 An electric motor is used to lift a 10.0 kg object 3.0 m in 5.0 s. If the potential difference across the motor is 12 V, then how much current is in the motor? Assume that the motor is 100% efficient.

  21. Solution 9 Work done by motor = D Energy Work = mgh = (10.0 kg) (9.8 m s-2) (3.0 m) W = 300 J Power = Work ÷ Time P = 300 J / 5.0 s = 60 W Electrical Power: P = VI I = P ÷ V = 60 / 12 = 5.0 A

  22. Problem 10 Three resistors are in series. One has a resistance of 2.0 W. The second has a resistance of 3.1 W, and the last has a resistance of 3.9 W. What is the equivalent resistance of these resistors?

  23. Solution 10 In series, the equivalent resistance is the sum of the individual resistances: Req = R1 + R2 + R3 Req = (2.0) + (3.1) + (3.9) Req = 9.0 W

  24. Problem 11 Three resistors are in parallel. One has a resistance of 2.0 W. The second has a resistance of 3.1 W, and the last has a resistance of 3.9 W. What is the equivalent resistance of these resistors?

  25. Solution 11 This is an important problem to do the calculations by hand. In parallel, the equivalent resistance is: 1/Req = 1/R1 + 1/R2 + 1/R3 1/Req = (2.0)-1 + (3.1) -1 + (3.9) -1 1/Req = 1.079 W Req = 0.93 W Notice that the equivalent resistance is less than the smallest resistance.

  26. 12) How much current is in the ammeters and in the Power Supply? 12 V - + A A 2 W

  27. 12) How much current is in the ammeters and in the Power Supply? 12 V - + 6 A 6 A 2 W

  28. 13) What is the reading of every meter?What is the equivalent Resistance? 6 V 2 W V 1 W V A

  29. 13) What is the reading of every meter?What is the equivalent Resistance? 6 V 2 W 2 V 1 W 4 V 2 A Req = 3 W

  30. 14) Find the reading on each meter. What is the equivalent resistance? V 1 W 12 V V 2 W A 3 W V

  31. 14) Find the reading on each meter. What is the equivalent resistance? 2 V 1 W 12 V 4 V 2 W 2 A 3 W 6 V Req = 6 W

  32. 15) Find the readings on all the meters, and find the equivalent resistance. A 4 W V V 4 W 12 V A A A

  33. 15) Find the readings on all the meters, and find the equivalent resistance. 6 A 4 W 12 V 12 V 4 W 12 V 3 A 3 A Req = 2 W 6 A

  34. 16) Find the readings on all the meters and the equivalent resistance. A A 1 W 2 W V V 12 V A

  35. 16) Find the readings on all the meters and the equivalent resistance. 12 A 6 A 1 W 2 W 12 V 12 V 12 V 18 A Req = 0.67 W

  36. 17) Voltage Divider. What voltage would be measured across the 12 W resistor? VIN 12V 12 W 4W VOUT

  37. Solution 17 Voltage Divider Equation: VOUT = VIN × { R2 / (R1+R2) } VOUT = (12) × { 12 / (4+12) } VOUT = 9 V If the researcher needed 9V, but only had a 12 V source, it is easy to build a “voltage divider” to get what is needed.

  38. Tonight’s HW: Go through the Currents section in your textbook and scrutinize the “Example Questions” and solutions. Bring in your questions to tomorrow’s class.

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