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Calculating Potential Changes in Circuits

Learn how to calculate potential changes, emf, terminal voltage, and internal resistance in circuits. Understand how to incorporate these quantities in circuit calculations and work-energy problems. Also, learn to calculate electric power in circuit components.

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Calculating Potential Changes in Circuits

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  1. Today’s agenda: Potential Changes Around a Circuit. You must be able to calculate potential changes around a closed loop. Emf, Terminal Voltage, and Internal Resistance. You must be able to incorporate all of the above quantities in your circuit calculations. Electric Power. You must be able to calculate the electric power dissipated in circuit components, and incorporate electric power in work-energy problems. Examples.

  2. circuit components in series In lecture 7 (the first capacitors lecture) I suggested using conservation of energy to show that the voltage drop across circuit components in series is the sum of the individual voltage drops: Vab C1 C2 C3 a b V2 V3 V1 + - V Vab = V = V1 + V2 + V3

  3. circuit components in series In general, the voltage drop across resistors in series (or other circuit components) is the sum of the individual voltage drops. R1 R2 R3 a b V1 V2 V3 - + V Vab = V = V1 + V2 + V3 I “derived” this in lecture 7. Here’s what your text means by Vab: Vab=Va-Vb=Vba You may use this in tomorrow’s homework. It “is”* on your starting equations sheet, and is a consequence of conservation of energy. Use this in combination with Ohm’s Law, V=IR. * V = 0 around closed loop

  4. Example: add the potential changes around the loop shown. Start at point a (or any other point) and follow the current in a clockwise path around the circuit and back to point a… R1 R2 R3 a b V1 V2 V3 - + I V - V1 - V2 - V3 + V = 0 - V1 - V2 - V3 + V = 0 - V1- V2- V3 + V = 0 - V1 - V2- V3+ V = 0 - V1 - V2 - V3+ V = 0 - V1 - V2 - V3 + V = 0 - IR1 - IR2 - IR3 + V = 0 For tomorrow’s homework, your path around the circuit should go in the same direction as your guessed current.

  5. Example: add the potential changes around the loop shown. Again, start at point a and follow the current in a clockwise path around the circuit and back to point a… R1 R2 R3 a b V1 V2 V3 - - + + I VB VA - V1- V2 - V3 + VA - VB = 0 - V1- V2- V3 + VA - VB = 0 - V1 - V2- V3+ VA - VB = 0 - V1 - V2 - V3+ VA - VB = 0 - V1 - V2 - V3 + VA- VB= 0 - V1 - V2 - V3 + VA- VB= 0 - V1 - V2 - V3 + VA- VB= 0 - IR1 - IR2 - IR3 + VA - VB = 0 For tomorrow’s homework, your path around the circuit should go in the same direction as your guessed current.

  6. Example:calculate I, Vab,and Vba for the circuit shown. To be worked at the blackboard in lecture. 5  b 10  - + a I 9 V +9 – 5 I – 10 I = 0 +9– 5 I – 10 I = 0 +9 – 5 I– 10 I = 0 +9 – 5 I – 10 I = 0 +9 – 5 I – 10 I = 0 15 I = +9 I = +9/15 = 0.6 A

  7. Example: calculate I, Vab,and Vba for the circuit shown. 5  b b 10  - + a a I = 0.6 A 9 V Va+ 9 – 5 (0.6) = Vb Va+ 9– 5 (0.6) = Vb Va + 9 – 5 (0.6)= Vb Va + 9 – 5 (0.6) = Vb Va + 9 – 5 (0.6) = Vb Vab = Va – Vb = – 9 + 5 (0.6) = -6 V

  8. Example: calculate I, Vab,and Vba for the circuit shown. 5  b b 10  - + a a I = 0.6 A 9 V Vb– 10 (0.6) = Va Vb – 10 (0.6) = Va Vb – 10 (0.6) = Va Vb – 10 (0.6) = Va Vba = Vb – Va = + 10 (0.6) = +6 V

  9. Example: calculate I, Vab,and Vba for the circuit shown. 5  b 10  - + a I = 0.6 A 9 V Note: Vba = +6 V = - Vab , as expected

  10. Example: calculate I, Vab,and Vba for the circuit shown. Graph the potential rises and drops in this circuit. 5  b 10  - + a I = 0.6 A 9 V 5 I = 3V b  = 9V 10 I = 6V a

  11. Example:calculate I, Vab,and Vba for the circuit shown. To be worked at the blackboard in lecture. 5  b 10  - - + + a I 9 V 6 V + 9 – 6 – 5 I – 10 I =0 + 9 – 6 – 5 I – 10 I = 0 + 9 – 6 – 5 I– 10 I = 0 + 9 – 6 – 5 I – 10 I= 0 + 9 – 6 – 5 I – 10 I = 0 + 9 – 6 – 5 I – 10 I = 0 15 I = 3 I = 0.2 A

  12. Example:calculate I,Vab,and Vba for the circuit shown. 5  b 10  - - + + a I = 0.2 A 9 V 6 V Va+ 9 – 6 – 5 (0.2) = Vb Va+ 9 – 6 – 5 (0.2) = Vb Va + 9 – 6– 5 (0.2) = Vb Va + 9– 6 – 5 (0.2) = Vb Va + 9– 6 – 5 (0.2) = Vb Va + 9– 6 – 5 (0.2) = Vb Vab = Va –Vb = – 9+ 6 + 5 (0.2) = – 2 V

  13. Example: calculate I, Vab,and Vba for the circuit shown. The smart way: Vba = - Vab = +2 V 5  b 10  - - + + a I = 0.2 A 9 V 6 V Vb – 10 (0.2) = Va Vb– 10 (0.2) = Va Vb– 10 (0.2) = Va Vb – 10 (0.2) = Va Vba = Vb – Va = + 10 (0.2) = + 2 V

  14. Example:calculate I, Vab,and Vba for the circuit shown. What if you guess the wrong current direction? 5  b 10  - - + + a I 9 V 6 V – 10 I – 5 I +6 – 9 = 0 – 10 I – 5 I+6 – 9 = 0 – 10 I – 5 I +6– 9 = 0 – 10 I – 5 I +6 – 9 = 0 – 10 I – 5 I +6 – 9 = 0 – 10 I – 5 I +6 – 9 = 0 15 I = – 3 I = – 0.2 A oops, guessed wrong direction, no big deal!

  15. DC Currents In Physics 2135, whenever you work with currents in circuits, you should assume (unless told otherwise) “direct current.” Current in a dc circuit flows in one direction, from + to -. We will not encounter ac circuits much in this course. For any calculations involving household current, which is ac, assuming dc will be “close enough” to give you “a feel” for the physics. If you need to learn about ac circuits, you’ll have courses devoted to them. The mathematical analysis is more complex. We have other things to explore this semester.

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