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Published byMargaret Sennott Modified over 3 years ago

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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.

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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: circuit components in series C1C1 C2C2 + - C3C3 a b V2V2 V1V1 V ab V V ab = V = V 1 + V 2 + V 3 V3V3

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In general, the voltage drop across resistors in series (or other circuit components) is the sum of the individual voltage drops. circuit components in series V ab = V = V 1 + V 2 + V 3 R3R3 R2R2 R1R1 + - V V1V1 V3V3 V2V2 a b 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. I “derived” this in lecture 7. Here’s what your text means by V ab : V ab =V a -V b =V b a * V = 0 around closed loop

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R3R3 R2R2 R1R1 + - V V1V1 V3V3 V2V2 a b Start at point a (or any other point) and follow the current in a clockwise path around the circuit and back to point a… - V 1 - V 2 - V 3 + V = 0 - IR 1 - IR 2 - IR 3 + V = 0 I Example: add the potential changes around the loop shown. - V 1 - V 2 - V 3 + V = 0 For tomorrow’s homework, your path around the circuit should go in the same direction as your guessed current.

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R3R3 R2R2 R1R1 + - VAVA V1V1 V3V3 V2V2 a b - V 1 - V 2 - V 3 + V A - V B = 0 + - VBVB Again, start at point a and follow the current in a clockwise path around the circuit and back to point a… - IR 1 - IR 2 - IR 3 + V A - V B = 0 For tomorrow’s homework, your path around the circuit should go in the same direction as your guessed current. Example: add the potential changes around the loop shown. I - V 1 - V 2 - V 3 + V A - V B = 0

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

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5 + - 9 V a b 10 Example: calculate I, V ab, and V ba for the circuit shown. I = 0.6 A V a + 9 – 5 (0.6) = V b V ab = V a – V b = – 9 + 5 (0.6) = -6 V V a + 9 – 5 (0.6) = V b a b

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5 + - 9 V a b 10 Example: calculate I, V ab, and V ba for the circuit shown. I = 0.6 A V ba = V b – V a = + 10 (0.6) = +6 V V b – 10 (0.6) = V a b a

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

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

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

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5 + - 9 V a b 10 + - 6 V Example: calculate I, V ab, and V ba for the circuit shown. I = 0.2 A V a + 9 – 6 – 5 (0.2) = V b V ab = V a – V b = – 9 + 6 + 5 (0.2) = – 2 V

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5 + - 9 V a b 10 + - 6 V Example: calculate I, V ab, and V ba for the circuit shown. I = 0.2 A V b – 10 (0.2) = V a V ba = V b – V a = + 10 (0.2) = + 2 V The smart way: V ba = - V ab = +2 V V b – 10 (0.2) = V a

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

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In Physics 2135, whenever you work with currents in circuits, you should assume (unless told otherwise) “direct current.” DC Currents 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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Physics 7B Lecture 320-Jan-2010 Slide 1 of 20 Physics 7B-1 (A/B) Professor Cebra Review of Circuits and the Linear Transport Model Winter 2010 Lecture.

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