Lecture_16

# Lecture_16 - Field and Current in a Simple Circuit...

• Notes
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Round-trip potential difference: 0 = Δ + Δ wire batt V V 0 ) ( = + EL emf L emf E = L emf enAu enAuE I = = Field and Current in a Simple Circuit

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Round-trip potential difference: Path 1 0 3 1 = Δ + Δ + Δ V V V batt 0 ) ( ) ( 3 3 1 1 = + + L E L E emf Path 2 0 3 2 = Δ + Δ + Δ V V V batt 0 ) ( ) ( 3 3 2 2 = + + L E L E emf E 1 L 1 = E 2 L 2 Field and Current in a Simple Circuit
The number or length of the connecting wires has little effect on the amount of current in the circuit. L V nAu nAuE i Δ = = Δ V wires + Δ V filament + Δ V battery = 0 Δ V battery emf i nAu L V = Δ u wires >> u filament filament wires V V Δ << Δ ) ( emf V filament Δ Work done by a battery goes mostly into energy dissipation in the bulb (heat). Δ V Across Connecting Wires

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Δ V 1 + Δ V 2 + Δ V 3 + Δ V 4 = 0 ( V B - V A )+ ( V C - V B )+ ( V F - V C )+ ( V A - V F )=0 General Use of the Loop Rule
Δ V 1 + Δ V 2 + Δ V 3 + Δ V batt = 0 (- E 1 L 1 ) + (- E 2 L 2 ) + (- E 3 L 3 ) + emf = 0 1. If battery emf =1.5V, what is the voltage drop on thin wire?

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• Winter '07
• k
• Electron, Current, Electric charge, Incandescent light bulb, identical light bulbs, Vbatt

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