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Transcript
PHYS219 Fall semester 2014 Lecture 08: Circuits, Fuses and Kirchhoff’s Laws Dimitrios Giannios Purdue University The Idea of a Circuit Diagram wire Load emf wire battery wire wire Assumption: Negligible voltage drop across wire Circuit Symbols Open Circuits continuous path open path Power Dissipation - the Technology Underlying Fuses and Circuit Breakers •In a fuse, current passes through a thin metal strip – The strip acts as a resistor with a small resistance – If a fault causes the current to become large, the increased power melts the strip to melt and the current stops – open circuit! •A circuit breaker “OPENS” when current exceeds a predetermined limit – The circuit breaker can be reset dc Circuit Analysis • An electric circuit is a combination of connected elements forming a continuous path through which charge is able to move • Calculating the current in a circuit is called circuit analysis • dc stands for direct current - the current is constant • The current is viewed as the motion of the positive charges traveling through the circuit • Exact values of current and voltage depend on where in the circuit you are measuring Kirchoff’s Loop Rule The change in electrical potential of a charge q as it moves around a closed loop MUST be zero q Recover Ohm’s Law: Power Dissipated in Resistors I q Energy Decrease = qΔV q = I δt Energy Decrease = (Iδt)ΔV = Energy Dissipated in R Power dissipated = Energy/δt = IΔV = IV using Ohm's Law : V = IR Power dissipated = IV = I2R = V2/R units : [Joule / sec] = [Watt] = [W] How Much Power is Dissipated in the Load? How Much Power is Dissipated in the Resistor? 9.0 V 360 mA Resistors in Series Change in potential around closed loop = + 𝜀 – IR1-IR2 = + 𝜀 – I(R1+R2) = +𝜀 – IRequiv = 0 junction or node Resistors in Parallel Kirchoff’s Junction Rule I1+I2=I The currents are NOT independent! 1/Requiv=1/R1 +1/R2 ε = I Requiv how can we calculate I1 and I2 ? ε - I1 R1 = 0, ε - I2 R2 =0 I = I1 +I2 junction or node Resistors in Parallel (more) I1+I2=I More complicated circuits I a) 10Ω 9 V 50Ω i1 i2 30Ω b) 10Ω 9 V 18.75Ω c) d) 9 V 28.75Ω Question: find Requiv in this circuit? a) Requiv =R1R2/(R1 +R2) I I3 I2 b) Requiv =R1 +R2 c) Requiv =R1 d) Requiv =R2 0 0 = - I2 R2 +I3 R3 I2 =0, I3=I Method II: ε - I R1 =0 Example – The Voltage Divider R1 R2 Vout Representative Values Pick α, control Vout Example - Automobile Application Source: http://www.smartdraw.com/examples/view/auto+wiring+diagram/ Example Question 1. Which is the Correct Circuit Diagram, a) or b)? a) fuse Current (I) Current (I) Total Resistance (RT) 12 V Load (R1) Load (R2) Ground Voltage (E) 2.If the resistance of each light bulb is 2Ω, estimate the current I? 3.What happens if one light bulb blows out? b) Summary – Kirchhoff’s Rules (circa 1845) • Kirchhoff’s Loop Rule –The total change in the electric potential around any closed circuit path must be zero • Kirchhoff’s Junction Rule –The current entering a circuit junction must equal the current leaving the junction • These circuit rules are just applications of fundamental laws of physics –Loop Rule – conservation of energy –Junction Rule – conservation of charge • The rules apply to ALL types of circuits involving ALL types of circuit elements