PHYS219 Fall semester 2014 Lecture 08: Circuits, Fuses and Kirchhoff’s Laws Dimitrios Giannios Purdue University

The Idea of a Circuit Diagram emf battery wire Assumption: Negligible voltage drop across wire wire Load

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:

Energy/ δt = IΔV = IV Power dissipated = using Ohm's Law :V = IR units : [Joule / sec] = [Watt] = [W] Energy Decrease = qΔV q = I δ t Energy Decrease = (Iδt)ΔV = Energy Dissipated in R I q Power Dissipated in Resistors Power dissipated = IV = I 2 R = V 2 /R

How Much Power is Dissipated in the Load? How Much Power is Dissipated in the Resistor? 360 mA 9.0 V

Resistors in Series Change in potential around closed loop

Resistors in Parallel The currents are NOT independent! junction or node 1/R equiv =1/R 1 +1/R 2 ε - I 1 R 1 = 0, ε - I 2 R 2 =0 ε = I R equiv how can we calculate I 1 and I 2 ? I 1 +I 2 =I I = I 1 +I 2 Kirchoff’s Junction Rule

Resistors in Parallel (more) I 1 +I 2 =I junction or node

10Ω 30Ω 50Ω 9 V9 V 10Ω 18.75Ω 28.75Ω 9 V9 V i 2 i 1 I 9 V9 V a) b)b) d)d) More complicated circuits c)

Question: find R equiv in this circuit? I3I3 I2I2 I 0 = - I 2 R 2 +I 3 R 3 0 a) R equiv =R 1 R 2 /(R 1 +R 2 ) b) R equiv =R 1 +R 2 c) R equiv =R 1 d) R equiv =R 2 Method II: ε - I R 1 =0 I 2 =0, I 3 =I

Example – The Voltage Divider R1R1 R2R2 V out Pick α, control V out Representative Values

Example - Automobile Application Source:

Example Question fuse 1. Which is the Correct Circuit Diagram, a) or b)? a) 2.If the resistance of each light bulb is 2Ω, estimate the current I? 3.What happens if one light bulb blows out? Current (I) Total Resistance (R T ) Voltage (E) Load (R 1 ) Load (R 2 ) Ground b) 12 V

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