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Transcript
Announcements • First lab this afternoon SHL 016 – Introduction to EWB http://www.physics.udel.edu/~jholder/Phys645/index.htm Tuesday: Matthew Bihler Gaurav Pandey Philip Zandona Trisha Wednesday: Joseph Brosch Erin Grace Ali Jafri Lei Chen Halise Celik Fatih Jingliang Zhang Eric Lecture 2 Overview • • • • • Kirchoff’s Laws Power and Resistance Practical sources Voltage/ Current Dividers DC circuit analysis – Element combination – Using Kirchoff’s Voltage Law – Using Kirchoff’s Current Law – Mesh Analysis Kirchhoff's three laws of spectroscopy 1. A hot solid object produces light with a continuous spectrum. 2. A hot tenuous gas produces light with spectral lines at discrete wavelengths (i.e. specific colors) which depend on the energy levels of the atoms in the gas. (emission spectrum) 3. A hot solid object surrounded by a cool tenuous gas (i.e. cooler than the hot object) produces light with an almost continuous spectrum which has gaps at discrete wavelengths depending on the energy levels of the atoms in the gas. (absorption spectrum) Gustav Robert Kirchhoff (12 March 1824 – 17 October 1887) Kirchoff’s Current Law The sum of the current at any node must equal zero: i.e., the current flowing into a node must equal the current flowing out of the node Conservation of Charge (Current = rate of change of charge) At Node 1: -i+i1+i2+i3=0 i=i1+i2+i3 Note convention: current flows from positive terminal In order for current to flow, there must exist a closed circuit Kirchoff’s Voltage Law The sum of the voltages around a closed loop is zero Potential: at a=va b=vb Potential difference: v2=va-vb Around the loop: -v1+v2=0 v1=v2 Note: potential measured relative to ground: true ground (earth) or chassis ground (enclosure) Circuit elements and their i-v characteristics: Resistor Ohm’s law: Current is proportional to applied Voltage, and inversely proportional to the Resistance: V=IR v L m R .m 2 A m 1 R = resistance: depends on materials and geometry ρ = resistivity: depends only on materials σ = conductivity Resistor Color code http://www.dannyg.com/examples/res2/resistor.htm 560KΩ ± 5% Electric Power Electric power = amount of work done/unit time Voltage V = work/unit charge So, to move charge Q, work done=VQ Power = VQ/t =VI P=IV =I2R =V2/R Units = Joules/sec = Watts (W) What do we pay for in the electricity bill? kWh=energy/time × time = energy More on Power and Resistance What do the power ratings of appliances mean? e.g. what does a 1000W hair dryer tell us? • Assume 120V (USA) if voltage not specified. P 1000 8.33A Max current V 120 V2 V 2 1202 P R 14.4 R P 1000 P iV i Never exceed the rated power Can you use this appliance on a 240V (UK) line? V 2 2402 P 4000W R 14.4 Destroys the appliance! Resistance Limits: Open and Short Circuits • Short Circuit: A wire! R=0, V=0 for any i. – Particularly bad for any voltage source • Open circuit: A break! R→, i=0 for any V. – Particularly bad for current source Series Resistors and the Voltage Divider Rule For N resistors in series: N N n 1 n 1 REQ Rn , i i1 i2 ... iN , v vn REQ>(R1,R2,…..,RN) Voltage Divider: R1 v1 (1.5V) REQ R v2 2 (1.5V) REQ v3 R3 (1.5V) REQ R3 Practical Voltage Sources Modelled with an ideal source and a series resistor Ideal voltage source: rS=0 Current is the same at all points vS vS iS REQ rS RL RL vL iS RL vS rS RL voltage divider So for a practical voltage source, the output voltage depends upon RL If rS<< RL , vL = vS , independent of RL Parallel Resistors and the Current Divider Rule For N resistors in parallel: N N 1 1 , v v1 ..... vN , iS in REQ n 1 R n n 1 REQ<(R1,R2,…..,RN) v REQ iS Current Divider: i1 R1 R1 REQ v i2 iS R2 R2 v REQ i3 iS R3 R3 Large current through smaller R. Advantage of a parallel circuit; a broken branch will not affect other branches Practical Current Sources Modelled with an ideal source and a parallel resistor Ideal current source: rS= iL REQ 1 REQ RL iS current divider rS RL 1 1 rS RL rS RL iL rS iS rS RL The output current now depends upon RL If rS>> RL , iL = iS , independent of RL Circuit analysis method 1: Apply element combination rules Series resistors Parallel resistors Series voltage sources Parallel current sources Circuit analysis method 1: element combination R=22Ω R1=10Ω R2=20Ω R3=30Ω V=10V Find the equivalent resistance and the current at I I= 10V/22Ω = 0.45A Circuit analysis method 2a: KVL and KCL Kirchoff’s Voltage Law The sum of the voltages around a closed loop must be zero • Draw the current direction (arbitrary) and label the voltage directions (determined by the defined current direction). Voltage on a voltage source is always from positive to negative end. • Define either clockwise or counterclockwise as positive direction for summing voltages. Once the direction is defined, use the same convention in every loop.Voltage across a resistor is +’ve if voltage direction the same as current direction, -’ve otherwise • Apply KVL Kirchoff’s Voltage Law: Multiloop The sum of the voltages around a closed loop must be zero • Draw the current direction (arbitrary) and label the voltage directions (determined by the defined current direction). R3 • Define either clockwise or counterclockwise as positive voltage direction. Once the direction is defined, use the same convention in every loop. • Apply KVL Vr V1 V2 0 V2 V3 0 Ir IR1 I 2 R2 I 2 R2 I 3 R3 0 Kirchoff’s Current Law The sum of the current at a node must be zero: Iin=Iout I=I2+I3 (1) ε=Ir+IR1+I2R2 (2) I3R3-I2R2 =0 (3) R3 Set r=1Ω, R1=3Ω, R2=5Ω, R3=10Ω, ε=3V I- I2- I3 = 0 (4) 4I+5I2+ 0I3= 3 (5) 0I- 5I2+10I3= 0 (6) Last note on KCL KVL analysis • If solutions to currents or voltages are negative, this means the real direction is opposite to what you originally defined • To deal with current sources: current is known, but assign a voltage across it which has to be solved Sample Problem