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LECTURE 2: BASIC PRINCIPLES OF ELECTRICITY REQUIRED READING: Kandel text, Appendix Chapter I Neurons transmit electrical currents Behavior of synaptically.

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Presentation on theme: "LECTURE 2: BASIC PRINCIPLES OF ELECTRICITY REQUIRED READING: Kandel text, Appendix Chapter I Neurons transmit electrical currents Behavior of synaptically."— Presentation transcript:

1 LECTURE 2: BASIC PRINCIPLES OF ELECTRICITY REQUIRED READING: Kandel text, Appendix Chapter I Neurons transmit electrical currents Behavior of synaptically linked neurons has similarities to behavior of solid-state electrical circuits Therefore, a fundamental appreciation of the nervous system requires understanding its electrical properties THIS LECTURE INTRODUCES BASIC CONCEPTS, TERMINOLOGY, AND EQUATIONS OF ELECTRICITY ESSENTIAL TO OUR TACKLING THE ELECTROPHYSIOLOGY OF NEURONS AND NEURAL CIRCUITS

2 CHARGED PARTICLES AND ELECTROSTATIC FORCE Some particles have electrical CHARGE; charge can be POSITIVE or NEGATIVE Charged particles exert FORCE on each other: LIKE charges REPEL OPPOSITE charges ATTRACT { examples of charged particles: electrons (-), ions (- OR +) } Force experienced by charged particle determined by the sum and distances of surrounding charges + + + + - - + - + - - + + + - NO FORCE REPULSIVE FORCE ATTRACTIVE FORCE NEUTRALPOSITIVENEGATIVE + + - - - + + + - POSITIVE

3 ELECTRICAL CONDUCTANCE AND RESISTANCE + + - - - + + + - NEGATIVE POSITIVE A B WHEN CHARGED PARTICLES ARE SUBJECT TO ELECTRICAL FORCE, THEIR ABILITY TO MOVE FROM POINT A TO B IS INFLUENCED BY CONDUCTIVE PROPERTY OF MATERIAL CONDUCTANCE (g) {units=siemens,S} measure of material’s ease in allowing movement of charged particles RESISTANCE (R) {units=Ohms  measure of material’s difficulty in allowing electrical conduction Resistance is the INVERSE of Conductance. I.e.: R = 1g1g OR g = 1R1R

4 VOLTAGE AND CURRENT + + - - - + + + - NEGATIVE POSITIVE A B VAVA VBVB VV I When there is a charge differential between two points, energy is stored. This stored energy is called ELECTRICAL POTENTIAL or VOLTAGE DIFFERENTIAL (  V) {units = volts, V}  V = V A - V B When there is a voltage differential between two points in a conductive material, charged particles will be forced to move. Movement of charge is an ELECTRICAL CURRENT CURRENT (I) {units = amperes, A} is the RATE of charge flow. I = dq / dt Where q = amount of charge {units = coulombs, Q} and t = time {units = seconds, s} NOTE: I > 0 means net flow of positive charge; I < 0 means net flow of negative charge

5 OHM’S LAW The amount of current flow is directly proportional to both the voltage differential and the conductance I  =  V x g Since g = 1 / R I =  V / R  V = I x R SCHEMATIC DIAGRAM R VAVA VBVB I  V = V A - V B = IR I =  V / R WATER PRESSURE ANALOGY PAPA PBPB FLOW RATE VALVE Water Pressure is analogous to Voltage Differential Valve Resistance is analogous to Electrical Resistance Flow Rate is analogous to Electrical Current Flow Rate = Water Pressure / R VALVE OR

6 THE “I-V PLOT” & OHM’S LAW I  =  V x g R VAVA VBVB I In a simple resistive circuit, the relationship between current and voltage is LINEAR 10 20 - 20- 101020 - 10 - 20 VV I 10 20 - 20- 101020 - 10 - 20 VV I HIGH CONDUCTANCE WEAKER CONDUCTANCE CONDUCTANCE ( g ) is SLOPE of line in I - V PLOT

7 MULTIPLE RESISTANCES IN SERIES V1V1 V2V2 R2R2 R1R1 I2I2 I1I1 acb IN SERIES RESISTANCES SUM TO GIVE OVERALL RESISTANCE R TOTAL (a,c) = R 1 (a,b) + R 2 (b,c) I TOTAL (a,c) = I 1 (a,b) = I 2 (b,c)  V TOTAL (a,c) =  V 1 (a,b) +  V 2 (b,c) Two resistances are summed to give the overall resistance between points a and c Currents are equal along the series By Ohm’s Law, the total voltage differential equals the sum of the component voltages POSITIVENEGATIVE

8 MULTIPLE RESISTANCES IN PARALLEL R1R1 POSITIVENEGATIVE R2R2 I2I2 I1I1 I TOTAL = I 1 + I 2 g TOTAL = g 1 + g 2  V TOTAL =  V 1 =  V 2 I 1 x R 1 = I 2 x R 2 Total current is the sum of individual parallel currents Total conductance is the sum of parallel conductances The voltage differential between two points is the same no matter what the path By Ohm’s Law, larger current travels thru the “path of least resistance” I TOTAL

9 POSITIVE NEGATIVE R1R1 R2R2 I2I2 I1I1 I TOTAL EQUIVALENT REPRESENTATIONS + - SYMBOL DESIGNATES A VOLTAGE GENERATOR (POWER SOURCE) WHICH MAINTAINS A CHARGE DIFFERENTIAL FROM ONE SIDE TO THE OTHER (e.g. A BATTERY) R1R1 R2R2 I2I2 I1I1 + - I TOTAL CIRCUIT DIAGRAM

10 R (10  ) + - I VV 10 V R (10  ) + - I VV 10 V SWITCH OPEN AT t = 0 sec SWITCH CLOSED AT t = 5 sec BEHAVIOR OF A SIMPLE RESISTIVE CIRCUIT CIRCUIT PROPERTIES t (sec) 0 5 t (sec) 0 5  V (volts) 0 10 I (Amps) 0 1

11 CAPACITANCE (C) {units = farads, F} is the measure of the AMOUNT OF CHARGE DIFFERENTIAL which builds up ACROSS a material when subjected to a voltage differential. q =  V x C or  V = q / C I.e. Larger capacitance ----> Larger charge stored A material that has capacitance is called a capacitor. The schematic symbol for a capacitor is: CAPACITANCE SOME MATERIALS CANNOT CONDUCT ELECTRICITY, BUT CAN ABSORB CHARGE WHEN SUBJECTED TO A CURRENT OR VOLTAGE C

12 BEHAVIOR OF A SIMPLE CAPACITIVE CIRCUIT C (10 F  ) + - I VV 10 V + - I VV SWITCH OPEN AT t = 0 sec SWITCH CLOSED AT t = 5 sec CIRCUIT PROPERTIES t (sec) 0 5 t (sec) 0 5  V (volts) 0 10 I (Amps) 0 C (10 F  ) t (sec) 0 5 Q (coulombs) 0 100

13 RELATIONSHIP OF CAPACITANCE AND CURRENT q = C x  V AS DESCRIBED BEFORE: SINCE I = d q / d t d q/ d t = I = C x d  V/ d t I.e. As current flows into a capacitor, the voltage across it increases

14 IN SERIES CIRCUIT WITH CAPACITANCE & RESISTANCE IN SERIES + - I 10 V SWITCH CLOSED AT t = 0 sec R (5  ) C (1 F  ) VAVA R (5  ) + - I VAVA 10 V SWITCH OPEN BEFORE t = 0 C (1 F  ) VBVB VBVB CIRCUIT PROPERTIES t (sec) -5 0 5 10  V A (volts) 0 10 ( REMEMBER: After switch closed,  V A +  V B =  V TOTAL = 10 V ) t (sec) -5 0 5 10  V B (volts) 0 10 t (sec) -5 0 5 10 I (amps) 0 2

15 LOGARHYTHMIC DECAY OF CURRENT THROUGH A IN SERIES CIRCUIT WITH CAPACITANCE & RESISTANCE IN SERIES + - I 10 V SWITCH CLOSED AT t = 0 sec R (5  ) C (1 F  ) VAVA R (5  ) + - I VAVA 10 V SWITCH OPEN BEFORE t = 0 C (1 F  ) VBVB VBVB I = C d V C d t I = = V TOT - V C R VR VR R Equ. AEqu. B Combine equations A & B and integrate V C (t) = V TOT ( 1 - e - t / RC ) V R (t) = V TOT ( e - t / RC ) As capacitor charges, V R and I decay logarhythmically

16 IN SERIES CIRCUIT WITH CAPACITANCE & RESISTANCE IN SERIES CONTROL OF CURRENT FLOW BY SIZE OF R AND C R + - I VAVA + - I SWITCH OPENSWITCH CLOSED C R C VAVA VBVB VBVB THE LARGER THE RESISTANCE (R) ----> THE SMALLER THE INITIAL CURRENT SIZE THE LONGER IT TAKES FOR CAPACITOR TO CHARGE THE SLOWER THE DECLINE IN CURRENT FLOW THE LARGER THE CAPACITANCE (C) ----> THE LONGER IT TAKES FOR CAPACITOR TO CHARGE THE SLOWER THE DECLINE IN CURRENT FLOW NO EFFECT ON INITIAL CURRENT SIZE t 1/2-max (sec) = 0.69 x R (  ) x C (F)

17 IN SERIES CIRCUIT WITH CAPACITANCE & RESISTANCE IN SERIES CHARGE AND DISCHARGE OF A CAPACITOR R (5  ) I + - VAVA 10 V CHARGE SWITCH CLOSED AT t = 0 sec C (1 F  ) VBVB CHARGE SWITCH OPENED AT t = 10 secDISCHARGE SWITCH CLOSED AT t = 10 sec CIRCUIT PROPERTIES t (sec) 0 5 10 15 20  V A (volts) 0 10 -10 t (sec) 0 5 10 15 20 I (amps) 0 2 -2 t (sec) 0 5 10 15 20  V B (volts) 0 10 -10 RESISTOR VOLTAGE CAPACITOR VOLTAGE

18 IN PARALLEL CIRCUIT WITH CAPACITANCE & RESISTANCE IN PARALLEL R B (5  ) R A (5  ) + - I TOT 10 V SWITCH OPEN BEFORE t = 0 sec SWITCH CLOSED AT t = 0 sec C (1 F  ) IBIB ICIC I TOT VAVA VBVB t (sec) -5 0 5 10  V A (volts) 0 10 5 t (sec) -5 0 5 10  V B (volts) 0 10 5 t (sec) -5 0 5 10 I C (amps) 0 2 1 t (sec) -5 0 5 10 I B (amps) 0 2 1 t (sec) -5 0 5 10 I TOT (amps) 0 2 1 CURRENT FLOW THROUGH PARALLEL RESISTOR IS DELAYED BY THE CAPACITOR {

19

20 IN PARALLEL CIRCUITS WITH TWO BATTERIES IN PARALLEL RBRB IBIB + - VAVA VBVB + - SWITCH CLOSED AT t = 0 sec t (sec) -5 0 5 10 I B (amps) 0 V A = V B + I B R B I B = (V A - V B ) / R B or RBRB IBIB + - VAVA VBVB + - RARA IAIA VCVC ICIC In this circuit, what is V C at steady state? I A = - I B therefore, (eq.2) V C = V A + I A R A = V B + I B R B (eq.1) I A + I B + I C = 0I C = 0 andalso Combining eq. 1 & 2, and converting R to g V C = V A g A + V B g B g A + g B V C is the weighted average of the two batteries, weighted by the conductance through each battery path CONCLUSION:

21 RESISTANCES & CAPACITANCES ALONG AN AXON MEMBRANE (C) ION CHANNEL (g) CYTOSOL (g) Lipid bilayer of plasma membrane is NONCONDUCTIVE, but has CAPACITANCE Ion channels in membrane provide sites through which selective ions flow, thereby giving some TRANSMEMBRANE CONDUCTANCE Flow of ions in cytosol only limited by diameter of axon; the WIDER the axon, the greater the AXIAL CONDUCTANCE

22 MODELLING THE AXON AS RESISTANCES & CAPACITANCES RMRM RMRM RMRM R AXON CMCM CMCM CMCM The axon can be thought of as a set of segments, each having an internal axon resistance in series with a transmembrane resistance and capacitance in parallel When a point along the axon experiences a voltage drop across the membrane, the SPEED and AMOUNT of current flow down the axon is limited by R AXON, R M, and C M. + - I A1 I A2 I M1 I C1 Axon current nearest the voltage source (I A1 ) does not all proceed down the axon (I A2 ). Some current is diverted through membrane conductance (I M1 ), and current propogation down axon is delayed by diversion into the membrane capacitance (I C1 ).

23 Next lecture: ION CHANNELS & THE RESTING MEMBRANE POTENTIAL REQUIRED READING: Kandel text, Chapters 7, pgs 105-139


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