Uploaded bygetaneh30
PPT, PDF12 views

chap-3lecture13.pptIEC 61850 Standard For Substation Automation

IEC 61850 Standard For Substation Automation

Embed presentation

Download to read offline
Today’s agenda: Measuring Instruments: ammeter, voltmeter, ohmmeter. You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments. RC Circuits. You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations.
Measuring current, voltage, and resistance A Ammeter: •measures current (A) •connected in series (current must go through instrument) I V a b Voltmeter: •measures potential difference (V) •connected in parallel Ohmmeter: •measures resistance of an isolated resistor (not in a working circuit) 
Effect of ammeter on circuit V R Measuring current in a simple circuit: A • connect ammeter in series Are we measuring the correct current? (the current in the circuit without ammeter)
• current without the ammeter would be V R . V I = R Measuring current in a simple circuit: . V I = R + r To minimize error, ammeter resistance r must be very small. (ideal ammeter would have zero resistance) • any ammeter has some resistance r. r • connect ammeter in series Are we measuring the correct current? (the current in the circuit without ammeter) • current in presence of ammeter is Effect of ammeter on circuit
Example: an ammeter of resistance 10 m is used to measure the current through a 10  resistor in series with a 3 V battery that has an internal resistance of 0.5 . What is the relative (percent) error caused by the ammeter? V=3 V R=10  r=0.5  Actual current without ammeter: V I = R + r 3 I = A 10+ 0.5 I = 0.2857 A = 285.7 mA You might see the symbol  used instead of V.
V=3 V R=10  r=0.5  Current with ammeter: A V I = R + r+ R 3 I = A 10+ 0.5+ 0.01 I = 0.2854 A = 285.4 mA RA  0.2857- 0.2854 % Error = 100 0.2857 % Error = 0.1 %
Designing an ammeter http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1 Galvanometer: •current flows through a coil in a magnetic field •coil experiences a torque, connected needle deflects (see later chapters of this class)
Designing an ammeter • ammeter can be based on galvanometer (for electronic instrument, use electronic sensor instead, analysis still applies) • simplest case: send current directly through galvanometer, observe deflection of needle Needle deflection is proportional to current. Each galvanometer has a certain maximum current corresponding to full needle deflection. What if you need to measure a larger current? • use shunt resistor
Ammeter uses a galvanometer and a shunt, connected in parallel: A I  Everything inside the green box is the ammeter. • Current I gets split into Ishunt and IG Homework hint: If your galvanometer reads 1A full scale but you want the ammeter to read 5A full scale, then RSHUNT must result in IG=1A when I=5A. What are ISHUNT and VSHUNT? G RG RSHUNT IG ISHUNT I A B galvanometer
G RG RSHUNT IG ISHUNT I A B Shunt also reduces resistance of the ammeter:   A G SHUNT 1 1 1 R R R   G SHUNT A G SHUNT R R R R R
A galvanometer-based ammeter uses a galvanometer and a shunt, connected in parallel: G RG RS IG IS I   A G S 1 1 1 R R R Example: what shunt resistance is required for an ammeter to have a resistance of 10 m, if the galvanometer resistance is 60 ?   S A G 1 1 1 R R R        G A S G A 60 .01 R R R 0.010 R -R 60-.01 (actually 0.010002 ) To achieve such a small resistance, the shunt is probably a large-diameter wire or solid piece of metal.
Web links: ammeter design, ammeter impact on circuit, clamp-on ammeter (based on principles we will soon be studying).
Effect of voltmeter on circuit Measuring voltage (potential difference) Vab in a simple circuit: • connect voltmeter in parallel Are we measuring the correct voltage? (the voltage in the circuit without voltmeter) =3 V R=10  r=0.5  V a b
• extra current changes voltage drop across r and thus Vab To minimize error, voltmeter resistance r must be very large. (ideal voltmeter would have infinite resistance) • voltmeter has some resistance RV • current IV flows through voltmeter Effect of voltmeter on circuit Measuring voltage (potential difference) Vab in a simple circuit: • connect voltmeter in parallel Are we measuring the correct voltage? (the voltage in the circuit without voltmeter) =3 V R=10  r=0.5  RV a b IV
Example: a galvanometer of resistance 60  is used to measure the voltage drop across a 10 k resistor in series with an ideal 6 V battery and a 5 k resistor. What is the relative error caused by the nonzero resistance of the galvanometer? Actual voltage drop without instrument: V=6 V R1=10 k R2=5 k a b    3 eq 1 2 R R + R = 15 10       -3 3 eq V 6 V I 0.4 10 A R 15 10         -3 3 ab V = IR 0.4 10 10 10 4 V
The measurement is made with the galvanometer. V=6 V R1=10 k R2=5 k G RG=60  a b 60  and 10 k resistors in parallel are equivalent to 59.6  resistor. Total equivalent resistance: 5059.6  Total current: I=1.186x10-3 A I=1.19 mA Vab = 6V – IR2 = 0.07 V. The relative error is:  4-.07 % Error = 100= 98% 4 Would you pay for this voltmeter? Would you pay for this voltmeter? We need a better instrument!
Example: a voltmeter of resistance 100 k is used to measure the voltage drop across a 10 k resistor in series with an ideal 6 V battery and a 5 k resistor. What is the percent error caused by the nonzero resistance of the voltmeter? We already calculated the actual voltage drop (2 slides back). V=6 V R1=10 k R2=5 k a b         -3 3 ab V = IR 0.4 10 10 10 4 V
The measurement is now made with the “better” voltmeter. V=6 V R1=10 k R2=5 k V RV=100 k a b 100 k and 10 k resistors in parallel are equivalent to an 9090  resistor. Total equivalent resistance: 14090  Total current: I=4.26x10-4 A I=.426 mA The voltage drop from a to b: 6-(4.26x10-4 )(5000)=3.87 V. The percent error is.  4- 3.87 % Error = 100= 3.25% 4 Not great, but much better.
• voltmeter must have a very large resistance • voltmeter can be made from galvanometer in series with a large resistance V G RG RSer  Everything inside the blue box is the voltmeter. a b Vab a b Vab Homework hints: “the galvanometer reads 1A full scale” would mean a current of IG=1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected RSer must result in IG=1A when Vab=10V. Homework hints: “the galvanometer reads 1A full scale” would mean a current of IG=1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected RSer must result in IG=1A when Vab=10V. Designing a voltmeter
• Ohmmeter measures resistance of isolated resistor • Ohmmeter can be made from a galvanometer, a series resistance, and a battery (active device). G RG RSer R=? • Terminals of ohmmeter are connected to unknown resistor • battery causes current to flow and galvanometer to deflect • V=I (Rser + RG + R) solve for unknown R  Measuring Instruments: Ohmmeter Everything inside the blue box is the ohmmeter. V
Alternatively: •separately measure current and voltage for resistor •Apply Ohm’s law Four-point probe: V A reference: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movcoil.html#c4
Today’s agenda: Measuring Instruments: ammeter, voltmeter, ohmmeter. You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments. RC Circuits. You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations.
Charging and discharging a capacitor • so far, we have assumed that charge instantly appears on capacitor • in reality, capacitor does not change instantaneously • charging speed depends on capacitance C and on resistance R between the battery and the capacitor Q t Q t What happens if we connect a capacitor to a voltage source? Charging and discharging are time-dependent phenomena!
Switch open, no current flows. RC circuit: Charging a Capacitor  R switch C t<0 Close switch, current flows. t>0 I Kirchoff’s loop rule* (green loop) at the time when charge on C is q. *Sign convention for capacitors is the same as for batteries: Voltage counts positive if going across from - to +. ε q - - IR = 0 C This equation is deceptively complex because I depends on q and both depend on time. - - + + -q +q
Empty capacitor: (right after closing the switch) q=0 VC=0, VR=  I=I0=/R  R switch C Full capacitor: (after a very long time) VC=, VR= 0 Q=C I=0 I ε q - - IR = 0 C - - + + q0=0 t=0 qF=C t= IF=0 Distinguish capacitor and resistor voltages VC and VR. They are not equal but VC + VR = . Limiting cases:
ε q - - IR = 0 C Arbitrary time: ε ε dq q C - q = - = dt R RC RC ε dq dt = C - q RC ε dq dt = - q- C RC dq I = dt • using gives • loop rule: ε q dq - -R = 0 C dt Differential equation for q(t) Solution: Separation of variables
More math: 0 ε   q t 0 dq' dt' = - q- C RC   0 0 ε q t 1 ln q'- C = - t' RC ε ε       q- C t ln = - -C RC ε ε t - RC q- C = -C e         t - RC final q t = Q 1- e ε  final Q C = RC: time constant of the circuit; it tells us “how fast” the capacitor charges and discharges.
  ε ε ε ε        t t t t - - - - RC RC RC dq 1 C I t = = C e = e = e = e dt RC RC R R Current as a function of time: • take derivative:   0 ε    t t - - I t = e I e R
Charging a capacitor; summary:         t - RC final q t = Q 1- e   ε t - RC I t = e R Charging Capacitor 0 0.002 0.004 0.006 0.008 0.01 0 0.2 0.4 0.6 0.8 1 t (s) q (C) Charging Capacitor 0 0.01 0.02 0.03 0.04 0.05 0 0.2 0.4 0.6 0.8 1 t (s) I (A) Sample plots with =10 V, R=200 , and C=1000 F. RC=0.2 s recall that this is I0, also called Imax
In a time t=RC, the capacitor charges to Qfinal(1-e-1 ) or 63% of its capacity… Charging Capacitor 0 0.002 0.004 0.006 0.008 0.01 0 0.2 0.4 0.6 0.8 1 t (s) q (C) Charging Capacitor 0 0.01 0.02 0.03 0.04 0.05 0 0.2 0.4 0.6 0.8 1 t (s) I (A) RC=0.2 s …and the current drops to Imax(e-1 ) or 37% of its maximum. =RC is called the time constant of the RC circuit
Capacitor charged, switch open, no current flows. Discharging a Capacitor R switch C t<0 Close switch, current flows. t>0 I Kirchoff’s loop rule* (green loop) at the time when charge on C is q. q - IR = 0 C +Q0 -Q0 -q +q *Sign convention for capacitors is the same as for batteries: Voltage counts positive if going across from - to +.
q - IR = 0 C dq dt = - q RC dq q -R = dt C negative because current decreases charge on C Arbitrary time: dq I = - dt • using gives • loop rule: Differential equation for q(t) q dq + R = 0 C dt Solve:
More math: 0    q t t Q 0 0 dq' dt' 1 = - = - dt' q' RC RC   0  t q Q 0 1 ln q' = - dt RC       0 q t ln = - Q RC t - RC 0 q(t) = Q e t t - - 0 RC RC 0 Q dq I(t) = - = e = I e dt RC same equation as for charging
Discharging a capacitor; summary:   0 t - RC I t = I e Sample plots with =10 V, R=200 , and C=1000 F. RC=0.2 s t - RC 0 q(t) = Q e Discharging Capacitor 0 0.002 0.004 0.006 0.008 0.01 0 0.2 0.4 0.6 0.8 1 t (s) q (C) Discharging Capacitor 0 0.01 0.02 0.03 0.04 0.05 0 0.2 0.4 0.6 0.8 1 t (s) I (A)
Discharging Capacitor 0 0.01 0.02 0.03 0.04 0.05 0 0.2 0.4 0.6 0.8 1 t (s) I (A) Discharging Capacitor 0 0.002 0.004 0.006 0.008 0.01 0 0.2 0.4 0.6 0.8 1 t (s) q (C) In a time t=RC, the capacitor discharges to Q0e-1 or 37% of its initial value… RC=0.2 s …and the current drops to Imax(e-1 ) or 37% of its maximum.
Charging Discharging Charge Q(t) Q(t) = Qfinal(1-e-t/ ) Q(t) = Q0 e-t/ Capacitor voltage VC(t) VC(t) = (1-e-t/ ) VC(t) = V0 e-t/ = (Q0/C) e-t/ Resistor voltage VR(t) VR(t) = -VC(t) =  e-t/ VR(t) = VC(t)=V0 e-t/ = (Q0/C) e-t/ Current I(t) I(t) = I0 e-t/ = (/R) e-t/ I(t) = I0 e-t/ = [Q0/(RC)] e-t/ Only the equations for the charge Q(t) are starting equations. You must be able to derive the other quantities. =RC
Homework Hints V = IR Q(t) = CV(t) This is always true for a capacitor. Ohm’s law applies to resistors, not capacitors. Can give you the current only if you know V across the resistor. In a series RC circuit, the same current I flows through both the capacitor and the resistor.
Example: For the circuit shown C = 8 μF and ΔV = 30 V. Initially the capacitor is uncharged. The switch S is then closed and the capacitor begins to charge. Determine the charge on the capacitor at time t = 0.693RC, after the switch is closed. (From a prior test.) Also determine the current through the capacitor and voltage across the capacitor terminals at that time. C ΔV R S To be worked at the blackboard in lecture.
Example: For the circuit shown C = 8 μF and ΔV = 30 V. Initially the capacitor is uncharged. The switch S is then closed and the capacitor begins to charge. Determine the charge on the capacitor at time t = 0.693RC, after the switch is closed. (From a prior test.) Also determine the current through the capacitor and voltage across the capacitor terminals at that time. C ΔV R S Highlighted text tells us this is a charging capacitor problem.
Example: For the circuit shown C = 8 μF and ΔV = 30 V. Determine the charge on the capacitor at time t = 0.693RC, after the switch is closed. C ΔV R S         t - RC final q t = Q 1- e          0.693RC - RC q 0.693RC = C V 1- e       -6 -0.693 q 0.693RC = 8x10 30 1- e     -6 q 0.693RC = 240x10 1- 0.5    q 0.693RC = 120 C Nuc E’s should recognize that e-0.693 = ½.
Example: For the circuit shown C = 8 μF and ΔV = 30 V. Determine the current through the capacitor at t = 0.693RC. You can’t use V = IR! (Why?) C ΔV R S     dq t I t = dt                       t t - - RC RC final final d d I t = Q 1- e = -Q e dt dt                 t t t - - - RC RC RC final final d d t 1 I t = e = -Q e = -Q e dt dt RC RC
Example: For the circuit shown C = 8 μF and ΔV = 30 V. Determine the current through the capacitor at t = 0.693RC. C ΔV R S    t t - - final RC RC Q C V I t = e = e RC RC    t - RC V I t = e R           0.693RC - RC V V 1 I 0.693RC = e = R R 2    0 1 V 1 I 0.693RC = = I 2 R 2 We can’t provide a numerical answer because R (and therefore I0) is not given.
Example: For the circuit shown C = 8 μF and ΔV = 30 V. Determine the voltage across the capacitor terminals at time t = 0.693RC, after the switch is closed. C ΔV R S         t - RC final q t = Q 1- e We just derived an equation for V across the capacitor terminals as a function of time! Handy!          t - RC C V t = C V 1- e          t - RC V t = V 1- e               t t - - RC RC 0 V t = 1- e = V 1- e V, , and V0 usually mean the same thing, but check the context!
Example: For the circuit shown C = 8 μF and ΔV = 30 V. Determine the voltage across the capacitor terminals at time t = 0.693RC, after the switch is closed. C ΔV R S          t - RC V t = V 1- e         0.693RC - RC V 0.693RC = 30 1- e         1 V 0.693RC = 30 1- = 15 V 2
Example: For the circuit shown C = 8 μF and ΔV = 30 V. Determine the voltage across the capacitor terminals at time t = 0.693RC, after the switch is closed. C ΔV R S   V 0.693RC = 15 V  R C Note that V + V = V, so      R C V 0.693RC = V - V 0.693RC   R V 0.693RC = 30 - 15= 15 V     V 0.693RC 15 I 0.693RC = = R R An alternative way to calculate I(0.693 RC), except we still don’t know R. Digression…
Example: For the circuit shown C = 8 μF and ΔV = 30 V. Determine the voltage across the capacitor terminals at time t = 0.693RC, after the switch is closed. C ΔV R S    q 0.693RC = 120 C  q(t) q(t) C= V(t) = V(t) C Easier! A different way to calculate V(t)…   -6 -6 120 10 V(0.693RC) = = 15 V 8 10
Demo Charging and discharging a capacitor. Instead of doing a physical demo, if I have time I will do a virtual demo using the applet linked on the next slide. The applet illustrates the same principles as the physical demo.
make your own capacitor circuits http://phet.colorado.edu/en/simulation/circuit-construction-kit-ac For a “pre-built” RC circuit that lets you both charge and discharge (through separate switches), download this file, put it in your “my documents” folder, run the circuit construction applet (link above), maximize it, then select “load” in the upper right. Click on the “capacitor_circuit” file and give the program permission to run it. You can put voltmeters and ammeters in your circuit. You can change values or R, C, and V. Also, click on the “current chart” button for a plot of current (you can have more than one in your applet) or the “voltage chart” button for a plot of voltage.
more applets http://webphysics.davidson.edu/physlet_resources/bu_semester2/c11_RC.html http://subaru.univ-lemans.fr/AccesLibre/UM/Pedago/physique/02/electri/condo2.html http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=31.0

More Related Content

PPT
lecture13.ppt
bykumarjha16
 
PPT
lecture13.ppt
byArpitMistry6
 
PPT
EM material eee.ppt
byVara Prasad
 
PPT
electrical measurements
byanu200770
 
PPT
Usage of meters
byAdithya Ballaji
 
PDF
Ammeter and voltmeter
byHemin Patel
 
PPT
5 ammeter and voltmeter
byNeil MacIntosh
 
PDF
3-phyl201-ammeters-v3.pdf Boun Lab ammeter exp
byasafgungordu
 
lecture13.ppt
bykumarjha16
 
lecture13.ppt
byArpitMistry6
 
EM material eee.ppt
byVara Prasad
 
electrical measurements
byanu200770
 
Usage of meters
byAdithya Ballaji
 
Ammeter and voltmeter
byHemin Patel
 
5 ammeter and voltmeter
byNeil MacIntosh
 
3-phyl201-ammeters-v3.pdf Boun Lab ammeter exp
byasafgungordu
 

Similar to chap-3lecture13.pptIEC 61850 Standard For Substation Automation

PPTX
Physics 28 - Electrical quantities 2.pptx
byJayrNaces
 
PPTX
Electric circuits
byjmemler
 
PPTX
ELECTROMECHANICAL-INSTRUMENTATION INSTRUMENTS.pptx
byJANUBASGLORIBETHL
 
PDF
eem-2020309.pdf
bySACHCHIDANAND JHA
 
PDF
Usage of meters - Notes
byAdithya Ballaji
 
DOCX
Lab report 2
byEvandro Conceicao
 
PPT
5.2
byPaula Mills
 
PPTX
FUNDAMENTAL OF VOLTAGE AND CURRENT
byAmolD8
 
PPTX
Current electricity
byNatasia Gouws
 
PPTX
Electric Circuits
byitutor
 
PPT
3059
byRockr Vasanth
 
DOCX
Voltmeter - Measurement and Instruments
byGowtham Cr
 
PPTX
Electrical Test Equipment and devices in Works
byAlejandroTPedria
 
PDF
Bai thi nghiem_linear_circuit_cttt
bykutyonljne
 
PDF
Electricity
bymrwoollaston
 
PPT
1_IMP_electricity.ppt
byYash93108
 
PPTX
AS Level Physics- Electric circuits
byEbin Robinson
 
PPTX
Current electricity.pptxyyyuujjjjjjkkkn
bySaulosiSaidi
 
PPT
Topic 1 a_basic_concepts_and_theorem
byGabriel O'Brien
 
PPTX
1 Conversion of Galvanometer into Ammeter and Voltmeter 2 Differences between...
byPriyanka Jakhar
 
Physics 28 - Electrical quantities 2.pptx
byJayrNaces
 
Electric circuits
byjmemler
 
ELECTROMECHANICAL-INSTRUMENTATION INSTRUMENTS.pptx
byJANUBASGLORIBETHL
 
eem-2020309.pdf
bySACHCHIDANAND JHA
 
Usage of meters - Notes
byAdithya Ballaji
 
Lab report 2
byEvandro Conceicao
 
5.2
byPaula Mills
 
FUNDAMENTAL OF VOLTAGE AND CURRENT
byAmolD8
 
Current electricity
byNatasia Gouws
 
Electric Circuits
byitutor
 
3059
byRockr Vasanth
 
Voltmeter - Measurement and Instruments
byGowtham Cr
 
Electrical Test Equipment and devices in Works
byAlejandroTPedria
 
Bai thi nghiem_linear_circuit_cttt
bykutyonljne
 
Electricity
bymrwoollaston
 
1_IMP_electricity.ppt
byYash93108
 
AS Level Physics- Electric circuits
byEbin Robinson
 
Current electricity.pptxyyyuujjjjjjkkkn
bySaulosiSaidi
 
Topic 1 a_basic_concepts_and_theorem
byGabriel O'Brien
 
1 Conversion of Galvanometer into Ammeter and Voltmeter 2 Differences between...
byPriyanka Jakhar
 

More from getaneh30

PDF
296713773-IEC-61850-Based-Substation-Automation-System-SAS-PPT-1.pdf
bygetaneh30
 
PPTX
best-Based-Substation-Automation-System-SAS-PPT-1.pptx
bygetaneh30
 
PPTX
296713773-IEC-61850-Based-Substation-Automation-System-SAS-PPT-1.pptx
bygetaneh30
 
PPT
chap-4,5,besfinal.ppt@ best presentation for chap-4 & 5,PLC2025.ppt
bygetaneh30
 
PPT
best presentation for chap-4 & 5,PLC2025.ppt
bygetaneh30
 
PPTX
C4A - Magnetic Field and Transformer.pptx
bygetaneh30
 
PPTX
Presentation__Computer_Maintenance_Basics.pptx
bygetaneh30
 
PDF
Chapter_2_Basic_Ladder_Logic_Programming (1).pdf
bygetaneh30
 
PPTX
General_overview_of_scada_system_in_powe.pptx
bygetaneh30
 
PPTX
BEST PPT FOR AGC.pptx and without timers
bygetaneh30
 
PPTX
Literature Survey.PPT.pptx4.3 Assemble Star-delta starting with timers
bygetaneh30
 
PDF
electric governor.pdf,MODELING,OFELECTRICAL
bygetaneh30
 
PDF
lecture11.pdf
bygetaneh30
 
296713773-IEC-61850-Based-Substation-Automation-System-SAS-PPT-1.pdf
bygetaneh30
 
best-Based-Substation-Automation-System-SAS-PPT-1.pptx
bygetaneh30
 
296713773-IEC-61850-Based-Substation-Automation-System-SAS-PPT-1.pptx
bygetaneh30
 
chap-4,5,besfinal.ppt@ best presentation for chap-4 & 5,PLC2025.ppt
bygetaneh30
 
best presentation for chap-4 & 5,PLC2025.ppt
bygetaneh30
 
C4A - Magnetic Field and Transformer.pptx
bygetaneh30
 
Presentation__Computer_Maintenance_Basics.pptx
bygetaneh30
 
Chapter_2_Basic_Ladder_Logic_Programming (1).pdf
bygetaneh30
 
General_overview_of_scada_system_in_powe.pptx
bygetaneh30
 
BEST PPT FOR AGC.pptx and without timers
bygetaneh30
 
Literature Survey.PPT.pptx4.3 Assemble Star-delta starting with timers
bygetaneh30
 
electric governor.pdf,MODELING,OFELECTRICAL
bygetaneh30
 
lecture11.pdf
bygetaneh30
 

Recently uploaded

PDF
ElyriaSoftware — Powering the Future with Blockchain Innovation
byElyria Software
 
PDF
DIGITAL FORENSICS - Notes for Everything.pdf
bypankajkumavatbeit
 
PPTX
Ethics in AI - Artificial Intelligence Fundamentals.pptx
byemenyiblessing
 
PDF
Unser Jahresrückblick – MarvelClient in 2025
bypanagenda
 
PPTX
From Backup to Resilience: How MSPs Are Preparing for 2026
byMSP360
 
PDF
Energy Storage Landscape Clean Energy Ministerial
bySurajitBanerjee38
 
PDF
Safeguarding AI-Based Financial Infrastructure
byYasir Naveed Riaz
 
PDF
Generative AI in 2026: Hype, Bubble, Winter or The Real Deal?
byDr. Tathagat Varma
 
PDF
How Mobile Apps Are Shaping the Future of Digital Innovation
byWilliam Taylor
 
PDF
The year in review - MarvelClient in 2025
bypanagenda
 
PDF
Accelerating Responsible AI Adoption in Public Sector and Private Organizations.
byYasir Naveed Riaz
 
PPTX
The Future of IT Service Management AI Automation & Beyond.pptx
byChetu
 
PDF
Day 1 - Cloud Security Strategy and Planning ~ 2nd Sight Lab ~ Cloud Security...
by2nd Sight Lab
 
PDF
The major tech developments for 2026 by Pluralsight, a research and training ...
byChris Skinner
 
PDF
Recursive Self Improvement vs Continuous Learning
byBob Marcus
 
PPTX
Unit-4-ARTIFICIAL NEURAL NETWORKS.pptx ANN ppt Artificial neural network
byssuserd4481a
 
PDF
Knowing and Doing: Knowledge graphs, AI, and work
bymarainglezakis1
 
PPTX
Cybersecurity Best Practices - Step by Step guidelines
byYasir Naveed Riaz
 
PDF
Eredità digitale sugli smartphone: cosa resta di noi nei dispositivi mobili
bySpeck&Tech
 
PDF
Generative AI in UiPath: Mastering the Generative Extractor for Intelligent D...
byDianaGray10
 
ElyriaSoftware — Powering the Future with Blockchain Innovation
byElyria Software
 
DIGITAL FORENSICS - Notes for Everything.pdf
bypankajkumavatbeit
 
Ethics in AI - Artificial Intelligence Fundamentals.pptx
byemenyiblessing
 
Unser Jahresrückblick – MarvelClient in 2025
bypanagenda
 
From Backup to Resilience: How MSPs Are Preparing for 2026
byMSP360
 
Energy Storage Landscape Clean Energy Ministerial
bySurajitBanerjee38
 
Safeguarding AI-Based Financial Infrastructure
byYasir Naveed Riaz
 
Generative AI in 2026: Hype, Bubble, Winter or The Real Deal?
byDr. Tathagat Varma
 
How Mobile Apps Are Shaping the Future of Digital Innovation
byWilliam Taylor
 
The year in review - MarvelClient in 2025
bypanagenda
 
Accelerating Responsible AI Adoption in Public Sector and Private Organizations.
byYasir Naveed Riaz
 
The Future of IT Service Management AI Automation & Beyond.pptx
byChetu
 
Day 1 - Cloud Security Strategy and Planning ~ 2nd Sight Lab ~ Cloud Security...
by2nd Sight Lab
 
The major tech developments for 2026 by Pluralsight, a research and training ...
byChris Skinner
 
Recursive Self Improvement vs Continuous Learning
byBob Marcus
 
Unit-4-ARTIFICIAL NEURAL NETWORKS.pptx ANN ppt Artificial neural network
byssuserd4481a
 
Knowing and Doing: Knowledge graphs, AI, and work
bymarainglezakis1
 
Cybersecurity Best Practices - Step by Step guidelines
byYasir Naveed Riaz
 
Eredità digitale sugli smartphone: cosa resta di noi nei dispositivi mobili
bySpeck&Tech
 
Generative AI in UiPath: Mastering the Generative Extractor for Intelligent D...
byDianaGray10
 

chap-3lecture13.pptIEC 61850 Standard For Substation Automation

  • 1.
    Today’s agenda: Measuring Instruments:ammeter, voltmeter, ohmmeter. You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments. RC Circuits. You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations.
  • 2.
    Measuring current, voltage,and resistance A Ammeter: •measures current (A) •connected in series (current must go through instrument) I V a b Voltmeter: •measures potential difference (V) •connected in parallel Ohmmeter: •measures resistance of an isolated resistor (not in a working circuit) 
  • 3.
    Effect of ammeteron circuit V R Measuring current in a simple circuit: A • connect ammeter in series Are we measuring the correct current? (the current in the circuit without ammeter)
  • 4.
    • current withoutthe ammeter would be V R . V I = R Measuring current in a simple circuit: . V I = R + r To minimize error, ammeter resistance r must be very small. (ideal ammeter would have zero resistance) • any ammeter has some resistance r. r • connect ammeter in series Are we measuring the correct current? (the current in the circuit without ammeter) • current in presence of ammeter is Effect of ammeter on circuit
  • 5.
    Example: an ammeterof resistance 10 m is used to measure the current through a 10  resistor in series with a 3 V battery that has an internal resistance of 0.5 . What is the relative (percent) error caused by the ammeter? V=3 V R=10  r=0.5  Actual current without ammeter: V I = R + r 3 I = A 10+ 0.5 I = 0.2857 A = 285.7 mA You might see the symbol  used instead of V.
  • 6.
    V=3 V R=10  r=0.5 Current with ammeter: A V I = R + r+ R 3 I = A 10+ 0.5+ 0.01 I = 0.2854 A = 285.4 mA RA  0.2857- 0.2854 % Error = 100 0.2857 % Error = 0.1 %
  • 7.
    Designing an ammeter http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1 Galvanometer: •currentflows through a coil in a magnetic field •coil experiences a torque, connected needle deflects (see later chapters of this class)
  • 8.
    Designing an ammeter •ammeter can be based on galvanometer (for electronic instrument, use electronic sensor instead, analysis still applies) • simplest case: send current directly through galvanometer, observe deflection of needle Needle deflection is proportional to current. Each galvanometer has a certain maximum current corresponding to full needle deflection. What if you need to measure a larger current? • use shunt resistor
  • 9.
    Ammeter uses agalvanometer and a shunt, connected in parallel: A I  Everything inside the green box is the ammeter. • Current I gets split into Ishunt and IG Homework hint: If your galvanometer reads 1A full scale but you want the ammeter to read 5A full scale, then RSHUNT must result in IG=1A when I=5A. What are ISHUNT and VSHUNT? G RG RSHUNT IG ISHUNT I A B galvanometer
  • 10.
    G RG RSHUNT IG ISHUNT I A B Shunt alsoreduces resistance of the ammeter:   A G SHUNT 1 1 1 R R R   G SHUNT A G SHUNT R R R R R
  • 11.
    A galvanometer-based ammeteruses a galvanometer and a shunt, connected in parallel: G RG RS IG IS I   A G S 1 1 1 R R R Example: what shunt resistance is required for an ammeter to have a resistance of 10 m, if the galvanometer resistance is 60 ?   S A G 1 1 1 R R R        G A S G A 60 .01 R R R 0.010 R -R 60-.01 (actually 0.010002 ) To achieve such a small resistance, the shunt is probably a large-diameter wire or solid piece of metal.
  • 12.
    Web links: ammeterdesign, ammeter impact on circuit, clamp-on ammeter (based on principles we will soon be studying).
  • 13.
    Effect of voltmeteron circuit Measuring voltage (potential difference) Vab in a simple circuit: • connect voltmeter in parallel Are we measuring the correct voltage? (the voltage in the circuit without voltmeter) =3 V R=10  r=0.5  V a b
  • 14.
    • extra currentchanges voltage drop across r and thus Vab To minimize error, voltmeter resistance r must be very large. (ideal voltmeter would have infinite resistance) • voltmeter has some resistance RV • current IV flows through voltmeter Effect of voltmeter on circuit Measuring voltage (potential difference) Vab in a simple circuit: • connect voltmeter in parallel Are we measuring the correct voltage? (the voltage in the circuit without voltmeter) =3 V R=10  r=0.5  RV a b IV
  • 15.
    Example: a galvanometerof resistance 60  is used to measure the voltage drop across a 10 k resistor in series with an ideal 6 V battery and a 5 k resistor. What is the relative error caused by the nonzero resistance of the galvanometer? Actual voltage drop without instrument: V=6 V R1=10 k R2=5 k a b    3 eq 1 2 R R + R = 15 10       -3 3 eq V 6 V I 0.4 10 A R 15 10         -3 3 ab V = IR 0.4 10 10 10 4 V
  • 16.
    The measurement ismade with the galvanometer. V=6 V R1=10 k R2=5 k G RG=60  a b 60  and 10 k resistors in parallel are equivalent to 59.6  resistor. Total equivalent resistance: 5059.6  Total current: I=1.186x10-3 A I=1.19 mA Vab = 6V – IR2 = 0.07 V. The relative error is:  4-.07 % Error = 100= 98% 4 Would you pay for this voltmeter? Would you pay for this voltmeter? We need a better instrument!
  • 17.
    Example: a voltmeterof resistance 100 k is used to measure the voltage drop across a 10 k resistor in series with an ideal 6 V battery and a 5 k resistor. What is the percent error caused by the nonzero resistance of the voltmeter? We already calculated the actual voltage drop (2 slides back). V=6 V R1=10 k R2=5 k a b         -3 3 ab V = IR 0.4 10 10 10 4 V
  • 18.
    The measurement isnow made with the “better” voltmeter. V=6 V R1=10 k R2=5 k V RV=100 k a b 100 k and 10 k resistors in parallel are equivalent to an 9090  resistor. Total equivalent resistance: 14090  Total current: I=4.26x10-4 A I=.426 mA The voltage drop from a to b: 6-(4.26x10-4 )(5000)=3.87 V. The percent error is.  4- 3.87 % Error = 100= 3.25% 4 Not great, but much better.
  • 19.
    • voltmeter musthave a very large resistance • voltmeter can be made from galvanometer in series with a large resistance V G RG RSer  Everything inside the blue box is the voltmeter. a b Vab a b Vab Homework hints: “the galvanometer reads 1A full scale” would mean a current of IG=1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected RSer must result in IG=1A when Vab=10V. Homework hints: “the galvanometer reads 1A full scale” would mean a current of IG=1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected RSer must result in IG=1A when Vab=10V. Designing a voltmeter
  • 20.
    • Ohmmeter measuresresistance of isolated resistor • Ohmmeter can be made from a galvanometer, a series resistance, and a battery (active device). G RG RSer R=? • Terminals of ohmmeter are connected to unknown resistor • battery causes current to flow and galvanometer to deflect • V=I (Rser + RG + R) solve for unknown R  Measuring Instruments: Ohmmeter Everything inside the blue box is the ohmmeter. V
  • 21.
    Alternatively: •separately measure currentand voltage for resistor •Apply Ohm’s law Four-point probe: V A reference: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movcoil.html#c4
  • 22.
    Today’s agenda: Measuring Instruments:ammeter, voltmeter, ohmmeter. You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments. RC Circuits. You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations.
  • 23.
    Charging and discharginga capacitor • so far, we have assumed that charge instantly appears on capacitor • in reality, capacitor does not change instantaneously • charging speed depends on capacitance C and on resistance R between the battery and the capacitor Q t Q t What happens if we connect a capacitor to a voltage source? Charging and discharging are time-dependent phenomena!
  • 24.
    Switch open, nocurrent flows. RC circuit: Charging a Capacitor  R switch C t<0 Close switch, current flows. t>0 I Kirchoff’s loop rule* (green loop) at the time when charge on C is q. *Sign convention for capacitors is the same as for batteries: Voltage counts positive if going across from - to +. ε q - - IR = 0 C This equation is deceptively complex because I depends on q and both depend on time. - - + + -q +q
  • 25.
    Empty capacitor: (right afterclosing the switch) q=0 VC=0, VR=  I=I0=/R  R switch C Full capacitor: (after a very long time) VC=, VR= 0 Q=C I=0 I ε q - - IR = 0 C - - + + q0=0 t=0 qF=C t= IF=0 Distinguish capacitor and resistor voltages VC and VR. They are not equal but VC + VR = . Limiting cases:
  • 26.
    ε q - - IR= 0 C Arbitrary time: ε ε dq q C - q = - = dt R RC RC ε dq dt = C - q RC ε dq dt = - q- C RC dq I = dt • using gives • loop rule: ε q dq - -R = 0 C dt Differential equation for q(t) Solution: Separation of variables
  • 27.
    More math: 0ε   q t 0 dq' dt' = - q- C RC   0 0 ε q t 1 ln q'- C = - t' RC ε ε       q- C t ln = - -C RC ε ε t - RC q- C = -C e         t - RC final q t = Q 1- e ε  final Q C = RC: time constant of the circuit; it tells us “how fast” the capacitor charges and discharges.
  • 28.
      ε εε ε        t t t t - - - - RC RC RC dq 1 C I t = = C e = e = e = e dt RC RC R R Current as a function of time: • take derivative:   0 ε    t t - - I t = e I e R
  • 29.
    Charging a capacitor;summary:         t - RC final q t = Q 1- e   ε t - RC I t = e R Charging Capacitor 0 0.002 0.004 0.006 0.008 0.01 0 0.2 0.4 0.6 0.8 1 t (s) q (C) Charging Capacitor 0 0.01 0.02 0.03 0.04 0.05 0 0.2 0.4 0.6 0.8 1 t (s) I (A) Sample plots with =10 V, R=200 , and C=1000 F. RC=0.2 s recall that this is I0, also called Imax
  • 30.
    In a timet=RC, the capacitor charges to Qfinal(1-e-1 ) or 63% of its capacity… Charging Capacitor 0 0.002 0.004 0.006 0.008 0.01 0 0.2 0.4 0.6 0.8 1 t (s) q (C) Charging Capacitor 0 0.01 0.02 0.03 0.04 0.05 0 0.2 0.4 0.6 0.8 1 t (s) I (A) RC=0.2 s …and the current drops to Imax(e-1 ) or 37% of its maximum. =RC is called the time constant of the RC circuit
  • 31.
    Capacitor charged, switch open,no current flows. Discharging a Capacitor R switch C t<0 Close switch, current flows. t>0 I Kirchoff’s loop rule* (green loop) at the time when charge on C is q. q - IR = 0 C +Q0 -Q0 -q +q *Sign convention for capacitors is the same as for batteries: Voltage counts positive if going across from - to +.
  • 32.
    q - IR =0 C dq dt = - q RC dq q -R = dt C negative because current decreases charge on C Arbitrary time: dq I = - dt • using gives • loop rule: Differential equation for q(t) q dq + R = 0 C dt Solve:
  • 33.
    More math: 0   q t t Q 0 0 dq' dt' 1 = - = - dt' q' RC RC   0  t q Q 0 1 ln q' = - dt RC       0 q t ln = - Q RC t - RC 0 q(t) = Q e t t - - 0 RC RC 0 Q dq I(t) = - = e = I e dt RC same equation as for charging
  • 34.
    Discharging a capacitor;summary:   0 t - RC I t = I e Sample plots with =10 V, R=200 , and C=1000 F. RC=0.2 s t - RC 0 q(t) = Q e Discharging Capacitor 0 0.002 0.004 0.006 0.008 0.01 0 0.2 0.4 0.6 0.8 1 t (s) q (C) Discharging Capacitor 0 0.01 0.02 0.03 0.04 0.05 0 0.2 0.4 0.6 0.8 1 t (s) I (A)
  • 35.
    Discharging Capacitor 0 0.01 0.02 0.03 0.04 0.05 0 0.20.4 0.6 0.8 1 t (s) I (A) Discharging Capacitor 0 0.002 0.004 0.006 0.008 0.01 0 0.2 0.4 0.6 0.8 1 t (s) q (C) In a time t=RC, the capacitor discharges to Q0e-1 or 37% of its initial value… RC=0.2 s …and the current drops to Imax(e-1 ) or 37% of its maximum.
  • 36.
    Charging Discharging Charge Q(t)Q(t) = Qfinal(1-e-t/ ) Q(t) = Q0 e-t/ Capacitor voltage VC(t) VC(t) = (1-e-t/ ) VC(t) = V0 e-t/ = (Q0/C) e-t/ Resistor voltage VR(t) VR(t) = -VC(t) =  e-t/ VR(t) = VC(t)=V0 e-t/ = (Q0/C) e-t/ Current I(t) I(t) = I0 e-t/ = (/R) e-t/ I(t) = I0 e-t/ = [Q0/(RC)] e-t/ Only the equations for the charge Q(t) are starting equations. You must be able to derive the other quantities. =RC
  • 37.
    Homework Hints V =IR Q(t) = CV(t) This is always true for a capacitor. Ohm’s law applies to resistors, not capacitors. Can give you the current only if you know V across the resistor. In a series RC circuit, the same current I flows through both the capacitor and the resistor.
  • 38.
    Example: For thecircuit shown C = 8 μF and ΔV = 30 V. Initially the capacitor is uncharged. The switch S is then closed and the capacitor begins to charge. Determine the charge on the capacitor at time t = 0.693RC, after the switch is closed. (From a prior test.) Also determine the current through the capacitor and voltage across the capacitor terminals at that time. C ΔV R S To be worked at the blackboard in lecture.
  • 39.
    Example: For thecircuit shown C = 8 μF and ΔV = 30 V. Initially the capacitor is uncharged. The switch S is then closed and the capacitor begins to charge. Determine the charge on the capacitor at time t = 0.693RC, after the switch is closed. (From a prior test.) Also determine the current through the capacitor and voltage across the capacitor terminals at that time. C ΔV R S Highlighted text tells us this is a charging capacitor problem.
  • 40.
    Example: For thecircuit shown C = 8 μF and ΔV = 30 V. Determine the charge on the capacitor at time t = 0.693RC, after the switch is closed. C ΔV R S         t - RC final q t = Q 1- e          0.693RC - RC q 0.693RC = C V 1- e       -6 -0.693 q 0.693RC = 8x10 30 1- e     -6 q 0.693RC = 240x10 1- 0.5    q 0.693RC = 120 C Nuc E’s should recognize that e-0.693 = ½.
  • 41.
    Example: For thecircuit shown C = 8 μF and ΔV = 30 V. Determine the current through the capacitor at t = 0.693RC. You can’t use V = IR! (Why?) C ΔV R S     dq t I t = dt                       t t - - RC RC final final d d I t = Q 1- e = -Q e dt dt                 t t t - - - RC RC RC final final d d t 1 I t = e = -Q e = -Q e dt dt RC RC
  • 42.
    Example: For thecircuit shown C = 8 μF and ΔV = 30 V. Determine the current through the capacitor at t = 0.693RC. C ΔV R S    t t - - final RC RC Q C V I t = e = e RC RC    t - RC V I t = e R           0.693RC - RC V V 1 I 0.693RC = e = R R 2    0 1 V 1 I 0.693RC = = I 2 R 2 We can’t provide a numerical answer because R (and therefore I0) is not given.
  • 43.
    Example: For thecircuit shown C = 8 μF and ΔV = 30 V. Determine the voltage across the capacitor terminals at time t = 0.693RC, after the switch is closed. C ΔV R S         t - RC final q t = Q 1- e We just derived an equation for V across the capacitor terminals as a function of time! Handy!          t - RC C V t = C V 1- e          t - RC V t = V 1- e               t t - - RC RC 0 V t = 1- e = V 1- e V, , and V0 usually mean the same thing, but check the context!
  • 44.
    Example: For thecircuit shown C = 8 μF and ΔV = 30 V. Determine the voltage across the capacitor terminals at time t = 0.693RC, after the switch is closed. C ΔV R S          t - RC V t = V 1- e         0.693RC - RC V 0.693RC = 30 1- e         1 V 0.693RC = 30 1- = 15 V 2
  • 45.
    Example: For thecircuit shown C = 8 μF and ΔV = 30 V. Determine the voltage across the capacitor terminals at time t = 0.693RC, after the switch is closed. C ΔV R S   V 0.693RC = 15 V  R C Note that V + V = V, so      R C V 0.693RC = V - V 0.693RC   R V 0.693RC = 30 - 15= 15 V     V 0.693RC 15 I 0.693RC = = R R An alternative way to calculate I(0.693 RC), except we still don’t know R. Digression…
  • 46.
    Example: For thecircuit shown C = 8 μF and ΔV = 30 V. Determine the voltage across the capacitor terminals at time t = 0.693RC, after the switch is closed. C ΔV R S    q 0.693RC = 120 C  q(t) q(t) C= V(t) = V(t) C Easier! A different way to calculate V(t)…   -6 -6 120 10 V(0.693RC) = = 15 V 8 10
  • 47.
    Demo Charging and discharging acapacitor. Instead of doing a physical demo, if I have time I will do a virtual demo using the applet linked on the next slide. The applet illustrates the same principles as the physical demo.
  • 48.
    make your owncapacitor circuits http://phet.colorado.edu/en/simulation/circuit-construction-kit-ac For a “pre-built” RC circuit that lets you both charge and discharge (through separate switches), download this file, put it in your “my documents” folder, run the circuit construction applet (link above), maximize it, then select “load” in the upper right. Click on the “capacitor_circuit” file and give the program permission to run it. You can put voltmeters and ammeters in your circuit. You can change values or R, C, and V. Also, click on the “current chart” button for a plot of current (you can have more than one in your applet) or the “voltage chart” button for a plot of voltage.
  • 49.