The document discusses measuring instruments, RC circuits, and charging/discharging capacitors. It begins with an agenda covering measuring instruments like ammeters, voltmeters, and ohmmeters. It then discusses calculating currents and voltages in RC circuits, and calculating the time constant of an RC circuit. The document provides details on how ammeters, voltmeters, and ohmmeters work, including the effects of their internal resistances on measurements. It also covers charging and discharging a capacitor in an RC circuit over time using differential equations and plots the capacitor voltage, charge, and current versus time.
To understand the basic working principle of a transformer.
To obtain the equivalent circuit parameters from Open circuit and Short circuit tests, and to estimate efficiency & regulation at various loads.
EEE 117L Network Analysis Laboratory Lab 1
1
EEE 117L Network Analysis Laboratory
Lab 1 – Voltage/Current Division and Filters
Lab Overview
The objective of Lab 1 is to familiarize students with a variety of basic applications of
passive R, C devices, and also how to measure the performance of these circuits using
both Spice simulations and the Digilent Analog Discovery 2 on the circuits constructed.
Prelab
Before coming to lab, students need to complete the following items for each of the
circuits studied in this lab :
• Any hand calculations needed to determine the values of components used in the
circuits such as resistors and capacitors, or specifications such as pole frequencies.
• A Spice simulation of each circuit to get familiar with how it works, and determine
what to expect when the circuit is built and its performance is measured.
Making connections on a Breadboard
Breadboards are used to easily construct circuits without the need to solder parts on a
printed circuit board. As seen in Figure 0 they have columns of pins that are connected
together internally, so that all the wires inserted in a column are shorted together. Note
that the columns on top and bottom are not connected together. There are also rows of
pins at the top and bottom that are connected together. These rows are intended for use
as the power supplies, and are typically labeled + and – and color coded red and blue for
the positive and negative power supplies. These rows are not connected in the middle.
Figure 0.
EEE 117L Network Analysis Laboratory Lab 1
2
Circuits to be studied
When choosing resistor and capacitor values use standard values available to you,
and keep all resistor values between 100 W and 100 kW.
1. Voltage and Current Dividers
One of the most commonly used circuits is a voltage divider
like the one shown in Figure 1.a. For example, if a signal is
too large to be input to a voltmeter or oscilloscope it can be
attenuated (reduced in size) using voltage division. The DC
voltage that an AC signal like a sine wave varies around can
also be reduced using this circuit.
For example, if all of the resistors in this circuit are the same
value, and the VS input source provides a DC voltage of 4V,
then the voltages in this circuit will be VA = 4V, VB = 3V,
VC = 2V, and VD = 1V. That is, voltage division will cause the voltage at node B to be
¾ of VS , the voltage at node C to be ½ of VS , and the voltage at node D to be ¼ of VS.
If a sine wave with an amplitude of 1V is then added so that VS = 4 + sin(wt) Volts, then
voltage division will cause the new values of VA , VB , VC and VD to be :
VA = 1.00*VS = 1.00*(4 + sin(wt)) = 4 + 1.00*sin(wt) Volts
VB = 0.75*VS = 0.75*(4 + sin(wt)) = 3 + 0.75*sin(wt) Volts
VC = 0.50*VS = 0.50*(4 + sin(wt)) = 2 + 0.50*sin(wt) Volts
VD = 0.25*VS = 0.25*(4 + sin(wt)) = 1 + 0.25*sin(wt) Volts
In this example both the amplitude of the ...
To understand the basic working principle of a transformer.
To obtain the equivalent circuit parameters from Open circuit and Short circuit tests, and to estimate efficiency & regulation at various loads.
EEE 117L Network Analysis Laboratory Lab 1
1
EEE 117L Network Analysis Laboratory
Lab 1 – Voltage/Current Division and Filters
Lab Overview
The objective of Lab 1 is to familiarize students with a variety of basic applications of
passive R, C devices, and also how to measure the performance of these circuits using
both Spice simulations and the Digilent Analog Discovery 2 on the circuits constructed.
Prelab
Before coming to lab, students need to complete the following items for each of the
circuits studied in this lab :
• Any hand calculations needed to determine the values of components used in the
circuits such as resistors and capacitors, or specifications such as pole frequencies.
• A Spice simulation of each circuit to get familiar with how it works, and determine
what to expect when the circuit is built and its performance is measured.
Making connections on a Breadboard
Breadboards are used to easily construct circuits without the need to solder parts on a
printed circuit board. As seen in Figure 0 they have columns of pins that are connected
together internally, so that all the wires inserted in a column are shorted together. Note
that the columns on top and bottom are not connected together. There are also rows of
pins at the top and bottom that are connected together. These rows are intended for use
as the power supplies, and are typically labeled + and – and color coded red and blue for
the positive and negative power supplies. These rows are not connected in the middle.
Figure 0.
EEE 117L Network Analysis Laboratory Lab 1
2
Circuits to be studied
When choosing resistor and capacitor values use standard values available to you,
and keep all resistor values between 100 W and 100 kW.
1. Voltage and Current Dividers
One of the most commonly used circuits is a voltage divider
like the one shown in Figure 1.a. For example, if a signal is
too large to be input to a voltmeter or oscilloscope it can be
attenuated (reduced in size) using voltage division. The DC
voltage that an AC signal like a sine wave varies around can
also be reduced using this circuit.
For example, if all of the resistors in this circuit are the same
value, and the VS input source provides a DC voltage of 4V,
then the voltages in this circuit will be VA = 4V, VB = 3V,
VC = 2V, and VD = 1V. That is, voltage division will cause the voltage at node B to be
¾ of VS , the voltage at node C to be ½ of VS , and the voltage at node D to be ¼ of VS.
If a sine wave with an amplitude of 1V is then added so that VS = 4 + sin(wt) Volts, then
voltage division will cause the new values of VA , VB , VC and VD to be :
VA = 1.00*VS = 1.00*(4 + sin(wt)) = 4 + 1.00*sin(wt) Volts
VB = 0.75*VS = 0.75*(4 + sin(wt)) = 3 + 0.75*sin(wt) Volts
VC = 0.50*VS = 0.50*(4 + sin(wt)) = 2 + 0.50*sin(wt) Volts
VD = 0.25*VS = 0.25*(4 + sin(wt)) = 1 + 0.25*sin(wt) Volts
In this example both the amplitude of the ...
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
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Final project report on grocery store management system..pdfKamal Acharya
In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
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EM material eee.ppt
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 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)
4. • 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
5. 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.
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%
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 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
11. 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.
12. Web links: ammeter design, ammeter impact on circuit,
clamp-on ammeter (based on principles we will soon be
studying).
13. 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
14. • 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
15. 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
16. 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!
17. 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
18. 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.
19. • 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
20. • 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
21. 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
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 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!
24. 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
25. 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:
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: 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.
28.
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
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 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
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 0.2 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.
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 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.
39. 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.
40. 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
0.693 RC
-
RC
q 0.693 RC =C V 1-e
-6 -0.693
q 0.693 RC = 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 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
42. 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
0.693 RC
-
RC
V V 1
I 0.693 RC = e =
R R 2
0
1 V 1
I 0.693 RC = = I
2 R 2
We can’t provide a numerical answer
because R (and therefore I0) is not given.
43. 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!
44. 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
0.693 RC
-
RC
V 0.693 RC =30 1-e
1
V 0.693 RC = 30 1- =15 V
2
45. 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 0.693 RC 15
I 0.693 RC = =
R R
An alternative way to calculate I(0.693 RC),
except we still don’t know R.
Digression…
46. 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(0.693 RC)= =15 V
8 10
47. 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.
48. 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.