1. The document discusses electric current (I), potential difference (V), resistance (R), and their relationships. Ohm's law states that I=V/R, where R is constant for a given resistor.
2. Resistance depends on the resistivity (ρ) of the material, its length (l), and cross-sectional area (A). R=ρl/A. Materials have different typical resistivity values.
3. In a circuit with both internal and external resistance, the total potential difference (E) across the circuit equals the potential drop across the external resistor (VR) plus the internal resistor (Vr). E=VR+Vr. Using Ohm's law,
DevOps and Testing slides at DASA ConnectKari Kakkonen
My and Rik Marselis slides at 30.5.2024 DASA Connect conference. We discuss about what is testing, then what is agile testing and finally what is Testing in DevOps. Finally we had lovely workshop with the participants trying to find out different ways to think about quality and testing in different parts of the DevOps infinity loop.
JMeter webinar - integration with InfluxDB and GrafanaRTTS
Watch this recorded webinar about real-time monitoring of application performance. See how to integrate Apache JMeter, the open-source leader in performance testing, with InfluxDB, the open-source time-series database, and Grafana, the open-source analytics and visualization application.
In this webinar, we will review the benefits of leveraging InfluxDB and Grafana when executing load tests and demonstrate how these tools are used to visualize performance metrics.
Length: 30 minutes
Session Overview
-------------------------------------------
During this webinar, we will cover the following topics while demonstrating the integrations of JMeter, InfluxDB and Grafana:
- What out-of-the-box solutions are available for real-time monitoring JMeter tests?
- What are the benefits of integrating InfluxDB and Grafana into the load testing stack?
- Which features are provided by Grafana?
- Demonstration of InfluxDB and Grafana using a practice web application
To view the webinar recording, go to:
https://www.rttsweb.com/jmeter-integration-webinar
Neuro-symbolic is not enough, we need neuro-*semantic*Frank van Harmelen
Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using Deplo...James Anderson
Effective Application Security in Software Delivery lifecycle using Deployment Firewall and DBOM
The modern software delivery process (or the CI/CD process) includes many tools, distributed teams, open-source code, and cloud platforms. Constant focus on speed to release software to market, along with the traditional slow and manual security checks has caused gaps in continuous security as an important piece in the software supply chain. Today organizations feel more susceptible to external and internal cyber threats due to the vast attack surface in their applications supply chain and the lack of end-to-end governance and risk management.
The software team must secure its software delivery process to avoid vulnerability and security breaches. This needs to be achieved with existing tool chains and without extensive rework of the delivery processes. This talk will present strategies and techniques for providing visibility into the true risk of the existing vulnerabilities, preventing the introduction of security issues in the software, resolving vulnerabilities in production environments quickly, and capturing the deployment bill of materials (DBOM).
Speakers:
Bob Boule
Robert Boule is a technology enthusiast with PASSION for technology and making things work along with a knack for helping others understand how things work. He comes with around 20 years of solution engineering experience in application security, software continuous delivery, and SaaS platforms. He is known for his dynamic presentations in CI/CD and application security integrated in software delivery lifecycle.
Gopinath Rebala
Gopinath Rebala is the CTO of OpsMx, where he has overall responsibility for the machine learning and data processing architectures for Secure Software Delivery. Gopi also has a strong connection with our customers, leading design and architecture for strategic implementations. Gopi is a frequent speaker and well-known leader in continuous delivery and integrating security into software delivery.
Dev Dives: Train smarter, not harder – active learning and UiPath LLMs for do...UiPathCommunity
💥 Speed, accuracy, and scaling – discover the superpowers of GenAI in action with UiPath Document Understanding and Communications Mining™:
See how to accelerate model training and optimize model performance with active learning
Learn about the latest enhancements to out-of-the-box document processing – with little to no training required
Get an exclusive demo of the new family of UiPath LLMs – GenAI models specialized for processing different types of documents and messages
This is a hands-on session specifically designed for automation developers and AI enthusiasts seeking to enhance their knowledge in leveraging the latest intelligent document processing capabilities offered by UiPath.
Speakers:
👨🏫 Andras Palfi, Senior Product Manager, UiPath
👩🏫 Lenka Dulovicova, Product Program Manager, UiPath
"Impact of front-end architecture on development cost", Viktor TurskyiFwdays
I have heard many times that architecture is not important for the front-end. Also, many times I have seen how developers implement features on the front-end just following the standard rules for a framework and think that this is enough to successfully launch the project, and then the project fails. How to prevent this and what approach to choose? I have launched dozens of complex projects and during the talk we will analyze which approaches have worked for me and which have not.
Search and Society: Reimagining Information Access for Radical FuturesBhaskar Mitra
The field of Information retrieval (IR) is currently undergoing a transformative shift, at least partly due to the emerging applications of generative AI to information access. In this talk, we will deliberate on the sociotechnical implications of generative AI for information access. We will argue that there is both a critical necessity and an exciting opportunity for the IR community to re-center our research agendas on societal needs while dismantling the artificial separation between the work on fairness, accountability, transparency, and ethics in IR and the rest of IR research. Instead of adopting a reactionary strategy of trying to mitigate potential social harms from emerging technologies, the community should aim to proactively set the research agenda for the kinds of systems we should build inspired by diverse explicitly stated sociotechnical imaginaries. The sociotechnical imaginaries that underpin the design and development of information access technologies needs to be explicitly articulated, and we need to develop theories of change in context of these diverse perspectives. Our guiding future imaginaries must be informed by other academic fields, such as democratic theory and critical theory, and should be co-developed with social science scholars, legal scholars, civil rights and social justice activists, and artists, among others.
2. 16
16.1 ˂
˂ (electric current) ˂
ˈ
. .
F ˂
ˈ ˂ . F F
ˈ F F F F
F F F F
ˈ . ˂
˂ F F F
F F ˂ F F F ˂
F F F ˂ F F F F F ˂ F F F
F ˂ F F F F ˂ F F
F ˂ F F F F F ˂ ˈ F
16.1.1 ˂
˂ ˂ F ˂
˂
ˈ ˂ (free electron) FF
F F F F F F (Brownian motion)
F ˈ F
F ˆ F www.schoolDD.com 1
3. F F F F ˂ F
˂ F F ˈ F
(drift velocity) F ˂ F ˂
VA VB VA VB
VA = VB VB > VA
F F
F F F F F F ˂
F F F ˂ (VA = VB) (VB > VA) ˂ F
16.1.2 ˂ ˂
˂ ˂ F F
˂
˂ F F ˂ F
F ˈ F F ˈ F F F (A)
˂
ۿ
I=
ܜ
I= ˂ (A)
Q= ( ) F (C)
t= F (s)
˂ I ˂ E
F 1
F ˂ 1.25 F F F F F F
F 5 F F
F ˆ F www.schoolDD.com 2
4. I = 1.25 A t=5,Q=?
୕
˂ I= ୲
Q = It
= 1.25x5.0
Q = 6.25 C Ans
F 2
F F F F ˂ F F 3.2 F
F F 5.0 ( ˂ 1.6 X10-19 )
F
I = 3.2 A ୕
˂ I= ୲
Q = It
I I
Q = 3.2x5.0 C
F 1 1.6 X10-19 C
= ଵ.௫ଵషభవ
ଷ.ଶ୶ହ.
= 1x1020 Ans
˂
F F ˂
I ˂ , F E
E
˂ (I)
F
s = vt
F ˆ F www.schoolDD.com 3
5. ୕
˂ I= ୲ ˂ Q = Ne = (nAs)e = (nAvt)e = nevtA
୬ୣ୴୲
I = ୲ .
I = nevA
n= F ( / m3)
e= (1.6 x 10- 19 C)
v= (m/s)
A= (m2)
F 3
F 5.0 x 1028 F F F 2.5 F
F F 0.30 F ˂ F
F
A = 2.5x10-6 m2
I=? I = nevA
= 5x1028x1.6x10-19x0.3x10-3x2.5x10-6
n = 5x1028 I = 6.0 A Ans
v = 0.3x10-3 m/s
16.2 F F ˂ (I) F F (V)
16.2.1 F F
˂ F F F F ˂ F
IןV
I = KV ( K= F )
ଵ
୍
=
ଵ
୍
=R (
= R)
V=IR F
R= F F ˈ F F F F (Ω)
F ˆ F www.schoolDD.com 4
6. - F F F ˈ F
I I I I
V V V V
F
F 4
F F F F ˂ ˈ F F
V (V)
3
2
1
I (A)
0 0.2 0.4 0.6
F 5
F F F 1.0 F F F F ˂ 1.0 F F
F F
R = 1x106 Ω
F V = IR
I = 1x10-3 A I
= 1x10-3x1x106
V = 1x103 V Ans
V=?
F (Resistor)
F ˈ F F F F ˂ F F ˂ F
F 2
1. F (Fixed resistor) F F ˈ F
F F F ˂ F F F
F ˆ F www.schoolDD.com 5
7. 1 2 3 4
4 F
F
1
0 0 1 -
2
1 1 101 േ1%
3
2 2 102 േ2%
F 3 3 103 -
4 ˈ F
4 4 104 -
F F
5 5 105 - 3
F 15x10 F
6 6 106 -
േ2 % F 15,000 F േ
F 7 7 - -
300 F F F 14,700 F
8 8 - -
15,300 F
9 9 - -
- - 10-1 േ5%
- - 10-2 േ 10 %
2. F F (Variable resistor) ˈ F F F F F
˂ F F F F F
F F 1 3 F F 2 F F ˂
F F
1 3 F F F 2
A
2 F 1 F 3 F
F F
(Diode)
ˈ F F F F F
˂ ˂ ˂
F ˂ 2
I A F F A F
F ˂ F F ˂
F ˆ F www.schoolDD.com 6
8. - F ˂ F F F ˂ ˈ ˂
F 6
F a F F b F F C
F F F F
F
F a F = 30x102 Ω േ 150 Ω
3 0 102 േ 5%
F
F b F = 18x103 Ω േ 1800 Ω
3
1 8 10 േ 10%
F c F =
16.2.2 F ˂
F ˂ ߩ ˈ F F
˂ F F F F F F ˂
F ˈ F
F F F F ˂ (V) ˂ (I) F
(R) F F
ℓ
ן
୍
. F
ℓ
R ן
࣋र
R= ߩ F F ˂
ۯ
F F ߩ = 1.6x10-8 Ω m , = 1.7x10-8 Ω m ߩ
ߩ F = 1010- 1014 Ω m , ߩ = 1014- 1018 Ω m
F 7
F 10 F F F F 1 F
F F F 2 F F F ˈ F F
F ( ߩ = 1.7x10-8 Ω m)
F ˆ F www.schoolDD.com 7
9. d = 1x10-3 m
F ℓ = 10 m గௗమ ଷ.ଵସ୶ሺଵ୶ଵషయ ሻమ
F A= = = 7.85x10-7 m2
ସ ସ
ఘℓ
R=
ଵ.୶ଵషఴ ୶ଵ
R1 = .଼ହ୶ଵషళ
R1 = 0.216 Ω Ans
d = 2x10-3 m
F ℓ = 10 m గௗమ ଷ.ଵସ୶ሺଶ୶ଵషయ ሻమ
A= = = 3.14x10-6 m2
F ସ ସ
ఘℓ
R=
ଵ.୶ଵషఴ ୶ଵ
R2 = ଷ.ଵସ୶ଵషల
R2 = 0.054 Ω Ans
ୖభ .ଶଵ
ୖమ
= .ହସ
=4
1 F F 2 F 4 F Ans
˂ G ˈ F F ˂ F ˈ ( F )-1 F (S)
ଵ
G= ୖ
˂ ߪ ˈ ˈ F F ˂ F ˈ
( F )-1 F F (S/m)
ଵ
ߪ=
ఘ
16.2.3 F F
ˈ F F F
T ---> R
F F F F F
T ---> R
ˈ F F
T ---> R
F ˆ F www.schoolDD.com 8
10. (Superconductivity) F ˂ ˈ F (ߩ = 0) F
F (critical temperature) F (Supper
conductor)
16.3 ˂
16.3.1 ˂ ( E) F F (V)
F F ˂ ( F F ) F F F F
˂ F F F ˂ ˂
F F ˂
F R
I
I I
E I
F ˂
F F F R F ˂
E F r F F ˂ I F F
R
VR
I
V
r E
Vr
˂ F ˂ ( ) F F F ˂ Q
˂ E F F F r R F F ˂ Q F F
F F
F WE = ˂ F
WR = ˂ F F R
Wr = ˂ F F r
F FF WE = W R + W r
QE = QVR + QVr
F E = VR + Vr
F ˆ F www.schoolDD.com 9
11. E = IR + Ir
F I = ሺୖା୰ሻ
F V F F F F V = VR
F V F F R F F F V؆E F F V=E
F 8
F ˂ 12.0 F F 2.0 F F F 7.0
F F F F F F F F F F F
F F F F F F
R = 20 Ω
VR
I
V=?
r = 2 E = 12 V
Vr
I E = VR + Vr
E = IR + Ir
ଵଶ ଵ
I = ሺୖା୰ሻ = ሺାଶሻ
=
A
V = VR = IR
ଵ
= x70
V = 11.67 V Ans
F F F F
V=?
F V ؆ E = 12 V Ans
r = 2 E = 12 V
F ˆ F www.schoolDD.com 10
12. 16.3.2 ˂
˂
R
I V I
E
FQ= ˂ F F t (C)
I= ˂ F F (A)
V= F F F (V)
W= ˂ F F ( J)
W = QV
୕
I= ୲
Q = It V = IR
మ ୲
W = ItV = I2Rt = ୖ
˂ P ˂ F F F ˈ F F (W)
܅ ItV
P= P= ୲
= ୲
= IV
ܜ
F 9
F 1.5 F2 F F F ˂ 20 F
F 20 F F 15 F ˂ F ˂ 1 F
t = 20 hr
I=20 mA
E = 1.5x2 = 3 V
˂ F W = QV
= ItV
F ˆ F www.schoolDD.com 11
13. = 20x10-3x(20x60x60)x3
W = 4320 J
ଵହ୶ଶ
F ˂ F ˂ 1 F = ସଷଶ = 0.007 / Ans
16.4 F F
16.4.1 F F
1. F
ˈ F F F F F
V1 V2 V3
R1 R2 R3
I1 I2 I3
V
I
E
F FF
- ˂ F F F
I = I1 = I2 = I3
- F F ˂ = F F ˂ F
V = V1 + V2 + V3
V = V1 + V2 + V3
V = IR ( F )
IR = I1 R1 + I2 R2 + I3R3
R = R1 + R2 + R3 (I = I1 = I2 = I3)
R = F R1,R2 R3
2. F
ˈ F F F ˈ F
F ˆ F www.schoolDD.com 12
14. V1 = V2 = V3
R1
I1 R2
I2
I3 R3
I I
V
E
F FF
- F F F F F = F F ˂
V = V1 = V2 = V3
- ˂ F = ˂ F F F
I = I1 + I2 + I3
I = I1 + I2 + I3
V = IR I=ୖ
భ
ୖ
= ୖభ
+ ୖమ + ୖయ
మ య
= +܀+ ܀ (V = V1 = V2 = V3)
܀ ܀
R = F R1 , R 2 R3
- F F F F F F F
˂ F F F F F ˂ F F
- F F ˈ F F F
F 10
F F F
. FS ʽ . FS ʽ
F ˆ F www.schoolDD.com 13
15. 1Ω 2Ω
A B
S
I
3Ω 4Ω
. FS ʽ 1Ω 2Ω
F F R1 R2
A B
R3 R4
I
3Ω 4Ω
F R1 R2 F
F F R 12 = R 1 + R 2 = 1 +2 = 3 Ω
F R3 R4 F
F F R 34 = R 3 + R 4 = 3 +4 = 7 Ω
F F
3Ω
R12
A B
R34
I
7Ω
F R 12 R 34 F
ଵ ଵ ଵ ଵ ଵ ାଷ
F F ୖ
=
ୖభమ
+
ୖయర
=
ଷ
+
=
ଶଵ
ଵ ଵ
=
ୖ ଶଵ
R = 2.1 Ω Ans
. FS ʽ
F F
1Ω 2Ω
R1 R2
A B
R3 R4
I
3Ω 4Ω
F R1 R3 F
F ˆ F www.schoolDD.com 14
16. ଵ ଵ ଵ ଵ ଵ ଷାଵ
F F ୖభయ
=
ୖభ
+
ୖయ
=
ଵ
+
ଷ
=
ଷ
ଵ ସ
=
ୖభయ ଷ
ଷ
R13 = ସ
Ω
F R2 R4 F
ଵ ଵ ଵ ଵ ଵ ଶାଵ
F F ୖమర
=
ୖమ
+
ୖర
=
ଶ
+
ସ
=
ସ
ଵ ଷ
=
ୖమర ସ
ସ
R24 = ଷ
Ω
F F
A 3/4 Ω 4/3 Ω B
R13 R24
I
F R 13 R 24 F
F F R = R13 + R24
ଷ ସ ଽ ାଵ
= ସ+ଷ = ଵଶ
ଶହ
=
ଵଶ
R = 2.08 Ω Ans
16.4.2 F
1. F
ˈ F F F F F
R
I
r1 E1 r2 E2 r3 E3
F FF
- ˂ = ˂ F
E = E1 + E2 + E3
F F F F
F ˆ F www.schoolDD.com 15
17. r = r1 + r2 + r3
2. F
ˈ F F F F
R
I r1 E1
r2 E2
r3 E3
F ˂ F F F E1 = E2 = E3
F FF
- ˂ = ˂ F F F
E = E1 = E2 = E3
- F F F F
ܚ
= ܚ
+ܚ+ ܚ
- F F ˂ F F ˂
F
F 11
F F ˂ 0.25 F F F 8 F
F F ˂ 0.16 F ˂ F
F ˂ F F F F E r
F ˆ F www.schoolDD.com 16
18. F
R=8Ω
I = 0.25 A V
r1 E1 r2 E2
˂ E = E1 + E2 = E + E = 2E
F r = r1 + r2 = r + r = 2r
F F
R=8Ω
I = 0.25 A
V
r E
V = E Ir
V = IR
F IR = 2E I(2r)
0.25x8 = 2E 0.25(2r)
E 0.25r = 1 ---1
F
R=8Ω
V
I = 0.16 A r1 E1
r2 E2
˂ E = E1 = E2 = E
ଵ ଵ ଵ ଵ ଵ ଶ
F ୰
=
୰భ
+
୰మ
=
୰
+ =
୰ ୰
୰
r = ଶ
F F
R=8Ω
I = 0.16 A
V
r E
F ˆ F www.schoolDD.com 17
19. V = E Ir
V = IR
୰
F IR = E I(ଶ)
୰
0.16x8 = E 0.16(ଶ)
E 0.08r = 1.28 ---2
2-1, -0.08r + 0.25r = 0.28
0.17r = 0.28
r = 1.7 Ω Ans
F r 2 F E 0.08(1.7) = 1.28
E = 1.41 V Ans
16.5 F ˂ F
˂ F F ˂ F F F
F F ˂
F 12
˂F F a,b c ˂
F E1 = 6 V , E2 = 6 V , r1 = 1 Ω , r2 = 1 Ω , Ra = 7 Ω , Rb = 4 Ω , Rc = 12 Ω
b
a
c
I
r1 E1 r2 E2
F
F b c F F F F a
ଵ ଵ ଵ ଵ ଵ
F b c ୖౘౙ
=
ୖౘ ୖౙ ସ ଵଶ
+ = +
ଷାଵ ସ ଵ
=
ଵଶ
= ଵଶ
= ଷ
Rbc = 3 Ω
F a bc R = Ra + Rbc
= 7+3
R = 10 Ω
E1 E2 F
˂ E = E1 + E2 = 6 + 6 = 12 V
F ˆ F www.schoolDD.com 18
20. F r = r1 + r2 = 1 + 1 = 2 Ω
F F
R
I
V
r E
V = E Ir
V = IR
F IR = E I r
Ix10 = 12 Ix2
I = 1.0 A
˂ F F a , Ia = I = 1.0 A Ans
F Ia = Ib + Ic
1.0 = Ib + Ic
Ic = 1.0 - Ib
Vb = Vc
IbRb = IcRc
IbRb = (1.0 - Ib)Rc
Ibx4 = (1.0 - Ib)x12
ଵଶ ଷ
Ib = ଵ = ସ = 0.75 A Ans
Ic = 1.0 - Ib = 1.0 - 0.75 = 0.25 A Ans
F 13
F F ˂ 10 F F F F
F F 1.2 F F F 6 F F F F
F F F F F F F
F F F ˂ 6 V.
F F F F F 1.2 V. ˂ F F 10
mA.
F ˆ F www.schoolDD.com 19
21. F F F F F F F F F
F R F
VR VD
R
I F
V
E,r=0
V = VR + VD
0
V = E Ir = E
F E = VR + VD
VR = E VD = 6 1.2
VR = 4.8 V
F
VR = I R
ସ.଼
R = ୍ = ଵ௫ଵషయ
R = 480 Ω Ans
F 14
˂ ˂ 2 F F F F F F
. ˂ F
. F F F F
. F
. ˂ F 10
. ˂ F
R = 0.5 Ω
I=2A
r=0 R1
E=6V
.
˂ F F ˂ F F F
I = I = 2 A Ans
F ˆ F www.schoolDD.com 20
22. . VR = ?
F V = IR
VR = 2x0.5
VR = 1.0 V Ans
. R1 = ?
E = I r + I R + I R1
6 = 2x0 + 2x0.5 + 2xR1
ିଵ
R1 = ଶ
R1 = 2.5 Ω Ans
. W=? t = 10 s
W = QV
= QE
୕
= ItE (I = ୲ )
= 2x10x6
W = 120 J Ans
. PR = ?
W = QV
P= ୲
QV
P= ୲
QVR ItሺIRሻ
PR = ୲ = ୲
= I2R = 22x0.5
PR = 2 W Ans
F 15
F F F a b
I I
R1 = 1 Ω R1 = 1 Ω
a a
E=9V E=9V
R2 = 2 Ω R2 = 2 Ω R3 = 2 Ω
b
b
. .
F ˆ F www.schoolDD.com 21
23. . V2 = ? I
V2 V2 = I R2
F R2 I
E V R
R1 R2 F r=0
R = R1 + R2 = 1 + 2 = 3 Ω
V = IR
0
V = E Ir
F E = IR
9 = Ix3
I = 3 A
V2 = I R2 = 3x2
V2 = 6 V Ans
. V2 = ?
R2 R3 F
ଵ ଵ ଵ ଵ ଵ
F ୖౘ
=
ୖమ
+
ୖయ
=
ଶ
+
ଶ
Rab = 1 Ω
R1 Rab F
F R = R1 + Rab = 1 + 1 = 2 Ω
V = IR 0
V = E Ir
F E = IR
9 = Ix2
I = 4.5 A
Vab = I Rab = 4.5x1
Vab = 4.5 V Ans
16.6 ˂
˂ F F F F F F F F
F (galvanometer) F F F
F
F ˆ F www.schoolDD.com 22
24. 16.6.1 F
ˈ ˂ F F F ˂
R R
F A
I I
E,r E,r
˂ F F F F
˂ F I
F ˈ F F F F F (shunt) F
Rs F F G F F I ˈ F F F
F F IG F F F F IS
I IG RG
G
IS
RS
F
F A
F FF F F F F F F
F F F
VS = VG
ISRS = IGRG
୍ ୖ
F RS = ృ୍ ృ
I = IG + IS
F 16
F F F F 1000 F ˂ 50 F
F ˈ F ˂ F 100 F
. F F F F
IG = 50ߤA RG = 1000 Ω
. F F ˈ F I = 100 mA
G
IS
RS = ?
F ˆ F www.schoolDD.com 23
25. . RS = ?
I = IG + IS
IS = I - IG
IS = 100x10-3 50x10-6 A
VS = VG (F )
ISRS = IGRG
୍ ୖ
RS = ృ୍ ృ
ହ୶ଵషల ୶ଵ ଵ
= ଵ୶ଵషయିହ୶ଵషల = ଶ୶ଵయ ିଵ
RS = 0.5 Ω Ans
. RA = ?
ଵ ଵ ଵ
= + ( F )
ୖఽ ୖృ ୖ
ଵ ଵ ଵ ଵାଶ ଵ
= + = =
ୖఽ ଵ .ହ ଵ .ହ
RA = 0.5 Ω Ans
- F F F F F F ˂
F
16.6.2 F F
ˈ F F ˂ ˂ F F F F ˂
F F
V
R a R
a b
b
I
I
E,r E,r
˂ F F F F V F R
F F ˂ F
F ˈ F F F F F F
(multiplier) F Rm F F G F F F F
F Vm F ˈ F F F F VG
F ˆ F www.schoolDD.com 24
26. IG Rm RG
G
F
Vm VG
F F V
V = Vm + VG
V = IGRm + IGRG
V = IG(Rm + RG)
Rm = ୍ృ
- RG
F FF F F F F F
F F F F F F F
F 17
F F F 16 F ˈ F F F F F 10 F
. F F F F F
. F F F F F F F F
. F F F F F
IG = 50 ߤA Rm = ? RG = 1000 Ω
G
Vm VG
V
. Rm = ?
V = Vm + VG
V = IGRm + IGRG
10 = 50x10-6xRm + 50x10-6x1000
ଵି.ହ
Rm = ହ୶ଵషల
Rm = 200,000 Ω Ans
F ˆ F www.schoolDD.com 25
27. . F Rm = 0 , VG = ?
VG = IGRG
= 50x10-6x1000
VG = 50x10-3 V Ans
. RV = ?
RV = Rm + RG ( F )
= 200.000 + 1000
RV = 201,000 Ω Ans
- F F F F F F ˂
F
16.6.3 F F
ˈ F F
F F
R O
x R y
F
x y
F F F
F F F F G F F F R0 E
F F Rx F x y F F
˂ F F F F F F F F ˈ F F
RG
E,r G
R0
I
x y
RX
F F O
ˈ ˂ FF
E = Vr + VRX + VRG + VR0
E = I r + I R X + I RG + I R0
F ˆ F www.schoolDD.com 26