quick summary of last lecture block a unit 1

EE2301: Basic Electronic Circuit

Quick Summary of Last LectureBlock A Unit 1

Three Basic Laws

EE2301: Block A Unit 2 1

Block A Unit 1 2

Fundamental law for charge

Current has to flow in closed loop No current flows if there is a break in the path Underlying physical law: Charge cannot be created or

destroyed This is the basis of Kirchhoff’s Current Law

i1

i4

i3

i2

Kirchhoff’s current law

Sum of currents at a node must equal to zero:

i1 + i2 + i3 + i4 = 0

Block A Unit 1 3

Fundamental law on voltage Energy is required to push electrons through a resistive element That same energy needs to be generated by a source Total energy generated in a circuit must equal total energy consumed in the circuit Energy cannot be created or destroyed Therefore, voltage rise = voltage drop

- V3 +

+ V1 -

-

V4

+

+

V2

-

Kirchhoff’s voltage law

Net voltage around a closed circuit is zero:

v1 + v2 + v3 + v4 = 0

Block A Unit 1 4

Resistance and Ohm’s Law

i

v

V

I

1/R

Ohm’s law: V = IR

Ideal RESISTOR shows linear resistance obeying Ohm’s law

When current flows through any circuit element, there will always be a resistance to its flow which results in a voltage drop across that circuit element

+

_

IMPORTANT: Positive current is defined here as flowing from higher to lower voltage (Remember)

Unit: Ohm (Ω)

A

L

ρ: resistivity (material property)

A: cross-sectional area

A

LR

Block A Unit 1 5

Parallel network (Highlights)

R1 R2IsRN

I1 I2 IN

RPIs

Equivalent Resistance

1/RP = 1/R1 + 1/R2 + …+ 1/RN

Current divider rule

SP

NN I

R

RI

1

1

Block A Unit 1 6

Series network (Highlights)

R1

R2

Vs

+-

RSVs

+-

Equivalent Resistance

RS = R1 + R2 + …+ RN

RN

Voltage divider rule

VN = VS(RN/RS)

+

-V1

+

-V2

+

-VN

Let’s have a look first



Block A Unit 2 outline

Applying the 3 laws to analyze DC circuits Systematic methods for analysis

> Nodal voltage analysis (application of KCL and Ohm’s law)

> Mesh current analysis (application of KVL and Ohm’s law)

> Superposition (can be a powerful tool)

G. Rizzoni, “Fundamental of EE” Chapter 3.2 – 3.5


Nodal voltage analysis

V1

X

R2

R3R1

V2

V3

Nodal voltage analysis is simply an application of Ohm’s law and KCL together. Here we express branch currents in terms of voltage and resistance using Ohm’s law

03

3

2

2

1

1

R

vv

R

vv

R

vv xxx

Applying NVA at node X:

This can be seen if we first consider currents in each branch arriving at X:

Current from V1 to VX via R1 = (V1 - VX)/R1



Summing these together, we obtain the above final expression


NVA example 1

Problem 3.1:

Use nodal voltage analysis to find the voltages V1 and V2

1 Ω

Apply KVL at V1:

123413

A4

21

211

VV

VVV

Apply KVL at V2:

21

2221

21121

VV

VVVV

Solve for V1 and V2:

V1 = 4.8 V, V2 = 2.4 V


NVA example 2Problem 3.4:

Use nodal voltage analysis to find the current through the voltage source

First, define and label the unknown nodes

Apply KCL at V1:

125.05.0

2

132

1312

VVV

VVVV

Apply KCL at V2:

iVV

iVVV

21

221

6225.05.0

Apply KCL at V3:

iVV

Vi

VV

13331

331

2533.05.0


NVA example 2 solution

This slide is meant to be blank

There are now 3 equations but 4 unknowns; we need one more equation!!

V3 - V2 = 3

Now, eliminate V3 from all 3 equations:

8.31AA8 299 i

V2 + (3 + V2) - 2V1 = 1 V1 - V2 = 1

iVV 12331 235 iVV 1233

1111 2515

Now, eliminate V2:iV 33

2133

1 103

iV 41

23

1


Mesh current analysis

vs

+_

R1

R2

R3

R4

0V

i1 i2

Mesh current analysis is simply an application of Ohm’s law and KVL together. Here we express branch voltages in terms of current and resistance using Ohm’s law

Around Mesh 1:

vs = i1(R1 + R2) - i2R2

Around Mesh 2:

i2(R3 + R4 + R2) - i1R2 = 0

Apply KVL to each mesh in turn

If the current direction is known, then the easiest choice is simply to follow it. This will avoid any confusion (e.g. in a voltage source, define the current as flowing in the direction of voltage gain)

Pay close attention to the direction of one mesh current to another (e.g. in this instance, i1 is flowing opposite to i2 hence we take the difference in R2


MCA example 1


Problem 3.30

Use mesh current analysis to find the current (i) through the 1/5 Ω resistor

I 1/5

1/2

i2

i3

i1

1/4

1/3

1

i


MCA example 1 solution


Since I = i1, so only 2 unknown meshes to solve for

KVL around mesh 2:

0

011

351

21017

51

351

21

2

Iii

Iii

KVL around mesh 3:

0

0

41

251

36047

41

51

251

31

41

3

Iii

Iii

Solving for I2 and I3:

I2 = 20/31 A, I3 = 15/31 A

Current through the 1/5 Ω resistor,

i = i3 - i2 = - 5/31 = - 0.161 A


MCA example 2

Problem 3.17

Use mesh current analysis to find the voltage across the current source

I3

Apply KVL around mesh 1:

2 = I1(2+3) - I2(3)

5I1 - 3I2 = 2

Apply KVL around mesh 2:

-V = I2(3+1) - I1(3)

4I2 - 3I1 = -V Apply KVL around mesh 3:

V = I3(3+2)

5I3 = V


MCA example 2


There are now 3 equations but 4 unknowns; we need one more equation!!

2 = I3 - I2

Substitute into mesh 3 to eliminate i3 and use this to eliminated i2 in mesh 2:

38

53

1

151 324

VI

VIV

Substitute into mesh 1 to find V:

V89.3

2235

935

51

38

53

V

VV


Principle of superposition

A B

Consider what happens when you throw a stone into a pool at A & B

AB

Both A and B together

Only at B

BAOnly at A

If the stones hit A and B together, the result will be a combination of the individual responses of A and B. This is the principle of superposition.


Superposition in circuits

B

A

B’

A’

INPUT OUTPUTSYSTEM

A+B A’+B’

+

-VG

IB

RBRG

I

Say we want to find the current through RB

If we apply superposition, this current is the sum of the individual currents corresponding to each of the sources in the circuit, i.e. current associated only with VG or IB alone.That is to say, we need to remove the effects from all the sources except one, and find the corresponding value of I for this source. We then repeat this for the other sources.

But how do we remove the effects of a certain source?


Disabling sources in superpositionVoltage source:

We want no voltage drop across

Therefore replace with a short

Current source:

We want no current flowing through

Therefore replace with an open circuit

IB

RBRG

I1

+

-VG

RBRG

I2

VG = 0 IB = 0

I = I1 + I2O

pe

n ci

rcu

it

Short circuit

Vs

+

-

IN


Superposition in Circuits

+

-VG

IB

RBRG

I

IB

RBRG

I1

+

-VG

RBRG

I2

VG = 0 IB = 0

I = I1 + I2O

pe

n ci

rcu

it

Short circuit

Find I


Superposition example 1Problem 3.40

Determine, using superposition, the current through R1 due only to the source VS2

8105.3560 321 RkRR

VS2 90 V


Superposition example 1 solution


Re-drawn circuit due to VS2 only

Note that now R1 || R2 are parallel

Suggested strategy:

1) Use voltage divider rule to find voltage across R1

2) Use ohm’s law to find current through R1

V61.33

||

||2

321

211

SR VRRR

RRV

mA60

1121

RVVI RSR


Superposition example 2


Problem 3.41

Determine, using superposition, the voltage across R 23.03.0V121A12 RRVRI GGBB


Superposition example 2


VR due to VG only:

R and RB are parallel, which together are in series with RG

Apply voltage divider rule:

V608.4

||

||

GGB

BGR V

RRR

RRVV

VR due to IB only:

All 3 resistor are in parallel

V382.1

||||

BGBBR IRRRIV

Adding the two solutions together:

VR = 5.99V

quick summary of last lecture block a unit 1

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