EEE (2110005) - ACTIVE LEARNING ASSIGNMENT

Presented by: Divyang Vadhvana(130120116086)

Branch: Information Technology

Electrical circuitsElectrical circuits often contain one or often contain one or more resistors grouped together and more resistors grouped together and attached to an energy source, such as attached to an energy source, such as a battery.a battery.The following symbols are often The following symbols are often

used:used:

+ - + -- + - + -

Ground Battery-+

Resistor

Resistors are said to be connected inResistors are said to be connected in series series when there is a when there is a single pathsingle path for the for the

current.current.The current The current I I is the same is the same for each resistor for each resistor RR11, R, R22 and and RR33..The energy gained The energy gained through through EE is lost through is lost through RR11, R, R22 and and RR33..The same is true for The same is true for

voltages:voltages:For series

connections:

For series connection

s:

I = I1 = I2 = I3

VT = V1 + V2 + V3

I = I1 = I2 = I3

VT = V1 + V2 + V3

R1

IVT

R2

R3

Only one current

The The equivalent resistance Requivalent resistance Ree of a of a number of resistors connected in number of resistors connected in series is equal to the series is equal to the sumsum of the of the individual resistances.individual resistances.

VVT T = V= V11 + V + V22 + V + V33 ; (V = ; (V = IR)IR)

IITTRRe e = I= I11RR11+ I+ I22RR22 + + II33RR33But . . . IBut . . . ITT = I = I11 = I = I22 = I = I33

Re = R1 + R2 + R3Re = R1 + R2 + R3

R1

IVT

R2

R3

Equivalent Resistance

The The output directionoutput direction from from a source of emf is from a source of emf is from ++ side:side:

E+-a b

Thus, from Thus, from aa to to b b the the potential increasespotential increases by by E; From ; From bb to to aa, the , the potential decreasespotential decreases by by E..

Example:Example: Find Find V V for path for path AB AB and then for path and then for path BABA.. R

3 V +-

+

-9 V

A

B

AB: AB: V = +9 V – 3 V = V = +9 V – 3 V = +6 +6 VVBA: BA: V = +3 V - 9 V = V = +3 V - 9 V = -6 -6 VV

Consider the simple Consider the simple series circuitseries circuit drawn drawn below:below:

2

3 V+-

+

-15 V

A

C B

D

4

Path ABCD: Energy and V increase through the 15-V source and decrease through the 3-V source.15 V - 3 V = 12 VE=

The net gain in potential is lost through The net gain in potential is lost through the two resistors: these voltage drops are the two resistors: these voltage drops are IRIR22 and and IRIR44, so that , so that the sum is zero for the the sum is zero for the entire loopentire loop..

Resistance Rule: Re = R

Voltage Rule: E = IR

: Current IR

E

R2

E1

E2R1

A A complexcomplex circuit is circuit is one containing more one containing more than a single loop than a single loop and different current and different current paths.paths.

R2 E1

R3 E2

R1

I1

I3

I2

m nAt junctions m and n:At junctions m and n:

II11 = I = I22 + I + I3 3 oror I I22 + I + I33 = I = I11

Junction Rule:

I (enter) = I (leaving)

Junction Rule:

I (enter) = I (leaving)

Resistors are said to be connected inResistors are said to be connected in parallel parallel when there is more than one path when there is more than one path for current.for current.

2 4 6

Series Connection:

For Series For Series Resistors:Resistors:II22 = I = I4 4 = I= I66 = I = ITT

VV22 + V + V44 + V + V66 = = VVTT

Parallel Connection:

6 2 4

For Parallel For Parallel Resistors:Resistors:VV22 = V = V4 4 = V= V66 = =

VVTTII22 + I + I44 + I + I66 = I = ITT

VVTT = V = V1 1 = V= V22 = = VV33IITT = I = I11 + I + I22 + I + I33

Ohm’s Ohm’s law:law:

VI

R

31 2

1 2 3

T

e

VV V V

R R R R

1 2 3

1 1 1 1

eR R R R

The equivalent resistance for Parallel resistors:

The equivalent resistance for Parallel resistors:

1

1 1N

ie iR R

Parallel Connection:

R3R2

VT

R1

The equivalent resistance The equivalent resistance RRee for for twotwo parallel resistors is the parallel resistors is the product divided product divided by the sumby the sum..

1 2

1 1 1;

eR R R 1 2

1 2e

R RR

R R

(3 )(6 )

3 6eR

Re = 2 Re = 2

ExamplExample:e: R2VT R1

6 3

In complex circuits resistors are often In complex circuits resistors are often connected inconnected in both both seriesseries and and parallelparallel. .

VTR2 R3

R1

In such cases, it’s best to use rules for series and parallel resistances to reduce the circuit to a simple circuit containing one source of emf and one equivalent resistance.

In such cases, it’s best to use rules for series and parallel resistances to reduce the circuit to a simple circuit containing one source of emf and one equivalent resistance.

VTRe

Kirchoff’s first law:Kirchoff’s first law: The sum of the The sum of the currents entering a junction is equal to currents entering a junction is equal to the sum of the currents leaving that the sum of the currents leaving that junction.junction.

Kirchoff’s first law:Kirchoff’s first law: The sum of the The sum of the currents entering a junction is equal to currents entering a junction is equal to the sum of the currents leaving that the sum of the currents leaving that junction.junction.

Kirchoff’s second law:Kirchoff’s second law: The sum of the The sum of the emf’s around any closed loop must emf’s around any closed loop must equal the sum of the IR drops around equal the sum of the IR drops around that same loop.that same loop.

Kirchoff’s second law:Kirchoff’s second law: The sum of the The sum of the emf’s around any closed loop must emf’s around any closed loop must equal the sum of the IR drops around equal the sum of the IR drops around that same loop.that same loop.

Junction Rule: I (enter) = I (leaving)

Junction Rule: I (enter) = I (leaving)

Voltage Rule: E = IR

Voltage Rule: E = IR

When applying Kirchoff’s laws you must When applying Kirchoff’s laws you must assume a consistent, positive assume a consistent, positive tracing tracing direction.direction. When applying the When applying the voltage rulevoltage rule, emf’s , emf’s are are positivepositive if normal output direction of if normal output direction of the emf is the emf is withwith the assumed tracing the assumed tracing direction.direction.

If tracing from If tracing from A to BA to B, , this emf is considered this emf is considered positivepositive..

EA B

++

If tracing from If tracing from B to AB to A, , this emf is considered this emf is considered negativenegative..

EA B

++

When applying the When applying the voltage rulevoltage rule, , IR dropsIR drops are are positivepositive if the assumed current if the assumed current direction is direction is withwith the assumed tracing the assumed tracing direction.direction.

If tracing from If tracing from A to BA to B, , this IR drop is this IR drop is positivepositive..

If tracing from If tracing from B to AB to A, , this IR drop is this IR drop is negativenegative..

IA B

++

IA B

++

R3

R1

R2E2

E1

E3

1. Assume possible 1. Assume possible consistent flow of consistent flow of currents.currents.2. Indicate positive output 2. Indicate positive output directions for emf’s.directions for emf’s.

3. Indicate consistent 3. Indicate consistent tracing direction. tracing direction. (clockwise)(clockwise)

+

Loop II1

I2

I3

Junction Rule: I2 = I1 + I3

Junction Rule: I2 = I1 + I3

Voltage Rule: E = IR

E1 + E2 = I1R1 + I2R2

Voltage Rule: E = IR

E1 + E2 = I1R1 + I2R2

4. Voltage rule for Loop II: 4. Voltage rule for Loop II: Assume Assume counterclockwise counterclockwise positive tracing positive tracing direction.direction.Voltage Rule: E = IR

E2 + E3 = I2R2 + I3R3

Voltage Rule: E = IR

E2 + E3 = I2R2 + I3R3

R3

R1

R2E2

E1

E3

Loop II1

I2

I3Loop II

Bottom Loop (II)

+

Would the same Would the same equation apply if traced equation apply if traced

clockwiseclockwise??- E2 - E3 = -I2R2 -

I3R3

- E2 - E3 = -I2R2 - I3R3

Yes!Yes!

5. Voltage rule for Loop III: 5. Voltage rule for Loop III: Assume Assume counterclockwise counterclockwise positive tracing positive tracing direction.direction.Voltage Rule: E = IR

E3 – E1 = -I1R1 + I3R3

Voltage Rule: E = IR

E3 – E1 = -I1R1 + I3R3

Would the same Would the same equation apply if traced equation apply if traced

clockwiseclockwise??E3 - E1 = I1R1 - I3R3E3 - E1 = I1R1 - I3R3YesYes

!!

R3

R1

R2E2

E1

E3

Loop II1

I2

I3Loop II

Outer Loop (III)

+

+

6. Thus, we now have four 6. Thus, we now have four independent equations independent equations from Kirchoff’s laws:from Kirchoff’s laws:

R3

R1

R2E2

E1

E3

Loop II1

I2

I3Loop II

Outer Loop (III)

+

+

II22 = I = I11 + I + I33

EE1 1 + + EE22 = I= I11RR11 + I + I22RR22

EE2 2 + + EE33 = I= I22RR22 + I + I33RR33

EE3 3 - - EE11 = -I= -I11RR11 + I + I33RR33

Resistance Rule: Re = R

Voltage Rule: E = IR

: Current IR

E

Rules for a simple, single loop circuit containing a source of emf and

resistors.

Rules for a simple, single loop circuit containing a source of emf and

resistors.

2

3 V+-

+

-18 V

A

C B

D

3

Single Loop

For resistors connected in series:

Re = R1 + R2 + R3Re = R1 + R2 + R3

For series connection

s:

For series connection

s:

I = I1 = I2 = I3

VT = V1 + V2 + V3

I = I1 = I2 = I3

VT = V1 + V2 + V3

Re = RRe = R

2

12 V

1 3

Resistors connected in parallel:

For parallel connection

s:

For parallel connection

s:

V = V1 = V2 = V3

IT = I1 + I2 + I3

V = V1 = V2 = V3

IT = I1 + I2 + I3

1 2

1 2e

R RR

R R

1

1 1N

ie iR R

R3R2

12 V

R1

2 4 6

VT

Parallel Connection

Kirchoff’s first law:Kirchoff’s first law: The sum of the The sum of the currents entering a junction is equal to currents entering a junction is equal to the sum of the currents leaving that the sum of the currents leaving that junction.junction.

Kirchoff’s first law:Kirchoff’s first law: The sum of the The sum of the currents entering a junction is equal to currents entering a junction is equal to the sum of the currents leaving that the sum of the currents leaving that junction.junction.

Kirchoff’s second law:Kirchoff’s second law: The sum of the The sum of the emf’s around any closed loop must emf’s around any closed loop must equal the sum of the IR drops around equal the sum of the IR drops around that same loop.that same loop.

Kirchoff’s second law:Kirchoff’s second law: The sum of the The sum of the emf’s around any closed loop must emf’s around any closed loop must equal the sum of the IR drops around equal the sum of the IR drops around that same loop.that same loop.

Junction Rule: I (enter) = I (leaving)

Junction Rule: I (enter) = I (leaving)

Voltage Rule: E = IR

Voltage Rule: E = IR