simple dc circuit pg 31-60 hughes electrical & electronic technology

Simple dc circuit

Pg 31-60

HUGHES

Electrical & Electronic Technology

A complete connection of a source and a load

Source : voltage source, current source such as battery

Load : resistor, capacitor, inductor

V

I

Load

A network is a combination of several circuits

V

I

Circuit 1

Circuit 3

Circuit 2

V

R 1

R 2

V1

V2

I

11 IRV tIRV

22 IRV

21 RRRt 21 IRIRIRt

21 VVV

therefore

11 IRV tIRV

22 IRV

321 RRRRt 321 IRIRIRIRt

33 IRV

321 VVVV

V

R 3

R 1

R 2

V1

V2

V3

I

therefore

nt RRRRR .........321

In general for n series of resistor, Rt

One man decides to connect two lamps of 60W@220V in series in order to get more light . However he found the lamps give out very litter light. Why? Can you explain this.

AR

VI

t

136.01614

220

AV

PI 27.0

220

60

R

VVIP

2

161480780721 RRRt

80760

22022

P

VR

To get full light , we must connect a single lamp to 220V source , thus we have

When connect two lamps in series , then

Since the current is less then the lamp cannot give light fully.

V

R 3=8

R 1=2

R 2=3

I=1.5A

V 1

V 2

V 3

VIRV 0.325.111

VVVVV 5.190.125.40.3321

VIRV 5.435.122

VIRV 0.1285.133

Calculate the voltage across each of resistors as in figure and hence calculate the supply voltage V

V=100V

R 3=70

R 1 =40

R 2 =50

I

V 1

V 2

V 3

Calculate the circuit’s current

160705040321 RRRR

AR

VI 625.0

160

100

V

R 1

R 2

V1

V2

I

221

22 RRR

VIRV

21 RRR

21

11

RR

R

V

V

21 RR

VI

121

11 RRR

VIRV

21

22

RR

R

V

V

V=30V

R 1

R 2

V1

10V

I

Given that R2=100, calculate R1 in order to obtain an output voltage 10V across R2

2001R

100

100

30

10

1

R

30010031001 R

21

22

RR

R

V

V

V

I

R 2R 1

I2I1

11 R

VI

22 R

VI

2211 RIRIV

21 R

V

R

VI

21 III

21

111

RRR

21

11

RRV

I

RV

I 1but then

V

I

R2

R1

I2I 1

R3

I3

321 R

V

R

V

R

VI

321

1111

RRRR

321

111

RRRV

I

then

nRRRRR

1.......

1111

321

In general

V

I

R 2R 1

I2I1

21

21

RR

RRR

21

21 RR

RII

21

111

RRR

21

21

RR

RRIIRV

2211 RIRIV

21

12 RR

RII

But

Therefore and

V=110V

I

R 244

R 122

I2I1

Calculate I1,I2 and I3

AR

VI 5.2

44

110

22

AR

VI 0.5

22

110

11

AIII 5.75.20.521

V=12V

I

R 32.2

R 16.8

I2I1

R 24.7

I3

815.0455.0213.0147.0

2.2

1

7.4

1

8.6

11111

321

RRRR

23.1815.0

1R A

R

VI 76.9

23.1

12

Calculate the effective resistance and the power supply

Hence

V

I=8A

R 22

R 1

I2I1

ARR

RII 0.4

22

28

21

12

ARR

RII 3.5

24

48

21

12

I1 = I - I2 = 8 - 4 = 4A

AIII 7.23.5821

I1 and I2 are not equal

Calculate the current in the 2 resistor, given that

(a)R1 =2 (b)R1 =4

I1 and I2 are equal

Current Law- At any instant the algebraic sum of the currents at a junction in a network is zero

I1

I4I3

I2

04321 IIII

0N

iiI

R 1 R 2

R 3

I1 I4

I2

I3I5

a

b

Determine the relationship between the currents I1 ,I2, I4 and I5.

413 III

0341 IIIAt junction a

05421 IIII

523 III

Hence

5241 IIII

0253 IIIAt junction b

Therefore

Then or

R 2

R 5

R 1R 4R 3

I3I1

I20321 III

AIII 0.15.15.2213

From Kirchoff’s law

Given that I1=2.5A and I2=-1.5A. Calculate the current I3.

I6=1A

I5

I4I3I1=3A I2

a

b

c

0321 III

AIII 121264

At junction a

Determine the current I2, I4 and I5.

At junction b

AIII 213312

0543 III

At junction c0642 III

AIII 211435

R 1=30

I1

I3=1AI2

R 3=30 R 2 60 132

23 I

RR

RI

AIII 5.015.1312

Use current divider concept

Determine the current I1

and I2.

AIR

RRI 5.11

60

30603

2

321

At junction a

0132 III

a

E

V 3

V 1

V 2

0n

iiV

321 VVVE

0321 EVVV

Total potential difference across connected components in a complete circuit is zero. The sign of potential difference (p.d) of the source (or e.m.f) is always in opposite sign of the passive components of the circuit

or

thus

E 2

E 1

V

E 3

321 EEEV

21 VVE

VVVV 628423

Loop A

Determine the Voltage V1 and V3.

Loop B

VVEV 481221

431 VVVE

Loop C

4320 VVV

1226412

E=12V

V 3V 1 V 2=8V

V 4=2VLoop BLoop A

Loop C

To check the result

20V A

V AB

R 1

25

R 4R 3

R 2

C

B

D

VVRR

RVAC 5.720

1525

15

31

3

ACBCABCABCAB VVVVVV 0

Branch A

Branch B

VVVV BCACAB 5.30.45.7

Applying Kirchoff’s law

VVRR

RVBC 0.420

1040

10

42

4

Calculate VAB for the network shown

E 1=10V

V 1

V 3=8V

V 2=6V

E 2211 VVE

3121 VVEE

Kirchoff’s law to left loop

841410

To check again for outside loop

322 VVE

Calculate V1 and the e.m.f E2

VVEV 4610211

VVVE 1486322

Kirchoff’s law to right loop