Transcript Cct component - Universiti Sains Malaysia

Simple dc circuit
Pg 31-60
HUGHES
Electrical & Electronic
Technology
I
V
Load
A complete connection of a source and a load
Source : voltage source, current source such as battery
Load : resistor, capacitor, inductor
Circuit 3
I
Circuit 1
Circuit 2
V
A network is a combination of several circuits
R1
V1
I
R2
V2
V
V  V1  V2
IRt  IR1  IR2
therefore
Rt  R1  R2
V  IRt
V1  IR1
V2  IR2
V1
R2
R3
V  IRt
V1  IR1
V3
V
R1
V2
I
V2  IR2
V3  IR3
V  V1  V2  V3
IRt  IR1  IR2  IR3
therefore
Rt  R1  R2  R3
In general for n series of resistor, Rt
Rt  R1  R2  R3  ......... Rn
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.
To get full light , we must connect a single lamp to 220V
source , thus we have
V2
P  VI 
R
P 60
I 
 0.27 A
V 220
V 2 2202
R

 807 
P
60
When connect two lamps in series , then
Rt  R1  R2  807 807  1614
V
220
I

 0.136 A
Rt 1614
Since the current is less then the
lamp cannot give light fully.
I=1.5A
Calculate the voltage across
each of resistors as in figure
and hence calculate the
supply voltage V
V
R1=2
V1
R2=3
V2
R3=8
V3
V1  IR1  1.5  2  3.0 V
V2  IR2  1.5  3  4.5 V
V3  IR3  1.5  8  12.0 V
V  V1  V2  V3  3.0  4.5  12.0  19.5 V
I
Calculate the circuit’s current
R1 =40 V
1
V=100V
R2 =50  V2
R3=70 V
3
R  R1  R2  R3  40  50  70  160 
V 100
I 
 0.625 A
R 160
V1
R1
R  R1  R2
V2
I
V
I
R1  R2
V
R2
V
V1  IR1 
R1
R1  R2
V1
R1

V R1  R2
V
V2  IR2 
R2
R1  R2
V2
R2

V R1  R2
V2
R2

V R1  R2
10
100

30 R1  100
R1  100  3 100  300
R1  200 
R1
V1
Given that R2=100, calculate
R1 in order to obtain an output
voltage 10V across R2
R2
10V
I
V=30V
I
V  I1R1  I 2 R2
I1
R1
V
I2
I1 
R2
V
R1
V
I2 
R2
I  I1  I 2
V V
I 
R1 R2
I
1
1
 
V R1 R2
I 1

V R
1 1 1
 
R R1 R2
but
then
I
I
V
1
R
1
V V V
I 

R1 R2 R3
then
In general
I3
I2
R
2
R
3
I
1
1
1
 

V R1 R2 R3
1 1
1
1
 

R R1 R2 R3
1 1
1
1
1
 
  .......
R R1 R2 R3
Rn
I
I1
V
R1
I2
1 1 1
 
R R1 R2
R2
R
R1 R2
R1  R2
R1R2
V  IR  I
R1  R2
But
Therefore
V  I1R1  I 2 R2
R2
I1  I
and
R1  R2
R1
I2  I
R1  R2
I
Calculate I1,I2 and I3
V 110
I1 

 5.0 A
R1 22
I1
V=110V
V 110
I2 

 2.5 A
R2 44
I  I1  I 2  5.0  2.5  7.5 A
R1
22
I2
R2
44
I
I1
Calculate the effective resistance
and the power supply
V=12V
R1
6.8
I2
R2
4.7
I3
R3
2.2
1 1
1
1
1
1
1
 




R R1 R2 R3 6.8 4.7 2.2
 0.147  0.213  0.455  0.815
1
R
 1.23 
0.815
Hence
V
12
I 
 9.76 A
R 1.23
I=8A
Calculate the current in the 2 
resistor, given that
I1
(a)R1 =2 
(b)R1 =4 
V
R1
I2
R2
2
R1
2
I2  I
 8
 4.0 A
R1  R2
22
I1 = I - I2 = 8 - 4 = 4A
I1 and I2 are equal
R1
4
I2  I
 8
 5.3 A
R1  R2
42
I1 and I2 are not equal
I1  I  I 2  8  5.3  2.7 A
Current Law- At any instant the
algebraic sum of the currents at a
junction in a network is zero
N
I
i
0
i
I1  I 2  I 3  I 4  0
I1
I4
I2
I3
At junction a
I1
R2
a
I4
I3
Determine the relationship
between the currents I1 ,I2, I4
and I5.
R1
R3
I2
I1  I 4  I 3  0
b
I5
I 3  I1  I 4
At junction b I 3  I 5  I 2  0
Hence
Therefore
Then
I3  I 2  I5
I1  I 4  I 2  I 5
or
I1  I 2  I 4  I 5  0
Given that I1=2.5A and I2=-1.5A.
Calculate the current I3.
R2
I1
I3
R3
R1
From Kirchoff’s law
R5
I1  I 2  I 3  0
I2
I 3  I1  I 2  2.5  1.5  1.0 A
R4
a
I1=3A
At junction a
I4
I 2  I1  I 3  3 1  2 A
I6=1A
I 2  I 4  I6  0
I 4  I 6  I 2  1  2  1A
c
I5
I1  I 2  I 3  0
At junction b
I2
I3
Determine the current
I2, I4 and I5.
b
At junction c
I3  I 4  I5  0
I5  I3  I 4  1 1  2 A
Determine the current I1
and I2.
I1
R1=30
Use current divider concept
R2
I3 
I1
R2  R3
R2  R3
60  30
I1 
I3 
1  1.5 A
R2
60
At junction a
I 2  I 3  I1  0
I 2  I1  I 3  1.5 1  0.5 A
a
R2 60
I2
R3=30
I3=1A
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
E  V1  V2  V3
or
thus
V1  V2  V3  E  0
n
V
i
i
0
V1
E
V2
V3
E1
E2
V
E3
V  E1  E2  E3
Loop C
Determine the Voltage
V1 and V3.
V1
E=12V
Loop A
V2=8V
Loop A
V3
Loop B
E  V1  V2
V1  E  V2  12  8  4V
Loop B
0  V2  V3  V4
V3  V2  V4  8  2  6V
To check the result
Loop C
E  V1  V3  V4
12  4  6  2  12
V4=2V
D
Calculate VAB for the
network shown
R2
R1

Branch A
VAC
R3
15

V
 20  7.5V
R1  R3
25  15
20V
25
VAB
A


R3
R4
Branch B
VBC
R4
10

V
 20  4.0V
R2  R4
40  10
Applying Kirchoff’s law
0  VAB  VBC  VCA  VAB  VBC  VAC
VAB  VAC  VBC  7.5  4.0  3.5V
B
C
Calculate V1 and the e.m.f E2
Kirchoff’s law to left loop
E1  V1  V2
E1=10V
Kirchoff’s law to right loop
 E2  V2  V3
E2  V2  V3  6  8  14V
To check again for outside loop
10  14  4  8
V3=8V
V2=6V
V1  E1  V2  10  6  4V
E1  E2  V1  V3
V1
E2