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Lecture 11
Ch27. Circuit Theory
University Physics: Waves and Electricity
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2013
Homework 9
(a) Find the equivalent resistance between points a and b in
the circuit diagram below.
(b) Calculate the current in each resistor if a potential difference
of 34 V is applied between points a and b.
Erwin Sitompul
University Physics: Wave and Electricity
11/2
Solution of Homework 9
(a) Req  4  (7 ||10)  9
7 10
9
7  10
 4  4.12  9
 17.12 
(b)
34
V
 1.986 A

i
Req 17.12
i4   i9   1.986 A
≡
 4
V4.12   V7   V10 
V4.12   i4.12   R4.12 
V4.12  (1.986)(4.12)
 8.182 V
Erwin Sitompul
University Physics: Wave and Electricity
11/3
Solution of Homework 9
V4.12   V7   V10 
i4.12   R4.12   i7   R7   i10   R10 
i7  
c
d
V7 
≡
R
8.182

7
 1.169 A
c
i10  
V10 
10 
8.182

10
 0.8182 A
Erwin Sitompul
d
V4.12   V7   V10   Vcd  8.182 V
University Physics: Wave and Electricity
11/4
Example: Single-Loop Circuit
For the circuit shown, determine Vab.
E1  12 V
• a  a, cw
E1  iR2  iR1  E2  0
12 1i  2i  6  0
3i  6
i2A
• a  b, ccw
Va +E2  iR1  Vb
Va  Vb  E2  iR1
Vab  6  (2)(2)
 10 V
Erwin Sitompul
i
b
a
E2  6 V
R1  2 
R2  1 
• Try again for i
assumed to be ccw
• Assume a current direction
• Perform the calculation
• i > 0 means that the current flows in
the assumed direction
• i < 0 means that the current flows in
opposite direction
University Physics: Wave and Electricity
11/5
Voltage Divider and Current Divider
Voltage Divider
i
i

V1

i1 
E
E
R1
Current Divider
R2
R1

V2

i2 

V1

R2
E
i
R1  R2
V1  iR1
R1
 V1 
E
R1  R2
V2  iR2
R2
 V2 
E
R1  R2
Erwin Sitompul

V2

E
i1 
R1
E
i2 
R2
E  iReq
Req
R2
i1 
i  i1 
i
R1
R1  R2
i2 
Req
R2
i  i2 
R1
i
R1  R2
University Physics: Wave and Electricity
11/6
Req of Two Resistance in Parallel
R1
R2
1
1 1 R2  R1

 
R1 R2
Req R1 R2
R1 R2
Req 
R1  R2
(3)(1)
 0.75 
Req 
3 1
2
18  5.33 V
V2  
2  0.75  4
4
18  10.67 V
V4  
2  0.75  4
V3   V1   18  5.33 10.67  2 V
0.75
18  2 V
V0.75  
2  0.75  4
Erwin Sitompul
University Physics: Wave and Electricity
11/7
Example: Finding Voltage and Current
i1  12 
 V1 
V2 
8
15  10 V
48
V1
5
 0.417 A

12 12
V2 10
 0.25 A
i2 

40 40
4
 V1 
i6  ? i10  ? i through the battery ?
Erwin Sitompul

V2

15 V
i1 
i2 
10 
4
15  5 V
48
≡
V1 
15 V
6
40 
For this circuit, determine
i1, i2, V1, and V2.

V2

University Physics: Wave and Electricity
11/8
Example: Two-Loop Circuit
E1
If E1 = 20 V, determine E2 and E3.
i 1  2
i 1A
1A 
R4  6 
1
R5  4 
• Loop 1, cw
E1  (1) R4  E2  (1) R5  0
20  (1)(6)  E2  (1)(4)  0
E2  18 V
• Loop 2, cw
2A
i
E2
2
E3
R6  2 
• Outer Loop, ccw
(1) R5  E2  (2) R6  E3  0
E3  (2) R6  (1) R4  E1  0
(1)(4)  18  (2)(2)  E3  0
E3  (2)(2)  (1)(6)  20  0
E3  10 V
Erwin Sitompul
E3  10 V
University Physics: Wave and Electricity
11/9
Multiloop Circuits
 Junction Rule: The sum of the currents entering any junction
must be equal to the sum of the currents leaving that junction
• Remember again Loop Rule,
Resistance Rule, and Emf Rule
 Our basic tools for solving complex circuits are the loop rule
and the junction rule.
• Junction Rule, at d
i1  i3  i2
. . . . . . . . . .(1)
• Loop Rule, left-hand loop,
b  b, ccw
E1  i1R1  i3 R3  0 . . . . . .(2)
• Loop Rule, right-hand loop,
b  b , ccw
i3 R3  i2 R2  E2  0 . . . . . .(3)
• Afterward, find the unknowns
from these three equations
Erwin Sitompul
University Physics: Wave and Electricity 11/10
Example: Multiloop Circuits
The value of the elements in the
following circuits are:
E1 = 3 V, E2 = 6 V,
R1 = 2 Ω, R2 = 4 Ω
Find the magnitude and
direction of the current in each
of the branches.
• Junction Rule, at a
. . . . . . . . . .(1)
i3  i1  i2
• Loop Rule, left-hand loop,
a  a, ccw
• Loop Rule, right-hand loop,
a  a , cw
i1R1  E1  i1R1  E2  i2 R2  0
2i1  3  2i1  6  4i2  0
4i1  4i2  3 . . .(2)
Erwin Sitompul
i3 R1  E2  i3 R1  E2  i2 R2  0
2i3  6  2i3  6  4i2  0
4i2  4i3  0 . . .(3)
University Physics: Wave and Electricity 11/11
Example: Multiloop Circuits
• Option B: Substitution Method
i3  i1  i2 . . . . . . .(1)
4i1  4i2  3 . . .(2)
4i2  4i3  0 . . .(3)
• From (2)
i1 
• Insert (1) into (3),
• Insert (5) into (4)
4i2  4(i1  i2 )  0
4i1  8i2  0. . .(4)
 3  4i2 
4
  8i2  0
 4 
3  4i2  8i2  0
• Option A: Elimination Method
4i1  4i2  3 1 4i1  4i2  3
4i1  8i2  0 1
4i1  8i2  0
12i2  3
i2  0.25 A

12i2  3
i2  0.25 A
8
i1   i2  0.5 A
4
i3  0.5  (0.25)  0.25 A
Erwin Sitompul
3  4i2
. . . . . . . .(5)
4
i1 
3  4(0.25)
 0.5 A
4
i3  0.5  (0.25)  0.25 A
University Physics: Wave and Electricity 11/12
Homework 10
For the circuit depicted below, determine the currents (i1, i2, i3)
and the potential difference across the resistors (V1, V2, V3).
2
i1
5V
 V1 
Erwin Sitompul
i3
i2

V2

8
3V

V3

4
University Physics: Wave and Electricity 11/13
Homework 10A
3
i1
 V1 
6V
1. For the circuit depicted on the right
side, determine the currents (i1, i2, i3)
and the potential difference across
the resistors (V1, V2, V3).
i3
i2

V2

7
4V

V3

5
2. For the circuit shown by the
figure on the right, determine
Vcb and i7.0 Ω.
Erwin Sitompul
University Physics: Wave and Electricity 11/14