text section 28.3 kirchhoff’s circuit rules
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Text section 28.3
Kirchhoff’s circuit rules
Practice: Chapter 28, problems 17, 19, 25, 26, 43
Junction Rule: total current in = total current out at each junction (from conservation of charge).
Loop Rule: Sum of emfs and potential differences around any closed loop is zero (from conservation of energy).
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Junction Rule: conservation of charge.
I1 I2
I3 I1 = I2 + I3
or
I1’ I2’ (= -I2)
I3’ (= -I3)
I1’ + I2’ + I3’ = 0
-Q +Q
C
- +
R
R I
I
changes going from left to right ΔV = -IR
ΔV = +IR
ΔV =
ΔV = Q/C
Loop Rule: conservation of energy. Follow a test charge q around a loop:
around any loop in circuit.
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18 Ω
6 Ω 3 Ω 9 V
3 V I1
I2
I3 a
b c
d
Find the current through each battery.
18 Ω
6 Ω 3 Ω 9 V
3 V I1
I2
I3 a
b c
d
The junction rule will give:
A) I1 + I2 + I3 = 0 B) -I1 + I2 + I3 = 0 C) I1 - I2 + I3 = 0 D) I1 + I2 - I3 = 0 E) none of these
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18 Ω
6 Ω 3 Ω 9 V
3 V I1
I2
I3 = - (I1 + I2) a
b c
d
Find the current through each battery.
18 Ω
6 Ω 3 Ω 9 V
3 V I1
I2
I3 a
b c
d
The junction rule will give:
A) I1 + I2 + I3 = 0 B) -I1 + I2 + I3 = 0 C) I1 - I2 + I3 = 0 D) I1 + I2 - I3 = 0 E) none of these
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18 Ω
6 Ω 3 Ω 9 V
3 V I1
I2
I3 a
b c
d
The loop rule applied to loop abcda will give:
A) 9A - 18I1 -3I3 = 0 B) 9A + 18I1 -3I3 = 0 C) 9A + 18I1 +3I3 = 0 D) 9A - 18I1 +3I3 = 0 E) none of these
18 Ω
6 Ω 3 Ω 9 V
3 V I1
I2
I3 a
b c
d
The loop rule applied to loop abda will give:
A) 12A - 18I1 + 6I2 = 0 B) 12A - 18I1 - 6I2 = 0 C) 6A - 18I1 - 6I2 = 0 D) 6A + 18I1 + 6I2 = 0 E) 6A - 18I1 + 6I2 = 0
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18 Ω
6 Ω 3 Ω 9 V
3 V I1
I2
I3 = - (I1 + I2) a
b c
d
Find the current through each battery.
Answers: I1 = 400 mA, I2 = 200 mA
18 Ω
6 Ω 3 Ω 9 V
3 V I1
I2
I3 = - (I1 + I2)
a
b c
d
Solution: Loop abcda:
1
2 Loop dbcd:
7× - : 60 I2 = 12A I2 = 200 mA 1 2
3 × - : 60 I1 = 24A I1 = 400 mA 1 2
Junction b: I1 + I2 + I3 = 0 so I3 = -(I1 + I2)
.....
.....
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10 Ω
20 Ω 40 Ω 6 V
3 V I1
I2
I3 = - (I1 + I2) a
b c
d
Find the current through each battery.
10 Ω
20 Ω 40 Ω 6 V
3 V I1
I2
I3 = - (I1 + I2) a
b c
d
Find the current through each battery.
Answers: I1 = 171 mA, I2 = -64.3 mA
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10 Ω
20 Ω 40 Ω 6 V
3 V I1
I2
I3 = - (I1 + I2)
a
b c
d Solution:
Loop abcda: 1
2 Loop dbcd:
4x - 5x : -140 I2 = 9 I2 = -64 mA 1 2
3x - 2x : 70 I1 = 12 I1 = 171 mA 1 2
Junction b: I1 + I2 + I3 = 0 so I3 = -(I1 + I2)
.....
.....
200 Ω
600 Ω
200Ω
300Ω
12 V a b
c
d
What is Vab (i.e., Va - Vb ) when the switch is open?
Exercise for fun: Find the current through the switch when it is closed.
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200 Ω
600 Ω
200Ω
300Ω
a b
c
d
The loop rule requires that Vab (i.e., Va - Vb )
obeys:
I1 I2
A) Vab = (200Ω)I1 + (200Ω)I2 B) Vab = (200Ω)I1 - (200Ω)I2 C) Vab = -(200Ω)I1 + (200Ω)I2 D) Vab = -(200Ω)I1 - (200Ω)I2
1 Ω
2 Ω
2Ω
1Ω
a b
c
d
1Ω Reff = ?
“Series and parallel“ rules don’t help in this case. You have to go back to the fundamentals—Kirchhoff’s Circuit Rules.
(Answer: Reff = 1.4Ω )
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1 Ω
2 Ω
2Ω
1Ω
a b
c
d
1Ω
Reff = VTOTAL/ITOTAL
ITOTAL
VTOTAL I3
I2 I1
I5
I4
1) Use Kirchhoff’s rules to write everything in terms of, one variable, e.g., I3.
2) Divide VTOTAL/ITOTAL .
Show that:
I1 = I5 = 3I3
I2 = I4 = 2I3
ITOTAL= 5I3
VTOTAL= (7Ω) I3
1 Ω
2 Ω
2Ω
1Ω
a b
c
d
1Ω
Reff = VTOTAL/ITOTAL
ITOTAL
VTOTAL I3
I2 I1
I5
I4
1) Use Kirchhoff’s rules to write everything in terms of, e.g., I3. (Results: I1 = I5 = 3I3, I2 = I4 = 2I3; ITOTAL= 5I3; VTOTAL= (7Ω) I3.)
2) Divide VTOTAL/ITOTAL .
Show that:
I1 = I5 = 3I3
I2 = I4 = 2I3
ITOTAL= 5I3
VTOTAL= (7Ω) I3
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Solution to part A:
250
250
100
400
12 V a b
c
d
I1
I1 + I3
I1 - I3
I2
I3
Solution to part b:
Loop adba:
1
Loop abca:
2
Then:
Solve…
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R1
R2
V1
V2
V
and
(from ; and etc)
1 Ω
2 Ω
3Ω
4Ω
a b
c
d
5Ω Reff = ?
Notice that “series and parallel“ don’t help in this case. You have to go back to Kirchhoff’s Rules.
(Answer: Reff = ( 155/74 ) Ω )