Study Guide
Kirchhoff's Rules
-
University:
Kentucky State University -
Course:
PHY 212 | General Physics II Academic year:
2019
-
Views:
96
Pages:
50
Author:
gaarmaakbn9h
A nontrivial circuit
Analyze this circuit for me, if you don’t mind!
Find the currents I1, I2, and I3.
h
30
I1
40
a
b
I3
1
2 = 45 V
c
d
20
I2
1 = 85 V 1
g
f
e Two sets of resistors in series.
This.
And this.
Further analysis is difficult. series1 seems to be in parallel with
the 30 resistor, but what about 2?
We need new tools to analyze that combination.
h
30
I1
40
a
I3
1
2 = 45 V
c
b
series1
I2
1 = 85 V 1
g
f
series2
e
d
20 Kirchhoff’s Rules
Kirchhoff’s Loop Rule:
The sum of potential changes around any closed path in a
circuit is zero.
V =0
around any closed loop
Energy conservation: a charge
ending up where it started neither
gains nor loses energy, i.e.
Ui = Uf ∆U = 0 = (q ∆V)
∆V = 0
Kirchhoff’s Junction Rule:
The sum of all currents entering a junction must equal the sum
of all currents leaving the junction
I = 0
at any junction
Charge conservation: charge in = charge out
(current in counts +, current out counts -) Recipe for problems that require Kirchhoff’s rules
1. Label the current in each branch of the circuit with a
symbol (I1, I2, I3, …) and an arrow (direction).
2. Apply Kirchhoff’s Junction Rule at each junction.
Current in is +, current out is -.
3. Apply Kirchhoff’s Loop Rule for as many independent loops
as necessary (pick travel direction and follow sign rules).
Resistor:
I
V is travel
Battery: + travel
4. Solve resulting system of linear equations.
V is + Loop rule in circuit of 3 resistors and 2 batteries
R1
a
R2
R3
b
V1
V2
- +
I
V3
+ -
VB
VA
1. Assign (guess) current direction
2. Pick travel direction: let’s say clockwise, i.e. with the flow
3. Add voltages
- V1 - V2 - V3 + VA - VB = 0
- IR1 - IR2 - IR3 + VA - VB = 0
Ohm’s law V=I R Example 7 : calculate I, Vab, and Vba for the circuit shown.
Presumably you can do this, if you recall last lecture.
5
b
10
+ -
I
a
9V
+9 – 5 I – 10 I = 0
15 I = +9
I = +9/15 = 0.6 A Example: calculate I, Vab, and Vba for the circuit shown.
5
b
10
+ -
I = 0.6 A
a
9V
Va + 9 – 5 (0.6) = Vb
It is just a convention how to define Vab
Vab = Va – Vb or Vab = Vb – Va
Vab = Va – Vb = – 9 + 5 (0.6) = -6 V Example: calculate I, Vab, and Vba for the circuit shown.
5
b
10
+ -
I = 0.6 A
a
9V
Vb – 10 (0.6) = Va
Vba = Vb – Va = + 10 (0.6) = +6 V Example: calculate I, Vab, and Vba for the circuit shown.
5
b
10
+ -
I = 0.6 A
Note:
a
9V
Vba = +6 V = - Vab , as expected Example: calculate I, Vab, and Vba for the circuit shown.
Graph the potential rises and drops in this circuit.
5
b
10
+ -
I = 0.6 A
a
9V
5 I = 3V
b
= 9V
10 I = 6V
a Example 8 : calculate I, Vab, and Vba for the circuit shown.
Presumably you can do this, if you recall last lecture.
5
b
10
I
- +
+-
6V
9V
a
+ 9 – 6 – 5 I – 10 I =
=00
15 I = 3
I = 0.2 A Example: calculate I, Vab, and Vba for the circuit shown.
5
b
10
I = 0.2 A
- +
+-
6V
9V
a
Va + 9 – 6 – 5 (0.2) = Vb
Vab = Va – Vb = – 9 + 6 + 5 (0.2) = – 2 V Example: calculate I, Vab, and Vba for the circuit shown.
The smart way: Vba = - Vab = +2 V
5
b
10
I = 0.2 A
- +
+-
6V
9V
a
Vb – 10 (0.2) = Va
Vba = Vb – Va = + 10 (0.2) = + 2 V Example: calculate I, Vab, and Vba for the circuit shown.
What if you guess the wrong current direction?
5
b
10
I
- +
+-
6V
9V
a
– 10 I – 5 I +6 – 9 = 0
15 I = – 3
I = – 0.2 A oops, guessed wrong direction, no big deal! h
30
I1
I3
40
a
1
b
2 = 45 V
c
dd
20
I2
1 = 85 V
1
g
e
f
Example 9 : calculate the 3 unknowns (I1, I2, and I3)
Back to our circuit: 3 unknowns (I1, I2, and I3), so we will need
3 equations. We begin with the junctions.
Junction a:
I3 – I1 – I 2 = 0
Junction d:
-I3 + I1 + I2 = 0
--eq. 1
Junction d gave no new information, so we still need two more equations. h
30
I1
I3
40
a
1
2 = 45 V
c
b
d
20
I2
1 = 85 V
g
There are three loops.
1
e
f
Loop 1.
Loop 2.
Loop 3.
Any two loops will produce independent equations.
Using the third loop will provide no new information. h
30
I1
40
a
I3
1
2 = 45 V
c
b
d
20
I2
1 = 85 V
g
1
e
f
The “green” loop (a-h-d-c-b-a):
(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0
I
V is travel
+-
V is +
travel h
30
I1
40
a
I3
1
2 = 45 V
c
b
d
20
I2
1 = 85 V
g
1
e
f
The “blue” loop (a-b-c-d-e-f-g):
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
I
V is travel
+-
V is +
travel After combining terms and simplifying, we now have three
equations, three unknowns; the rest is “just algebra.”
Junction a:
I3 – I1 – I 2 = 0
--eq. 1
The “green” loop
- 30 I1 + 45 - 41 I3 = 0
--eq. 2
The “blue” loop
41 I3 -130 + 21 I2 = 0
--eq. 3
I will skip the algebra!
Make sure to use voltages in V and resistances in . Then currents will be in A. Collect our three equations:
I3 – I1 – I2 = 0
- 30 I1 + 45 - 41 I3 = 0
41 I3 -130 + 21 I2 = 0
Rearrange to get variables in “right” order:
– I1 – I2 + I3 = 0
- 30 I1 - 41 I3 + 45 = 0
21 I2 + 41 I3 -130 = 0
Use the middle equation to eliminate I1:
I1 = (41 I3 – 45)/(-30)
There are many valid sets of steps to solving a system of equations. Any that works is acceptable. Two equations left to solve:
– (41 I3 – 45)/(-30) – I2 + I3 = 0
21 I2 + 41 I3 -130 = 0
Might as well work out the numbers:
1.37 I3 – 1.5 – I2 + I3 = 0
21 I2 + 41 I3 -130 = 0
– I2 + 2.37 I3 – 1.5 = 0
21 I2 + 41 I3 -130 = 0
Multiply the top equation by 21:
– 21 I2 + 49.8 I3 – 31.5 = 0
21 I2 + 41 I3 -130 = 0 Add the two equations to eliminate I2:
– 21 I2 + 49.8 I3 – 31.5 = 0
+ ( 21 I2 + 41 I3 -130 = 0 )
90.8 I3 – 161.5 = 0
Solve for I3:
I3 = 161.5 / 90.8
I3 = 1.78
Go back to the “middle equation” two slides ago for I1:
I1 = (41 I3 – 45)/(-30)
I1 = - 1.37 I3 + 1.5
I1 = - (1.37) (1.78) + 1.5
I1 = - 0.94 Go back two slides to get an equation that gives I2:
– I2 + 2.37 I3 – 1.5 = 0
I2 = 2.37 I3 – 1.5
I2 = (2.37) (1.78) – 1.5
I2 = 2.72
Summarize answers (don’t forget to show units):
I1 = - 0.94 A
I2 = 2.72 A
I3 = 1.78 A
Are these currents correct? How could you tell?
We’d better check our results. I3 – I1 – I2 = 0
- 30 I1 + 45 - 41 I3 = 0
41 I3 -130 + 21 I2 = 0
I1 = - 0.94 A
I2 = 2.72 A
I3 = 1.78 A
Check:
1.78 – (-0.94) – 2.72 = 0
- 30 (-0.94) + 45 - 41 (1.78) = 0.22
?
41 (1.78) -130 + 21 (2.72) = 0.10
?
Are the last two indication of a mistake or just roundoff error? Recalculating
while retaining 2 more digits gives I1=-0.933, I2=2.714, I3=1.7806, and the
last two results are 0.01 or less roundoff was the culprit! A nontrivial circuit
Analyze this circuit for me, if you don’t mind!
Find the currents I1, I2, and I3.
h
30
I1
40
a
b
I3
1
2 = 45 V
c
d
20
I2
1 = 85 V 1
g
f
e Two sets of resistors in series.
This.
And this.
Further analysis is difficult. series1 seems to be in parallel with
the 30 resistor, but what about 2?
We need new tools to analyze that combination.
h
30
I1
40
a
I3
1
2 = 45 V
c
b
series1
I2
1 = 85 V 1
g
f
series2
e
d
20 Kirchhoff’s Rules
Kirchhoff’s Loop Rule:
The sum of potential changes around any closed path in a
circuit is zero.
V =0
around any closed loop
Energy conservation principle:
a charge ending up where it started
neither gains nor loses energy, i.e.
Ui = Uf ∆U = 0 = (q ∆V)
∆V = 0
Kirchhoff’s Junction Rule:
The sum of all currents entering a junction must equal the sum
of all currents leaving the junction
I = 0
at any junction
Charge conservation principle:
charge in = charge out
(current in counts +, current out counts -) Recipe for problems that require Kirchhoff’s rules
1. Label the current in each branch of the circuit with a
symbol (I1, I2, I3, …) and an arrow (direction).
2. Apply Kirchhoff’s Junction Rule at each junction.
Current in is +, current out is -.
3. Apply Kirchhoff’s Loop Rule for as many independent loops
as necessary (pick travel direction and follow sign rules).
Resistor:
I
V is travel
Battery: + travel
4. Solve resulting system of linear equations.
V is + Loop rule in circuit of 3 resistors and 2 batteries
R1
a
R2
R3
b
V1
V2
- +
I
V3
+ -
VB
VA
1. Assign (guess) current direction
2. Pick travel direction: let’s say clockwise, i.e. with the flow
3. Add voltages
- V1 - V2 - V3 + VA - VB = 0
- IR1 - IR2 - IR3 + VA - VB = 0
Ohm’s law V=I R Example 7 : calculate I, Vab, and Vba for the circuit shown.
Presumably you can do this, if you recall last lecture.
5
b
10
+ -
I
a
9V
+9 – 5 I – 10 I = 0
15 I = +9
I = +9/15 = 0.6 A Example: calculate I, Vab, and Vba for the circuit shown.
5
b
10
+ -
I = 0.6 A
a
9V
Va + 9 – 5 (0.6) = Vb
It is just a convention how to define Vab
Vab = Va – Vb or Vab = Vb – Va
Vab = Va – Vb = – 9 + 5 (0.6) = -6 V Example: calculate I, Vab, and Vba for the circuit shown.
5
b
10
+ -
I = 0.6 A
a
9V
Vb – 10 (0.6) = Va
Vba = Vb – Va = + 10 (0.6) = +6 V Example: calculate I, Vab, and Vba for the circuit shown.
5
b
10
+ -
I = 0.6 A
Note:
a
9V
Vba = +6 V = - Vab , as expected Example: calculate I, Vab, and Vba for the circuit shown.
Graph the potential rises and drops in this circuit.
5
b
10
+ -
I = 0.6 A
a
9V
5 I = 3V
b
= 9V
10 I = 6V
a Example 8 : calculate I, Vab, and Vba for the circuit shown.
Presumably you can do this, if you recall last lecture.
5
b
10
I
- +
+-
6V
9V
a
+ 9 – 6 – 5 I – 10 I =
=00
15 I = 3
I = 0.2 A Example: calculate I, Vab, and Vba for the circuit shown.
5
b
10
I = 0.2 A
- +
+-
6V
9V
a
Va + 9 – 6 – 5 (0.2) = Vb
Vab = Va – Vb = – 9 + 6 + 5 (0.2) = – 2 V Example: calculate I, Vab, and Vba for the circuit shown.
The smart way: Vba = - Vab = +2 V
5
b
10
I = 0.2 A
- +
+-
6V
9V
a
Vb – 10 (0.2) = Va
Vba = Vb – Va = + 10 (0.2) = + 2 V Example: calculate I, Vab, and Vba for the circuit shown.
What if you guess the wrong current direction?
5
b
10
I
- +
+-
6V
9V
a
– 10 I – 5 I +6 – 9 = 0
15 I = – 3
I = – 0.2 A oops, guessed wrong direction, no big deal! h
30
I1
I3
40
a
1
b
2 = 45 V
c
dd
20
I2
1 = 85 V
1
g
e
f
Example 9 : calculate the 3 unknowns (I1, I2, and I3)
Back to our circuit: 3 unknowns (I1, I2, and I3), so we will need
3 equations. We begin with the junctions.
Junction a:
I3 – I1 – I 2 = 0
Junction d:
-I3 + I1 + I2 = 0
--eq. 1
Junction d gave no new information, so we still need two more equations. h
30
I1
I3
40
a
1
2 = 45 V
c
b
d
20
I2
1 = 85 V
g
There are three loops.
1
e
f
Loop 1.
Loop 2.
Loop 3.
Any two loops will produce independent equations.
Using the third loop will provide no new information. h
30
I1
40
a
I3
1
2 = 45 V
c
b
d
20
I2
1 = 85 V
g
1
e
f
The “green” loop (a-h-d-c-b-a):
(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0
I
V is travel
+-
V is +
travel h
30
I1
40
a
I3
1
2 = 45 V
c
b
d
20
I2
1 = 85 V
g
1
e
f
The “blue” loop (a-b-c-d-e-f-g-a):
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
I
V is travel
+-
V is +
travel After combining terms and simplifying, we now have three
equations, three unknowns; the rest is “just algebra.”
Junction a:
I3 – I1 – I 2 = 0
--eq. 1
The “green” loop
- 30 I1 + 45 - 41 I3 = 0
--eq. 2
The “blue” loop
41 I3 -130 + 21 I2 = 0
--eq. 3
I will skip the algebra!
Make sure to use voltages in V and resistances in . Then currents will be in A. Collect our three equations:
I3 – I1 – I2 = 0
- 30 I1 + 45 - 41 I3 = 0
41 I3 -130 + 21 I2 = 0
Rearrange to get variables in “right” order:
– I1 – I2 + I3 = 0
- 30 I1 - 41 I3 + 45 = 0
21 I2 + 41 I3 -130 = 0
Use the middle equation to eliminate I1:
I1 = (41 I3 – 45)/(-30)
There are many valid sets of steps to solving a system of equations. Any that works is acceptable. Two equations left to solve:
– (41 I3 – 45)/(-30) – I2 + I3 = 0
21 I2 + 41 I3 -130 = 0
Might as well work out the numbers:
1.37 I3 – 1.5 – I2 + I3 = 0
21 I2 + 41 I3 -130 = 0
– I2 + 2.37 I3 – 1.5 = 0
21 I2 + 41 I3 -130 = 0
Multiply the top equation by 21:
– 21 I2 + 49.8 I3 – 31.5 = 0
21 I2 + 41 I3 -130 = 0 Add the two equations to eliminate I2:
– 21 I2 + 49.8 I3 – 31.5 = 0
+ ( 21 I2 + 41 I3 -130 = 0 )
90.8 I3 – 161.5 = 0
Solve for I3:
I3 = 161.5 / 90.8
I3 = 1.78
Go back to the “middle equation” two slides ago for I1:
I1 = (41 I3 – 45)/(-30)
I1 = - 1.37 I3 + 1.5
I1 = - (1.37) (1.78) + 1.5
I1 = - 0.94 Go back two slides to get an equation that gives I2:
– I2 + 2.37 I3 – 1.5 = 0
I2 = 2.37 I3 – 1.5
I2 = (2.37) (1.78) – 1.5
I2 = 2.72
Summarize answers (don’t forget to show units):
I1 = - 0.94 A
I2 = 2.72 A
I3 = 1.78 A
Are these currents correct? How could you tell?
We’d better check our results. I3 – I1 – I2 = 0
- 30 I1 + 45 - 41 I3 = 0
41 I3 -130 + 21 I2 = 0
I1 = - 0.94 A
I2 = 2.72 A
I3 = 1.78 A
Check:
1.78 – (-0.94) – 2.72 = 0
- 30 (-0.94) + 45 - 41 (1.78) = 0.22
?
41 (1.78) -130 + 21 (2.72) = 0.10
?
Are the last two indication of a mistake or just roundoff error? Recalculating
while retaining 2 more digits gives I1=-0.933, I2=2.714, I3=1.7806, and the
last two results are 0.01 or less roundoff was the culprit!
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