Upload
theodore-welch
View
214
Download
1
Embed Size (px)
Citation preview
Kirchhoff’s Laws
SPH4UW
Last Time1 2 3 ...effectiveR R R R • Resistors in series:
• Resistors in parallel:1 2 3
1 1 1 1...
effectiveR R R R
Current thru is same; Voltage drop across is IRi
Voltage drop across is same; Current thru is V/Ri
Last
Lect
ure
• Solved Circuits
• What about this one?To
day
Kirchhoff’s Rules
Kirchhoff’s Voltage Rule (KVR):Sum of voltage drops around a loop is zero.
Kirchhoff’s Current Rule (KCR):Current going in equals current coming out.
Kirchhoff’s Rules
I
a b =-IR
I
a b =IR
=+Ea b
a b =-E
Between a and b
Kirchhoff’s Laws
(1) Label all currents Choose any direction
(2) Choose loop and directionMust start on wire, not element.
(3) Write down voltage drops -Batteries increase or decrease according to which end you encounter first.-Resistors drop if going with current.-Resistors increase if gong against current.
R4
I1
I3I2 I4
R1
E1
R2
R3E2
E3
R5
A
B
e1- I1R1- I2R2-e2=0
For inner loop
PracticeR1=5 W
I
ε1- IR1 - ε2 - IR2 = 050 - 5 I - 10 -15 I = 0I = +2 Amps
e1= 50V
R2=15 We2= 10V
A
B
What if only went from A to B?, Find VB-VA
VB - VA = +IR2 + e2
= 215 + 10 = +40 Volts
VB - VA = e1 - IR1
= 50 - 25 = 40 Volts
Find I:
e2= 10V
R1=5 W
I
e1= 50V
R2=15 WA
B
Label currentsChoose loopWrite KVR
or
Therefore B is 40V higher than A
UnderstandingR1=10 W
E1 = 10 V
IB
I1
E2 = 5 VR2=10 WI2
Resistors R1 and R2 are:
1) in parallel 2) in series 3) neither
+ -
Definition of parallel:
Two elements are in parallel if (and only if) you can make a loop that contains only those two elements.
Definition of series:Two elements are in series if (and only if) every loop that Contains R1 also contains R2
PracticeR1=10 W
E1 = 10 V
IB
I1
R2=10 WI2
1) I1 = 0.5 A
2) I1 = 1.0 A
3) I1 = 1.5 A
How would I1 change if the switch was closed?
E2 = 5 V
1) Increase 2) No change 3) Decrease
Calculate the current through resistor 1.
Understanding: Voltage Law
1 1 1
11
1
0
10
101
I R
IR
V
A
<--Start
slide 7
Understanding
R1=10 W
E1 = 10 V
IB
I1
E2 = 5 VR2=10 WI2
1) I2 = 0.5 A
2) I2 = 1.0 A
3) I2 = 1.5 A
Calculate the current through resistor 2.
Starting at Star and move clockwise around loop
1 2 2 2 0E E I R V
2
2
10 5 10 0
5
100.5
V V I V
VI
A
slide 7
Kirchhoff’s Junction Rule
Current Entering = Current Leaving
I1 I2
I3
I1 = I2 + I3
1) IB = 0.5 A
2) IB = 1.0 A
3) IB = 1.5 A
R=10 W
E1 = 10 V
IB
I1=1.0A
E = 5 V R=10 W
I2=0.5
+ -
Understanding
IB = I1 + I2
= 1.0A + 0.5 A
= 1.5 A
slide 7
Kirchhoff’s Laws
(1) Label all currents Choose any direction
(2) Choose loop and directionYour choice!
(3) Write down voltage dropsFollow any loops
(4) Write down node equationIin = Iout R4
R1
E1
R2
R3E2
E3
I1
I3I2 I4
R5
A
B
You try it!
R1
R2 R3
I1 I3
I2
Loop 1:
1. Label all currents
2. Choose loop and direction
3. Write down voltage drops
Loop 2:
e1
4. Write down node equation
Node:
e2
In the circuit below you are given ε1, ε2, R1, R2 and R3. Find I1, I2 and I3.
(Choose any direction)(Current goes + - for resistor)
(Your choice! Include all circuit elements!)
Loop 1
Loop 2
+ e1 - I1R1 + I2R2 = 0
- I2R2 - I3R3 - e2 = 0
I1 + I2 = I3
3 Equations, 3 unknowns the rest is math!
Calculations
R1
R2 R3
I1 I3
I2
e1
Loop 1
Loop 2
Loop 1:
Loop 2:
+ e1 - I1R1 + I2R2 = 0
Node:
- I2R2 - I3R3 - e2 = 0
I1 + I2 = I3
1
1
2
3
2
10
50
25
100
5
V
R
R
R
V
1
1 2 2 3
1 2 2 1 2
2 1 2
10 50 25 0 25 100 5 0
50 25 10 25 10
10 5 2 20 25
0 5 0
1
I I I I
I I I I
I I
I
I I
1 2
1 2
2
20 25 1
10 5
1
2
7
I I
I I
I A
1
1
25 110
7
9
5
70
0
I
I
A
3 2
3
1
9 1
70 7
1
70I A
I I I
The negatives only indicate that our current direction
choice was wrong.
Practice Circuits
R1=25
R2=100 R3=50
I1
I3I2
e1=12V
In the circuit below you are given ε1, R1, R2 and R3.
a) Determine the total resistance of the circuit
b) Find I1, I2 and I3.
Since R2 and R3 are in parallel
3 2
1 1 1
1 1
50 1003
100100
3
P
P
R R R
R
Now RP and R1 are in serial
1
10025
358.3
T PR R R
This circuit can be broken down into a simple circuit, no need for Kirchhoff
Now: T
VI
R
120.206
58.3
VA
This is the current of I1
The potential, V across R2 and R3 is
1 1
12 25 0.206
6.85
P TV V R I
V A
V
22
6.85
1000.0685
PVIR
V
A
33
6.85
500.137
PVIR
V
A
Therefore:
Practice
1. Label all currents
In the circuit below, find ε1, I2, I3
(Directions are given)
4Ω 6Ω
I1=0.5 I3
I2
e1
Loop 1
Loop 2
12V 4V
2Ω
2. Choose loop and direction (Your choice!)
PracticeIn the circuit below, find ε1, I2, I3
4Ω 6Ω
I1=0.5 I3
I2
e1
Loop 1
Loop 2
12V 4V
2Ω
3. Write down voltage drops
Loop 1: + (0.5A)(2Ω) + ε1- 12V- I2(4Ω) = 0Loop 2: + I2(4Ω) + 12V-4V + I3(6 Ω )= 0
5. Write down node equation
Node: 0.5A + I2 = I3
PracticeIn the circuit below, find ε1, I2, I3
4Ω 6Ω
I1=0.5 I3
I2
Loop 1Loop 2
12V 4V
2Ω
1) + (0.5A)(2Ω) + ε1- 12V - I2(4Ω) = 02) + I2(4Ω) + 12V - 4V + I3(6 Ω )= 03) 0.5A + I2 = I3
3 Equations, 3 Unknownse1
1 2
3
3 2
28 6 4 0
4 0
0
11
.5
V I
V I
I
I A I
220.8 4 056 A IV I
2 28 3 6 4 0V V I I
211 10 0V I 2 1.1I A
111 4 1.1 0V V
1
1
4.4 11
6.6
V V
V
3 1.10.5
0.6
AI A
A
The “-” on the currents indicate that our original direction guess was wrong
PracticeIn the circuit below, find the current in each resistor and the equivalent resistance of the network of five resistors.
2Ω
13V
1Ω
c
1Ω 1Ω
1Ω
I2
I5I4
I3
I1
a b
d
PracticeThis “bridge” network cannot be represented in terms of series and parallel combinations. There are five different currents to determine, but by applying the junction rule to junctions a and b, we can determine then in terms of three unknown currents.
2ΩI1+ I2Loop 1
Loop 3
13V
1Ω
cLoop 2
1Ω 1Ω
1Ω
I2
I2 + I3I1 – I3
I3
I1
a b
d
I5I4
Using the current directions as guides, we will define 3 loops (3 equations for the 3 unknowns)
Practice
2ΩI1+ I2Loop 1
Loop 3+
13V
1Ω
cLoop 2
1Ω 1Ω
1Ω
I2
I2 + I3I1 – I3
I3
I1
a b
d
I5I4
Loop 1: 1 1 313 1 1 0V I I I
Loop 2:
Loop 3:
2 2 313 1 2 0V I I I
1 3 21 1 1 0I I I
This is a set of 3 equations and three unknowns. So let’s solve
PracticeLoop 1: 1 1 313 1 1 0V I I I
Loop 2:
Loop 3:
2 2 313 1 2 0V I I I
1 3 21 1 1 0I I I
From loop 3: 2 1 3I I I
Substitute this into loop 1 and loop 2 (to eliminate I2) 1 313 2 1V I I Loop 1:
Loop 2: 1 313 3 5V I I
Multiply loop 1 by 5 and adding to loop 2 and solving for I1 1
1
78 13
6
V I
I A
thus 2 5I A and 3 1I A
PracticeThe total current is: 1 2 6 5 11I I A A A
The potential drop across this is equal to the battery emf, namely 13V. Therefore the equivalent resistance of the network is:13
1.211eq
VR
A
2ΩI1+ I2Loop 1
Loop 3+
13V
1Ω
cLoop 2
1Ω 1Ω
1Ω
I2
I2 + I3I1 – I3
I3
I1
a b
d
I5I4