Transcript SPH4UW Kirchhoff’s Laws Last Time • Resistors in series: Reffective  R1  R2  R3  ... Last Lecture Current thru is same; • Resistors in.

SPH4UW
Kirchhoff’s Laws
Last Time
• Resistors
in series: Reffective  R1  R2  R3  ...
Last Lecture
Current thru is same;
• Resistors
1
in parallel: R
effective

1 1 1
   ...
R1 R2 R3
Voltage drop across is same;
• Solved
Today
Voltage drop across is IRi
• What
Circuits
about this one?
Current thru is V/Ri
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
Between a and b
a
b
=-IR
b
=IR
b
=+E
I
a
I
a
a
b
=-E
Kirchhoff’s Laws
R1
(1) Label all currents
I1
A
Choose any direction
R2
(2) Choose loop and direction
B
E1
E3
I2
I3
I4
R3
R4
Must start on wire, not element.
E2
(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.
e1- I1R1- I2R2-e2=0
For inner loop
R5
Label KVR
Choose
Write
currents
loop
Practice
R1=5 W
Find I:
ε1- IR1 - ε2 - IR2 = 0
I
B
e1= 50V
50 - 5 I - 10 -15 I = 0
I = +2 Amps
A
R2=15 W
e2= 10V
What if only went from A to B?, Find VB-VA
R1=5 W
or
VB - VA = e1 - IR1
= 50 - 25 = 40 Volts
I
B
e1= 50V
VB - VA = +IR2 + e2
= 215 + 10 = +40 Volts
Therefore B is 40V higher than A
A
R2=15 W
e2= 10V
Understanding
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.
R1=10 W
I1
E2 = 5 V
I2
R2=10 W
IB
+ E1 = 10 V
Definition of series:
Two elements are in series if (and only if) every loop t
Contains R1 also contains R2
slide 7
Practice
Calculate the current through resistor 1.
1) I1 = 0.5 A
2) I1 = 1.0 A
3) I1 = 1.5 A
e1  I1 R1  0
e1
I1 
R1=10 W
I1
E2 = 5 V
I2
R2=10 W
R1
10V
10W
 1A

IB
<-Start
E1 = 10 V
Understanding: Voltage Law
How would I1 change if the switch was closed?
1) Increase
2) No change
3) Decrease
slide 7
Understanding
Calculate the current through resistor 2.
R1=10 W
I1
1) I2 = 0.5 A
E2 = 5 V
I2
2) I2 = 1.0 A
R2=10 W
3) I2 = 1.5 A
IB
E1  E2  I 2 R2  0V
E1 = 10 V
10V  5V  10W  I 2  0V
5V
10W
 0.5 A
I2 
Starting at Star and move
clockwise around loop
Kirchhoff’s Junction Rule
Current Entering = Current Leaving
I1 = I2 + I3
I1
I2
I3
Understanding
1) IB = 0.5 A
IB = I 1 + I 2
2) IB = 1.0 A
= 1.0A + 0.5 A
3) IB = 1.5 A
= 1.5 A
I1=1.0A R=10 W
E=5V
R=10 W
I2=0.5
IB
+ E1 = 10 V
slide 7
Kirchhoff’s Laws
(1) Label all currents
Choose any direction
(2) Choose loop and direction
Your choice!
R1
I1
A
(3) Write down voltage drops
R2
Follow any loops
(4) Write down node equation
B
E1
E3
I2
I3
Iin = Iout
R3
E2
R5
I4
R4
You try it!
In the circuit below you are given ε1, ε2, R1, R2 and R3. Find I1, I2 and I3.
1.
Label all currents
(Choose any direction)
(Current goes +  - for resistor)
Choose loop and direction (Your choice! Include all circuit
3. Write down voltage drops elements!)
2.
4.
Loop 1:
+ e1 - I1R1 + I2R2 = 0
Loop 2:
- I2R2 - I3R3 - e2 = 0
Write down node equation
R1
I3
I1
I2
e1
Node: I1 + I2 = I3
3 Equations, 3 unknowns the rest is math!
Loop 1
R2
R3
Loop 2
e2
Calculations
Loop 1:
+ e1 - I1R1 + I2R2 = 0
Loop 2:
- I2R2 - I3R3 - e2 = 0
Node:
R1
I1 + I2 = I 3
10  50 I1  25I 2  0
25I 2  100 I 3  5  0
25I 2  100  I1  I 2   5  0
10 I1  5 I 2  2
20 I1  25 I 2  1
10 I1  5 I 2  2

1
I2   A
7
I3
I1
I2
50 I1  25I 2  10
20 I1  25 I 2  1
e 1  10V
R1  50W
R2  25W
R3  100W
e 2  5V
I 3  I1  I 2
25  1

9 1
I1   10  


7  50

70 7
 
1
9
I3   A
I1 
A
70
70
e1
Loop 1
R2
R3
Loop 2
The negatives only
indicate that our
current direction
choice was wrong.
Practice Circuits
In the circuit below you are given ε1, R1, R2 and R3.
R1=25
a) Determine the total resistance of the circuit
I1
R2=100
b) Find I1, I2 and I3.
This circuit can be broken down into a
simple circuit, no need for Kirchhoff
Since R2 and R3 are
in parallel
1
1
1
 
RP R3 R2
1
1

50W 100W
3

100W
100
RP 
W
3
e1=12V
Now RP and R1 are in serial
Now: I  V
RT
RT  RP  R1

100

W  25W
3
 58.3W

The potential, V across R2 and R3 is
VP  VT  R1I1
 12V   25W  0.206 A
 6.85V
I2 R3=50
Therefore:
12V
 0.206 A
58.3W
This is the current of I1
I2 
VP
R2
6.85V
100W
 0.0685 A

I3 
VP
R3
6.85V
50W
 0.137 A

I3
Practice
In the circuit below, find ε1, I2, I3
1.
2.
(Directions are given)
Label all currents
Choose loop and direction (Your choice!)
I2
e1
12V
4V
Loop 1
4Ω
Loop 2
2Ω
I1=0.5
I3
6Ω
Practice
In the circuit below, find ε1, I2, I3
3. Write down voltage drops
Loop 1: + (0.5A)(2Ω) + ε1- 12V- I2(4Ω) = 0
Loop 2: + I2(4Ω) + 12V-4V + I3(6 Ω )= 0
I2
e1
12V
4V
Loop 1
4Ω
Loop 2
2Ω
5.
I1=0.5
Write down node equation
Node: 0.5A + I2 = I3
I3
6Ω
Practice
In the circuit below, find ε1, I2, I3
3 Equations, 3 Unknowns
I2
e1
1) + (0.5A)(2Ω) + ε1- 12V - I2(4Ω) = 0
2) + I2(4Ω) + 12V - 4V + I3(6 Ω )= 0 2Ω
3) 0.5A + I2 = I3
11V  e1  4 I 2  0
8V  6 I 3  4 I 2  0
I 3  0.5 A  I 2
8V  6  0.5 A  I2   4I2  0
8V  3V  6I 2  4I 2  0
11V  10I 2  0
I 2  1.1A
I 3  0.5 A  1.1A
 0.6 A
12V
4V
4Ω
Loop 2
Loop 1
6Ω
I3
I1=0.5
11V  e1  4 1.1V   0
e1  4.4V  11V
e1  6.6V
The “-” on the
currents indicate that
our original direction
guess was wrong
Practice
In the circuit below, find the current in each resistor and the equivalent
resistance of the network of five resistors.
c
I1
I2
1Ω
1Ω
13V
1Ω
a
1Ω
I3
b
2Ω
I5
I4
d
Practice
This “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.
Loop 2
c
I1
13V
a
I1+ I2
I2
1Ω Loop 3 1Ω
1Ω
Loop 1
1Ω
I3
b
2Ω
I25 + I3
I1 –I4 I3
d
Using the
current
directions as
guides, we will
define 3 loops
(3 equations
for the 3
unknowns)
Practice
Loop 2
c
I1
+
13V
a
I1+ I2
I2
1Ω Loop 3 1Ω
1Ω
Loop 1
1Ω
I3
b
2Ω
I52 + I3
I1 –I4 I3
d
Loop 1: 13V  I1 1W   I1  I3 1W  0
Loop 2: 13V  I2 1W   I2  I3  2W  0
Loop 3:
I1 1W  I3 1W  I2 1W  0
This is a set of 3
equations and
three unknowns.
So let’s solve
Practice
Loop 1: 13V  I1 1W   I1  I3 1W  0
Loop 2: 13V  I2 1W   I2  I3  2W  0
Loop 3:
I1 1W  I3 1W  I2 1W  0
From loop 3: I 2  I1  I3
Substitute this into loop 1 and loop 2 (to eliminate I2)
Loop 1: 13V  I1  2W  I3 1W
Loop 2: 13V  I1 3W  I3 5W
Multiply loop 1 by 5 and adding to loop 2 and solving for I1
78V  I1 13W 
I1  6 A
thus I 2  5 A and
I3  1A
Practice
The total current is: I1  I 2  6 A  5 A  11A
The potential drop across this is equal to the battery
emf, namely 13V. Therefore the equivalent resistance of
13V
the network is:
Req 
11A
 1.2W
Loop 2
c
I1
+
13V
a
I1+ I2
I2
1Ω Loop 3 1Ω
1Ω
Loop 1
1Ω
I3
b
2Ω
I52 + I3
I1 –I4 I3
d