Download Kirchhoff`s Rules

Physics 102: Lecture 06
Kirchhoff’s Laws
• Today’s lecture will cover Textbook
Sections 18.5, 7
Conflict Exam sign up available in grade book
Be Careful with round off errors in homework 3!
Physics 102: Lecture 6, Slide 1
Last Time
• Resistors
in series: Reffective  R1  R2  R3  ...
Last Lecture
Current thru is same; Voltage drop across is IRi
• Resistors
Reffective
1 1
1
    ...
R1 R2 R3
Voltage drop across is same; Current thru is V/Ri
• Solved
Today
in parallel:
1
• What
Circuits
about this one?
Physics 102: Lecture 6, Slide 2
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.
Physics 102: Lecture 6, Slide 3
Kirchhoff’s Laws
(1) Label all currents
Choose any direction
(2) Label +/- for all elements
R1
A
+
I1
-
+
Current goes +  - (for resistors)
(3) Choose loop and direction
Must start on wire, not element.
(4) Write down voltage drops
First sign you hit is sign to use.
Physics 102: Lecture 6, Slide 4
+
R2
+
B
E1
-
E2
E3
I2
I3
I4
R3
R4
-
+
R5
-
-
+
+
-
+
KVR Practice
R1=5 W
Find I:
B
I
-
+
+
e1= 50V
-
A
What if only went from A to B?
VA – VB =
Physics 102: Lecture 6, Slide 5
-
R2=15 W
R1=5 W
+
e2= 10V
B
I
-
+
VA - VB =
+
+
e1= 50V
-
A
-
+
R2=15 W
-
+
e2= 10V
22
KVR
Practice
Label currents
Find I:
Label elements +/Choose loop
Write KVR
–e1+IR1 + e2 + IR2 = 0
-50 + 5 I + 10 +15 I = 0
I = +2 Amps
R1=5 W
VA - VB = -IR2 – E2
= -215 - 10 = -40 Volts
Physics 102: Lecture 6, Slide 6
-
+
+
e1= 50V
-
A
What if only went from A to B?
VA - VB = –E1 + IR1
= -50 + 25 = -40 Volts
B
I
+
-
R2=15 W
R1=5 W
+
e2= 10V
B
I
-
+
+
e1= 50V
-
A
-
+
R2=15 W
-
+
e2= 10V
ACT: KVR
Resistors R1 and R2 are:
1) in parallel
2) in series 3) neither
R1=10 W
I1
E2 = 5 V
I2
R2=10 W
IB
+ E1 = 10 V
Physics 102: Lecture 6, Slide 7
ACT: KVR
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
Physics 102: Lecture 6, Slide 8
Preflight 6.1
I1
Calculate the current through resistor 1.
1) I1 = 0.5 A
R1=10 W
+
-
E2 = 5 V
I2
2) I1 = 1.0 A
R2=10 W
3) I1 = 1.5 A
IB
+ E1 = 10 V
ACT: Voltage Law
How would I1 change if the switch was closed?
1) Increase
Physics 102: Lecture 6, Slide 9
2) No change
3) Decrease
Preflight 6.1
Calculate the current through resistor 1.
1) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A
I1
R1=10 W
+
-
E2 = 5 V
I2
-E1 + I1R1 = 0  I1 = E1 /R1 = 1A
R2=10 W
IB
+ E1 = 10 V
ACT: Voltage Law
How would I1 change if the switch was opened?
1) Increase
Physics 102: Lecture 6, Slide 10
2) No change
3) Decrease
Preflight 6.2
Calculate the current through resistor 2.
R1=10 W
I1
1) I2 = 0.5 A
2) I2 = 1.0 A
E2 = 5 V
I2
R2=10 W
+
3) I2 = 1.5 A
IB
+ E1 = 10 V
Physics 102: Lecture 6, Slide 11
Preflight 6.2
Calculate the current through resistor 2.
R1=10 W
I1
1) I2 = 0.5 A 2) I2 = 1.0 A 3) I2 = 1.5 A
E2 = 5 V
I2
-E1 +E2 + I2R2 = 0
 I2 = 0.5A
Physics 102: Lecture 6, Slide 12
R2=10 W
+
IB
+ E1 = 10 V
Kirchhoff’s Junction Rule
Current Entering = Current Leaving
I1 = I2 + I3
I1
I2
I3
Preflight 6.3
1) IB = 0.5 A
R=10 W
I1
E=5V
I2
R=10 W
2) IB = 1.0 A
3) IB = 1.5 A
IB
+ E1 = 10 V
Physics 102: Lecture 6, Slide 13
Kirchhoff’s Junction Rule
Current Entering = Current Leaving
I1 = I2 + I3
I1
I2
I3
Preflight 6.3
1) IB = 0.5 A 2) IB = 1.0 A 3) IB = 1.5 A
IB = I1 + I2 = 1.5 A
R=10 W
I1
E=5V
I2
R=10 W
IB
+ E1 = 10 V
Physics 102: Lecture 6, Slide 14
You try it!
In the circuit below you are given e1, e2, R1, R2 and R3. Find I1, I2 and I3.
1.
2.
Label all currents
Label +/- for all elements
3.
Choose loop and direction
4. Write down voltage drops
Loop 1:
+ R1 - I1
Loop 2:
I3
I2
5.
Write down node equation
Node:
+
e1
-
R2
+
R3
Physics 102: Lecture 6, Slide 15
-
+
e2
+
Kirchhoff’s Laws
(1) Label all currents
Choose any direction
(2) Label +/- for all elements
R1
I1
A
Current goes +  - (for resistors)
(3) Choose loop and direction
Your choice!
(4) Write down voltage drops
R2
B
E1
E3
I2
I3
E2
R3
I4
R4
R5
Follow any loops
(5) Write down node equation
Iin = Iout
Physics 102: Lecture 6, Slide 16
36
You try it!
In the circuit below you are given e1, e2, R1, R2 and R3. Find I1, I2 and I3.
1.2.
3.
4.

Label all currents
Label +/- for all elements
(Choose any direction)
(Current goes +  - for resistor)
Choose loop and direction (Your choice!)
Write down voltage drops (First sign you hit is sign to use!)
Loop 1: – e1+I1R1 – I2R2 = 0
Loop 2: + I2R2 + I3R3 + e2 = 0
5.
+ R1 - I1
I3
I2
Write down node equation
+
e1
-
Loop 1
Node: I1 + I2 = I3
3 Equations, 3 unknowns the rest is math!
Physics 102: Lecture 6, Slide 17
R2
+
R3
+
-
Loop 2
-
e2
+
See you next lecture!
• Read Sections 18.10,11
Physics 102: Lecture 6, Slide 18