physics 102: lecture 6, slide 1 kirchhoffs laws todays lecture will cover textbook sections 18.5, 7...
DESCRIPTION
Physics 102: Lecture 6, Slide 3 Kirchhoff’s Rules Kirchhoff’s Voltage Rule (KVR):KVR –Sum of voltage drops around a loop is zero. Kirchhoff’s Current Rule (KCR): –Current going in equals current coming out.TRANSCRIPT
Physics 102: Lecture 6, Slide 1
Kirchhoff’s Laws
• Today’s lecture will cover Textbook Sections 18.5, 7
Physics 102: Lecture 06
Conflict Exam sign up available in grade book Be Careful with round off errors in homework 3!
Physics 102: Lecture 6, Slide 2
Last Time ...321 RRRReffective• Resistors in series:
• Resistors in parallel: ...1111
321
RRRReffective
Current thru is same; Voltage drop across is IRi
Voltage drop across is same; Current thru is V/Ri
Last
Lec
ture
• Solved Circuits
• What about this one?
Toda
y
Physics 102: Lecture 6, Slide 3
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 4
Kirchhoff’s Laws(1) Label all currents
Choose any direction
(2) Label +/- for all elementsCurrent goes + - (for resistors)
(3) Choose loop and directionMust start on wire, not element.
(4) Write down voltage dropsFirst sign you hit is sign to use.
R4
I1
I3I2 I4
+
+
+ +
+
-
- -
-
-+
+
+
-
-
-
R1
E1
R2
R3E2
E3
R5
A
B
Physics 102: Lecture 6, Slide 5
KVR PracticeR1=5 I
1= 50V+
-
+ -
+
-+-
R2=15 2= 10V
A
B
What if only went from A to B?
VA – VB =
VA - VB =
Find I:
2= 10V
R1=5 I
+
-
+ -
1= 50V
+
-+-
R2=15 A
B
22
Physics 102: Lecture 6, Slide 6
KVR PracticeR1=5 I
–1+IR1 + 2 + IR2 = 0-50 + 5 I + 10 +15 I = 0I = +2 Amps
1= 50V+
-
+ -
+
-+-
R2=15 2= 10V
A
B
What if only went from A to B?
VA - VB = -IR2 – E2
= -215 - 10 = -40 Volts
VA - VB = –E1 + IR1 = -50 + 25 = -40 Volts
Find I:
2= 10V
R1=5 I
+
-
+ -
1= 50V
+
-+-
R2=15 A
B
Label currentsLabel elements +/-Choose loopWrite KVR
Physics 102: Lecture 6, Slide 7
ACT: KVRR1=10
E1 = 10 V
IB
I1
E2 = 5 VR2=10 I2
Resistors R1 and R2 are:
1) in parallel 2) in series 3) neither
+ -
Physics 102: Lecture 6, Slide 8
ACT: KVRR1=10
E1 = 10 V
IB
I1
E2 = 5 VR2=10 I2
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
Physics 102: Lecture 6, Slide 9
Preflight 6.1R1=10
E1 = 10 V
IB
I1
R2=10 I2
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.
ACT: Voltage Law
Physics 102: Lecture 6, Slide 10
Preflight 6.1R1=10
E1 = 10 V
IB
I1
R2=10 I21) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A
+ -
+ -
-E1 + I1R1 = 0 I1 = E1 /R1 = 1A
How would I1 change if the switch was opened?
E2 = 5 V
1) Increase 2) No change 3) Decrease
Calculate the current through resistor 1.
ACT: Voltage Law
Physics 102: Lecture 6, Slide 11
Preflight 6.2
R1=10
E1 = 10 V
IB
I1
E2 = 5 VR2=10 I2
1) I2 = 0.5 A
2) I2 = 1.0 A
3) I2 = 1.5 A
+ -
+ -
Calculate the current through resistor 2.
Physics 102: Lecture 6, Slide 12
Preflight 6.2
R1=10
E1 = 10 V
IB
I1
E2 = 5 VR2=10 I2
1) I2 = 0.5 A 2) I2 = 1.0 A 3) I2 = 1.5 A
+ -
+ -
-E1 +E2 + I2R2 = 0
I2 = 0.5A
Calculate the current through resistor 2.
Physics 102: Lecture 6, Slide 13
Kirchhoff’s Junction RuleCurrent 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
E1 = 10 V
IB
I1
E = 5 V R=10 I2
+ -
Preflight 6.3
Physics 102: Lecture 6, Slide 14
Kirchhoff’s Junction RuleCurrent Entering = Current Leaving
I1 I2
I3
I1 = I2 + I3
1) IB = 0.5 A 2) IB = 1.0 A 3) IB = 1.5 A
IB = I1 + I2 = 1.5 A
R=10
E1 = 10 V
IB
I1
E = 5 V R=10 I2
+ -
Preflight 6.3
Physics 102: Lecture 6, Slide 15
You try it!
R1
R2 R3
I1 I3
I2
+-
+
+
+
Loop 1:
1. Label all currents 2. Label +/- for all elements 3. Choose loop and direction 4. Write down voltage drops
-
-
-
Loop 2:
1
5. Write down node equation
Node:
2
In the circuit below you are given 1, 2, R1, R2 and R3. Find I1, I2 and I3.
+-
Physics 102: Lecture 6, Slide 16
Kirchhoff’s Laws(1) Label all currents
Choose any direction
(2) Label +/- for all elementsCurrent goes + - (for resistors)
(3) Choose loop and directionYour choice!
(4) Write down voltage dropsFollow any loops
(5) Write down node equationIin = Iout
R4
R1
E1
R2
R3E2
E3
I1
I3I2 I4
R5
A
B
36
Physics 102: Lecture 6, Slide 17
You try it!
R1
R2 R3
I1 I3
I2
+-
+
+
+
Loop 1: – 1+I1R1 – I2R2 = 0
1. Label all currents (Choose any direction)2. Label +/- for all elements (Current goes + - for resistor)3. Choose loop and direction (Your choice!)4. Write down voltage drops (First sign you hit is sign to use!)
-
-
-
Loop 2:
1
5. Write down node equation
Node: I1 + I2 = I3
23 Equations, 3 unknowns the rest is math!
In the circuit below you are given 1, 2, R1, R2 and R3. Find I1, I2 and I3.
Loop 1
Loop 2
+-
+ I2R2 + I3R3 + 2 = 0
Physics 102: Lecture 6, Slide 18
See you next lecture!
• Read Sections 18.10,11