Download Kirchhoff`s Rules

Physics 102: Lecture 06
Exam I
Kirchhoff’s Laws
PRACTICE EXAM IS AVAILABLE-CLICK ON
“COURSE INFO” ON WEB PAGE
Conflict Exam sign up available in grade book
Be Careful with round off errors in homework 3!
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?
5
Good News and Bad News
Good News: You already know everything
you need to get 100% on circuit problems
for Hour Exam 1!
Bad News: You don’t realize it!
So…. Let’s see what you know!
10
Kirchhoff’s Rules
• Kirchhoff’s Junction Rule (KJR):
– Current going in equals current coming out.
• Kirchhoff’s Loop Rule (KLR):
– Sum of voltage drops around a loop is zero.
12
Using Kirchhoff’s Rules
(1) Label all currents
Choose any direction
(2) Label +/- for all elements
R1
A
+
I1
-
+
Current goes +  - (for resistors)
(3) Choose loop and direction
+
B
E1
-
E2
(4) Write down voltage drops
Be careful about signs
+
R2
E3
I2
I3
I4
R3
R4
-
+
R5
-
-
+
-
+
+
17
Loop Rule Practice
R1=5 W
Find I:
I
B
e1= 50V
A
R2=15 W
e2= 10V
22
Loop Rule Practice
R1=5 W
Find I:
Label currents
Label elements +/Choose loop
Write KLR
B
I
-
+
+
e1= 50V
-
A
-
+
R2=15 W
-
+
e2= 10V
–e1+IR1 + e2 + IR2 = 0
-50 + 5 I + 10 +15 I = 0
I = +2 Amps
22
ACT: KLR
Resistors R1 and R2 are
1) in parallel
2) in series 3) neither
R1=10 W
I1
E2 = 5 V
I2
Definition of parallel:
Two elements are in parallel if
(and only if) you can make a loop
that contains only those two
elements.
R2=10 W
IB
+ E1 = 10 V
Upper loop contains R1 and R2 but also E2.
25
Preflight 6.1
Calculate the current through resistor 1.
28%
62%
10%
1) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A
-E1 + I1R = 0  I1 = E1 /R = 1A
I1
R=10 W
+
-
E2 = 5 V
I2
R=10 W
IB
+ E1 = 10 V
27
Preflight 6.1
Calculate the current through resistor 1.
20%
63%
17%
1) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A
I1
R=10 W
+
-
E2 = 5 V
I2
-E1 + I1R = 0  I1 = E1 /R = 1A
R=10 W
IB
+ E1 = 10 V
ACT: Voltage Law
How would I1 change if the switch was opened?
1) Increase
2) No change
3) Decrease
32
Preflight 6.2
Calculate the current through resistor 2.
53%
31%
16%
1) I2 = 0.5 A 2) I2 = 1.0 A 3) I2 = 1.5 A
R=10 W
I1
E2 = 5 V
I2
-E1 +E2 + I2R = 0
 I2 = 0.5A
R=10 W
+
IB
+ E1 = 10 V
35
Preflight 6.2
How do I know the direction of I2?
It doesn’t matter. Choose whatever direction
you like. Then solve the equations to find I2.
If the result is positive, then your initial guess
was correct. If result is negative, then actual
direction is opposite to your initial guess.
R=10 W
I1
E2 = 5 V
I2
Work through preflight with opposite
sign for I2?
R=10 W
+
-
IB
+ E1 = 10 V
-E1 +E2 - I2R = 0 Note the sign change from last slide
 I2 = -0.5A Answer has same magnitude as before
but opposite sign. That means current goes to the
left, as we found before.
35
Kirchhoff’s Junction Rule
Current Entering = Current Leaving
I1
I1 = I2 + I3
I2
I3
Preflight 6.3
18%
37%
44%
R=10 W
I1
E=5V
I2
R=10 W
1) IB = 0.5 A 2) IB = 1.0 A 3) IB = 1.5 A
IB = I1 + I2 = 1.5 A
“The first two can be calculated using V=IR because the
voltage and resistance is given, and the current through E1
can be calculated with the help of Kirchhoff's Junction
rule, that states whatever current flows into the junction
IB
+ E1 = 10 V
38
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 junction equation
Iin = Iout
39
36
You try it!
In the circuit below you are given e1, e2, R1, R2 and R3. Find I1, I2 and I3.
+ R1 - I1
I3
I2
+
e1
-
R2
R3
+
-
e2
+
+
-
45
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 junction equation
+
e1
-
Loop 1
Node: I1 + I2 = I3
3 Equations, 3 unknowns the rest is math!
R2
+
R3
+
-
Loop 2
-
e2
+
45
Let’s put in real numbers
In the circuit below you are given e1, e2, R1, R2 and R3. Find I1, I2 and I3.
+ 5 - I1
I3
I2
+
20
-
10
+
10
-
+
-
2
1. left loop: -20+5I1-10I2 = 0
2. outer loop: -20 +5I1+10I3+2=0
3. junction: I3=I1+I2
+
solution: substitute Eq.3 for I3 in Eq. 2:
-20 + 5I1 + 10(I1+I2) + 2 = 0
rearrange:
15I1+10I2 = 18
rearrange Eq. 1:
5I1-10I2 = 20
Now we have 2 eq., 2 unknowns. Continue on next slide
45
15I1+10I2 = 18
5I1 - 10I2 = 20
Now we have 2 eq., 2 unknowns.
Add the equations together:
20I1=38 I1=1.90 A
U′
Plug into bottom equation:
5(1.90)-10I2 = 20 I2=-1.05 A
note that this means direction of I2 is opposite to
that shown on the previous slide
Use junction equation
I3=I1+I2 = 1.90-1.05
I3 = 0.85 A
We are done!
Some practice from the book:
18.52, 18.53, 18.54, 18.55, 18.67
See you Wednesday