# Physics 102 Lecture 06 Kirchhoff’s Laws Physics 102 Lecture 6 Slide 1 Last Time • Resistors in s

Document Sample

```					Physics 102: Lecture 06

Kirchhoff’s Laws

Physics 102: Lecture 6, Slide 1
Last Time
• Resistors   in series:
Current thru is same; Voltage drop across is IRi
Last Lecture

• Resistors   in parallel:
Voltage drop across is same; Current thru is V/Ri

• Solved   Circuits
Today

Physics 102: Lecture 6, Slide 2
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.

Physics 102: Lecture 6, Slide 3
Physics 102: Lecture 6, Slide 4
Physics 102: Lecture 6, Slide 5
Physics 102: Lecture 6, Slide 6
Physics 102: Lecture 6, Slide 7
Physics 102: Lecture 6, Slide 8
Using Kirchhoff’s Rules
(1) Label all currents
Choose any direction                             R1       I1
A                -
(2) Label +/- for all elements                       +
+                   +
Current goes +  - (for resistors)                                  R2
E3
-                   -
+                     B
(3) Choose loop and direction                    E1                         I2            I3    I4
-                                         -        -
+
E2                      R3        R4
-
(4) Write down voltage drops                                                               +        +
R5
Be careful about signs                                                   -         +

Physics 102: Lecture 6, Slide 9
Loop Rule Practice
R1=5 W         I            B
Find I:
e1= 50V

A
R2=15 W
e2= 10V

Physics 102: Lecture 6, Slide 10
Loop Rule Practice
R1=5 W          I           B
Find I:                                     +       -
+
Label currents            e1= 50V
Label elements +/-                  -
Choose loop                                 -       +
Write KLR                       A                              -    +
R2=15 W
e2= 10V

–e1+IR1 + e2 + IR2 = 0
-50 + 5 I + 10 +15 I = 0
I = +2 Amps

Physics 102: Lecture 6, Slide 11
Physics 102: Lecture 6, Slide 12
ACT: KLR
I1        R1=10 W
Resistors R1 and R2 are
1) in parallel   2) in series 3) neither   E2 = 5 V
I2     R2=10 W

IB
+ -
E1 = 10 V

13
Physics 102: Lecture 6, Slide25
Preflight 6.1
Calculate the current through resistor 1.   I1     R=10 W
+       -

E2 = 5 V
I2    R=10 W
1) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A

-E1 + I1R = 0  I1 = E1 /R = 1A                  IB
+ -
E1 = 10 V

14
Physics 102: Lecture 6, Slide27
Preflight 6.1
Calculate the current through resistor 1.    I1     R=10 W
+       -

E2 = 5 V
I2    R=10 W
1) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A

-E1 + I1R = 0  I1 = E1 /R = 1A                   IB
+ -
E1 = 10 V

ACT: Voltage Law
How would I1 change if the switch was opened?
1) Increase                2) No change     3) Decrease

15
Physics 102: Lecture 6, Slide32
Physics 102: Lecture 6, Slide 16
Preflight 6.2
Calculate the current through resistor 2.
I1        R=10 W

1) I2 = 0.5 A 2) I2 = 1.0 A 3) I2 = 1.5 A
E2 = 5 V
I2     R=10 W
+      -
-E1 +E2 + I2R = 0

 I2 = 0.5A                                      IB
+ -
E1 = 10 V

17
Physics 102: Lecture 6, Slide35
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.                I1        R=10 W

If the result is positive, then your initial guess
was correct. If result is negative, then actual               E2 = 5 V
I2      R=10 W
direction is opposite to your initial guess.                               -       +

Work through preflight with opposite                            IB

sign for I2?                                                           + -
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.                                                       18
Physics 102: Lecture 6, Slide35
Kirchhoff’s Junction Rule
Current Entering = Current Leaving
I1 = I 2 + I 3                   I1             I2

I3

Preflight 6.3                                        I1        R=10 W

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                 IB
voltage and resistance is given, and the current through E1                    + -
can be calculated with the help of Kirchhoff's Junction                       E1 = 10 V
rule, that states whatever current flows into the junction
must flow out. So I1 and I2 are added together.”                                                      19
Physics 102: Lecture 6, Slide38
Physics 102: Lecture 6, Slide 20
Kirchhoff’s Laws
(1) Label all currents
Choose any direction                          R1   I1
A
(2) Label +/- for all elements
Current goes +  - (for resistors)                           R2
E3

B
(3) Choose loop and direction                    E1                  I2         I3       I4
E2

(4) Write down voltage drops                                              R5

(5) Write down junction equation
Iin = Iout

21
Physics 102: Lecture 6, Slide36
39
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

+

-

22
Physics 102: Lecture 6, Slide45
You try it!
In the circuit below you are given e1, e2, R1, R2 and R3. Find I1, I2 and I3.

1.    Label all currents                (Choose any direction)

2.    Label +/- for all elements        (Current goes +  - for resistor)

4.
3.    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                      + R1 - I1             I3
I2
+
5.     Write down junction equation
e1
+
-
Loop 1    R2
-
R3
Loop 2             -
Node: I1 + I2 = I3                                                  +

-        +
3 Equations, 3 unknowns the rest is math!                                                  e2
23
Physics 102: Lecture 6, Slide45
Physics 102: Lecture 6, Slide 24
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
+   1. left loop: -20+5I1-10I2 = 0
-
20
+               10                 10           2. outer loop: -20 +5I1+10I3+2=0
-
+
-   3. junction: I3=I1+I2

-        +
2

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
25
Physics 102: Lecture 6, Slide45
15I1+10I2 = 18
5I1 - 10I2 = 20

Now we have 2 eq., 2 unknowns.

20I1=38 I1=1.90 A

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 (eq. 3 from previous page)
I3=I1+I2 = 1.90-1.05
I3 = 0.85 A
We are done!                        Physics 102: Lecture 6, Slide 26
Physics 102: Lecture 6, Slide 27

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 49 posted: 11/25/2011 language: English pages: 27
How are you planning on using Docstoc?