• Cartoon -Kirchhoff's Laws
• Topics
– Direct Current Circuits
– Kirchhoff's Two Rules
– Analysis of Circuits Examples
– Ammeter and voltmeter
– RC circuits
• Demos
– Three bulbs in a circuit
– Power loss in transmission lines
– Resistivity of a pencil
– Blowing a fuse
• Elmo

Transmission line demo

1.
The sum of the potential drops around a closed loop is zero.
This follows from energy conservation and the fact that the electric field is a conservative force.
2. The sum of currents into any junction of a closed circuit must equal the sum of currents out of the junction. This follows from charge conservation.

Example (Single Loop Circuit)
No junction so we don’t need that rule.
How do we apply Kirchhoff’s rule?
Must assume the direction of the current – assume clockwise.
Choose a starting point and apply Ohm’s Law as you go around the circuit.
a.
Potential across resistors is negative b.
Sign of E for a battery depends on assumed current flow c.
If you guessed wrong on the sign, your answer will be negative i
Start in the upper left hand corner.
!
iR
1
!
iR
2
!
E
2
!
ir
2
!
iR
3
+ E
1
!
ir
1
= 0
=
R
1
+ R
2
E
1
!
+ R
3
E
2
+ r
1
+ r
2

i i =
R
1
+ R
2
E
1
!
+ R
3
E
2
+ r
1
+ r
2
=
Now let us put in numbers.
Suppose: R
1
= R
2
= R
3 r
1
= r
2
= 1 !
E
1
= 10 V
10 +
E
10
10 +
" 5
10
2
+
=
1 +
5 V
1
V
!
=
= 10
5
32 amp
!
Note that we could have simply added all resistors and get the R eq. to get the E eq.
and added the EMFs
And simply divided.
i =
E eq .
R e q .
=
5 ( V )
32 ( !
)
=
5
32 amp
Sign of EMF
Suppose: E
E
2
1
= 5 V
= 10 V i =
( 5 !
10
32 "
) V
=
!
5
32 amp
We get a minus sign. It means our assumed direction of current must be reversed.
Battery 1 current flows from - to + in battery +E
1
Battery 2 current flows from + to in battery -E
2
In 1 the electrical potential energy increases
In 2 the electrical potential energy decreases

Example with numbers
Quick solution: i
3 #
= 1
E i
= 12 V " 4 V
I i
6 #
= 1
R i
=
= 16 !
E eq .
R e q .
=
10
16
A
+ 2 V = 10 V
Question: What is the current in the circuit?
Write down Kirchhoff’s loop equation.
Loop equation
Assume current flow is clockwise.
Do the batteries first – Then the current.
i
( +
=
12
10
16
" 4
V
!
+ 2 ) V
= 0 .
" i ( 1 + 5
625 amps
+ 5
=
+ 1 + 1 +
0 .
625 A
3 ) !
= 0

Example with numbers (continued)
Question: What are the terminal voltages of each battery?
12V:
2V:
4V:
V
V
V
=
=
=
$
$
$
#
#
ir ir ir
=
=
=
12
2
4
V
V
V
#
+
#
0
0
0 .
625
.
625
.
625
A
A
A
"
" 1 !
1 !
1
=
=
=
1
4
11 .
375
.
375
.
625
V
V
V

Kirchoff’s Rules
1.
!
i
V = 0
2.
!
i in
= !
i out
Multiloop Circuits
Rule 1 – Apply to 2 loops (2 inner loops) a.
b.
12
!
2 i
!
2
4
!
i
1
5
!
+
3 i
4 i
1
=
=
0
0
Rule 2 a.
i = i
1
+ i
2
Find i , i
1
, and i
2
We now have 3 equations with 3 unknowns.
12
12
!
5
!
!
+
4 i
1
7
4 i i
1
1
!
!
!
3 (
3
2 i i i
2
2
1
+
=
= i
0
2
0
) = 0 multiply by 2 multiply by 3
24
!
!
15
14
+ i
1
12 i
!
1
6 i
!
2
6 i
2
= 0
= 0 subtract them i i i
39
1
=
!
26 i = 0
2
=
=
2 .
0 A
1
39
26
0 .
5
=
A
1 .
5 A
Find the Joule heating in each resistor P=i 2 R.
Is the 5V battery being charged?

Method of determinants for solving simultaneous equations i
!
!
0
3
+ i i
1
4 i
!
!
1 i
4
2
!
i
=
+
0
1
2 i
2
0
=
=
5
!
12 i
1
= d
1 d
2 d
3 a a
1 a
2
3
Cramer’s Rule says if : a
1 i a
2 i
1
1 a
3 i
1
+
+
+ b
1 i b
2 i
2
2 b
3 i
2
+
+ c
1 i c
2 i
+ c
3 i
3
3
3
=
= d
1 d
= d
3
2
Then, b b
1 b
2
3 b b
1 b
2
3 c
1 c
2 c
3 c
1 c
2 c
3 i
2
= a a
1 a
2
3 a a
1 a
2
3 d
1 d
2 d
3 b b
1 b
2
3 c
1 c c
3
2 c
1 c c
3
2 i
3
= a a
1 a
2
3 a a
1 a
2
3 b b
1 b
2
3 b b
1 b
2
3 d
1 d
2 d
3 c
1 c
2 c
3

Method of determinants using Cramers Rule and cofactors
Also use this to remember how to evaluate cross products of two vectors.
For example solve for i i
=
0 !
1 !
1
!
12
!
4 0
5
+
4
!
2
1 !
1 !
1
!
3
!
4 0
=
0
#$
" !
4 0
4
!
2
%
&' !
1
#$
"
0
!
12
!
2 5
%
&' !
1
#$
" !
12
!
4
5 4
&'
%
1 #$
" !
4 0
4
!
2
%
&' !
1
"
#$
0 !
3
%
!
2 0
&' !
1
#$
" !
3 !
4
0 4
&'
%
=
24 + 48 !
20
8
+
6
+
12
=
52
=
2 A
26
0 + 4 !
2
You try it for i
1
and i
2
.
See inside of front cover in your book on how to use Cramer’s Rule.

Another example
Find all the currents including directions.
Loop 2 i i i
1 i
2 i
Loop 1 i i
2
Loop 1
0
0
0
=
=
=
+ 8 V
8
8
!
!
3 i
5 i
+
1
1
4 V
!
!
3
3 i i
2
2
!
4 V
!
2 i
1
!
3 i !
2 i
1
Loop 2
!
6 i
2
+ 4 + 2 i
1
= 0
Multiply eqn of loop 1 by
2 and subtract from the eqn of loop 2
!
!
6
6
0 i
2 i
2
!
+
+
12
4
16
+
+
!
12 i
2 i
1
10
=
1 i
=
1
0
=
0
0 i = i
1
+ i
2 i
1
= 1 A
!
6 i
2
+ 4 + 2 ( 1 A ) = 0 i i
2
=
= 1 A
2 A

Rules for solving multiloop circuits
1.
Replace series resistors or batteries with their equivalent values.
2.
Choose a direction for i in each loop and label diagram.
3.
Write the junction rule equation for each junction.
4.
Apply the loop rule n times for n interior loops.
5.
Solve the equations for the unknowns. Use Cramer’s Rule if necessary.
6.
Check your results by evaluating potential differences.

The circuit above shows three identical light bulbs attached to an ideal battery. If the bulb#2 burns out, which of the following will occur?
a) Bulbs 1 and 3 are unaffected. The total light emitted by the circuit decreases.
b) Bulbs 1 and 3 get brighter. The total light emitted by the circuit is unchanged.
c) Bulbs 1 and 3 get dimmer. The total light emitted by the circuit decreases.
d) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit is unchanged.
e) Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit is unchanged.
f) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit decreases.
g) Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit decreases.
h) Bulb 1 is unaffected, but bulb 3 gets brighter. The total light emitted by the circuit increases.
i) None of the above.

R eq
= R +
R
2
=
3
2
R
Power , P = I 2 R I =
V
R
When the bulb #2 is not burnt out:
I
1
I
2
=
I
2
1
For Bulb #1
I
1
=
3
2
V
R
=
2 V
3 R
For Bulb #2
I
2
=
I
2
1 =
V
3 R
For Bulb #3
I
3
=
I
2
1 =
V
3 R
P
1
= I
1
2 R =
4 V 2
9 R
= .
44
V
R
2
P
2
P
3
= I 2
2
R
= I 2
3
R
=
V
9 R
2
=
V 2
9 R
=
=
.
11
V
R
2
.
11
V
R
2
I
3
=
I
2
1

When the bulb #2 is burnt out:
I
1
I
3
= I
1
R eq
= R + R = 2 R
Power , P = I 2 R I =
V
R
For Bulb #1
I
1
=
V
2 R
P
1
= I
1
2 R =
V 2
4 R
= .
25
V 2
R
For Bulb #2
I
2
= 0 P
2
= I 2
2
R = 0
For Bulb #3
I
3
= I
1
=
V
2 R
P
3
= I 2
3
R =
V 2
4 R
= .
25
V
R
2
Before total power was P b
=
V
R
2 eq
=
3
2
V 2
R
= .
66
V
R
2
After total power is P a
=
V
R eq
2
=
V
2 R
2
= .
50
V
R
2
So, Bulb #1 gets dimmer and bulb #3 gets brighter. And the total power decreases.
f) is the answer.

Charge capacitor with 9V battery with switch open, then remove battery.
Now close the switch. What happens?

Potential across capacitor = V =
Q o
C just before you throw switch at time t = 0.
Potential across Resistor = i R
Q o
C
= i o
R !
i o
=
Q o
RC
V(t)
What is the current I at time t?
or
(i t ) = i =
Q ( t )
RC
Q
RC

So , i =
Q
RC
, but i = !
dQ dt
!
!
dQ dt dQ
Q
=
=
Q
RC dt
RC
Integrating both the sides
" !
dQ
Q
= !
dt
RC
!
ln Q = t
RC
+ A ln Q = !
t
RC
!
A
So , Q = e
!
t
RC
!
A
= e
!
t
RC e !
A
Q
Q
0
Time constant =RC t = RC
Q e
=
Q
2 .
7
At t=0, Q=Q
0
So , Q
0
= e
!
0
RC
!
A
= e !
A " Q = Q
0 e
!
t
RC t

Q = Q
0 e
!
t
RC i = dQ dt
= !
Q
0
RC e
!
t
RC = !
V
0
R e
!
t
RC i
V
0
R
Ignore - sign
RC t

How the charge on a capacitor varies with time as it is being charged
What about charging the capacitor?
CV
0
= Q
0
Q t
Q = CV
0
( 1 !
e
!
t
RC ) i =
V
0
R e
" t
!
Note that the current is zero when either the capacitor is fully charged or uncharged. But the second you start to charge it or discharge it, the current is maximum.
i
Same as before t

Galvanometers:
Ammeters:
Voltmeter:
Ohmmeters:
Multimeters:
a coil in a magnetic field that senses current.
measures current.
measures voltage.
measures resistance.
one device that does all the above.
Galvanometer is a needle mounted to a coil that rotates in a magnetic field.
The amount of rotation is proportional to the current that flows through the coil.
Symbolically we write
R g
Usually when
I g
R g
= 20
= 0 !
0 .
5 milliAmp
!

i =
R +
V
R s
+ R g
Adjust R s so when R=0 the galvanometer read full scale.

10 V
5 A
2 !
!
I = 5 A
The idea is to find the value of R
S that will give a full scale reading in the galvanometer for 5A
I g
R g
R s
I s
I = 5 A
I = I g
+ I s
= 5 A
I g
R g
= I s
R s
R g
= 20 $ and I g
= 0 .
5 # 10 " 3 A , So, I s
= 5 A " .
0005 A !
5 A
So , R s
=
I g
I s
R g
=
0 .
5 "
5
10
A
# 3 A
( 20 !
) = 0 .
002 !
Very small
Ammeters have very low resistance when put in series in a circuit.
You need a very stable shunt resistor.

10 V
Use the same galvanometer to construct a voltmeter for which full scale reading in 10 Volts.
I
2 !
!
R s
R g
I g
= 0 .
R g
5 " 10 # 3
= 20 !
A
What is the value of R
S now?
We need
10
R
R s s
V
+
+
=
R g
R g
I g
( R
=
= s
10
I g
+
V
R g
=
)
5
20 , 000 !
"
10 V
10 !
4 A
R s
= 19 , 980 !
So, the shunt resistor needs to be about 20K Ω .
Note: the voltmeter is in parallel with the battery.

In Figure 27-34, R
1
= 100 , R
2
= 30 , and the ideal batteries have emfs script e
1
= 6.0 V, script e
2
= 5.0 V, script e
3
= 3.0 V.
Fig. 27-34
(a) Find the current in resistor 1.
Fig. 27-34
(b) Find the current in resistor 2.
(c) Find the potential difference between points a and b.

In Figure 27-40, the resistances are R1 = 0.5 , R2 = 1.7 , and the ideal batteries have emfs script e1 = 2.0 V, and script e2 = script e3
= 3 V.
Fig. 27-40
(a) What is the current through each battery? (Take upward to be positive.) battery 1 battery 2 battery 3
(b) What is the potential difference Va - Vb?

A simple ohmmeter is made by connecting a 4.0 V battery in series with a resistance R and an ammeter that reads from 0 to 1.00
mA, as shown in Figure 27-47. Resistance R is adjusted so that when the clip leads are shorted together, the meter deflects to its full-scale value of 1.00 mA.
Fig. 27-47
(a) What external resistance across the leads results in a deflection of 10% of full scale?
(b) What resistance results in a deflection of 50% of full scale?
(c) What resistance results in a deflection of 90% of full scale?
(d) If the ammeter has a resistance of 40.0 and the internal resistance of the battery is negligible, what is the value of R?