Transcript Lecture - Galileo

Lecture 7 Circuits Ch. 27
• Cartoon -Kirchhoff's Laws
• Warm-up problems
• 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
Transmission line demo
Direct Current Circuits
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 Kirchoff’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.
b.
c.
Potential across resistors is negative
Sign of E for a battery depends on assumed
current flow
If you guessed wrong on the sign, your
answer will be negative
Start in the upper left hand corner.
 iR1  iR 2  E 2  ir 2  iR 3  E1  ir 1  0
E1  E 2
i
R1  R 2  R3  r1  r 2
i
E1  E 2
R1  R 2  R 3  r1  r 2
Now let us put in numbers.
Suppose: R1  R 2  R 3  10
r1  r 2  1
E1  10V
E 2  5V
i
10  5
V
5 amp

10  10  10  1  1  32
Suppose:
E1  5V
E 2  10V
(5  10)V  5
amp
i

32
32
We get a minus sign. It means our
assumed direction of current must be
reversed.
Note that we could have simply added all
resistors and get the Req. and added the EMFs
to get the Eeq. And simply divided.
i
Eeq.
5(V )
5


amp
Re q. 32() 32
Sign of EMF
Battery 1 current flows from - to + in battery +E1
Battery 2 current flows from + to - in battery -E2
In 1 the electrical potential energy increases
In 2 the electrical potential energy decreases
Example with numbers
Quick solution:
3
 E  12V  4V  2V  10V
i
i 1
6
 R  16
i
i 1
I
Eeq . 10
 A
Re q. 16
Question: What is the current in the circuit?
Write down Kirchoff’s loop equation.
Loop equation
Assume current flow is clockwise.
Do the batteries first – Then the current.
( 12  4  2)V  i (1  5  5  1  1  3)  0
10 V
i
 0.625amps  0.625 A
16 
Example with numbers (continued)
Question: What are the terminal voltages of each battery?
12V: V 
2V: V 
4V: V 
  ir  12V  0.625A 1  11.375V
  ir  2V  0.625A 1  1.375V
  ir  4V  0.625A 1  4.625V
Multiloop Circuits
Find i, i1, and i2
We now have 3 equations with 3
unknowns.
Kirchoff’s Rules
1.
Vi  0 in any loop

i
2.
 i in   i outat any junction
Rule 1 – Apply to 2 loops (2 inner loops)
a. 12  4i1  3i  0
b.  2i 2  5  4i1  0
Rule 2
a.
i  i1  i 2
12  4i1  3(i1  i 2 )  0
12  7i1  3i 2  0 multiply by 2
 5  4i1  2i 2  0 multiply by 3
24  14i1  6i 2  0
subtract them
 15  12i1  6i 2  0
39  26i 1  0
39
i1 
 1 .5 A
26
i 2  0 .5 A
i  2 .0 A
Find the Joule
heating in each
resistor P=i2R.
Is the 5V battery
being charged?
Method of determinants for solving simultaneous equations
i  i1  i 2  0
 3i  4i1  0  12
0  4i1  2i 2  5
Cramer’s Rule says if :
a1i1  b1i 2  c1i 3  d1
a2i1  b2i 2  c2i 3  d 2
a3 i1  b3 i 2  c3 i 3  d 3
Then,
d1 b1 c1
d 2 b2 c 2
d 3 b3 c3
i1 
a1 b1 c1
a2 b2 c 2
a3 b3 c3
a1 d1
a2 d 2
a3 d 3
i2 
a1 b1
a2 b2
a3 b3
c1
c2
c3
c1
c2
c3
a1 b1
a2 b2
a3 b3
i3 
a1 b1
a2 b2
a3 b3
d1
d2
d3
c1
c2
c3
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
2

 4
0
4

0
  1
2
2
 4



1


2 
2
1

4
0
0
0
12 
5
 12
  1
5
3
 3
  1
0
0
4 
4
4 
4



24  48  20
8  6  12
You try it for i1 and i2.
See inside of front cover in your book on how to use Cramer’s Rule.

52
26
 2A
Another example
Find all the currents including directions.
Loop 2
i
i
i2
i1
i
i
Loop 1
i  i1  i 2
Loop 1
i2
Multiply eqn of loop 1 by
0  8V  4V  4V  3i  2i1 2 and subtract from the
0  8  3i1  3i 2  2i1
eqn of loop 2
0  8  5i1  3i 2
Loop 2
 6i 2  4  2i1  0
 6i 2  4  2i1  0
 6i 2  16  10i1  0
0  12  12i1  0
i1  1A
 6i 2  4  2(1A)  0
i 2  1A
i  2A
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.
3 bulb question
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)
b)
c)
d)
e)
f)
g)
h)
i)
Bulbs 1 and 3 are unaffected. The total light emitted by the circuit decreases.
Bulbs 1 and 3 get brighter. The total light emitted by the circuit is unchanged.
Bulbs 1 and 3 get dimmer. The total light emitted by the circuit decreases.
Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit is
unchanged.
Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit is
unchanged.
Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit
decreases.
Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit
decreases.
Bulb 1 is unaffected, but bulb 3 gets brighter. The total light emitted by the circuit
increases.
None of the above.
When the bulb #2 is not burnt out:
I1
R eq  R 
R 3
 R
2 2
Power , P  I 2 R
I
I2  1
2
V
R
I
For Bulb #1
V 2V
I1  3 
3R
2R
4V 2
V2
P1  I R 
 .44
9R
R
For Bulb #2
I
V
I2  1 
2 3R
V2
V2
P2  I R 
 .11
9R
R
2
1
2
2
For Bulb #3
I
V
I3  1 
2 3R
V2
V2
P3  I R 
 .11
9R
R
2
3
I3 
I1
2
When the bulb #2 is burnt out:
I1
I 3  I1
R eq  R  R  2R
Power , P  I 2 R
I
V
R
For Bulb #1
I1 
V
2R
V2
V2
P1  I R 
 .25
4R
R
2
1
For Bulb #2
I2  0
After total power is
V2 V2
V2
Pa 

 .50
R eq 2R
R
P2  I 22 R  0
So, Bulb #1 gets dimmer and bulb #3 gets
brighter. And the total power decreases.
For Bulb #3
V
I 3  I1 
2R
2
2
2
Before total power was Pb  V  V  .66 V
R eq 32 R
R
V2
V2
P3  I R 
 .25
4R
R
2
3
f) is the answer.
How does a capacitor behave in a circuit with a resistor?
Charge capacitor with 9V battery with switch open, then remove battery.
Now close the switch. What happens?
Discharging a capacitor through a resistor
Qo
C
just before you throw switch at time t = 0.
Potential across capacitor = V =
V(t)
Potential across Resistor = iR
Qo
Qo
 i oR  i o 
C
RC
at t > 0.
What is the current I at time t?
Q( t )
RC
Q
or i 
RC
i( t ) 
What is the current I at time t?
So, i 
Q
dQ
, but i  
RC
dt
dQ
Q


dt RC
dQ dt


Q RC
Time constant =RC
Integrating both the sides
Q
dQ
dt

Q
RC
t
 ln Q 
A
RC
t
ln Q  
A
RC

So,
Qe

t
A
RC
e

t
RC
e
Q0
A
Q Q

e 2.7
t
t  RC
At t=0, Q=Q0
So,
Q0  e

0
A
RC
e
A
 Q  Q0e

t
RC
What is the current?
Q  Q0 e

t
RC
t
t


Q
V
dQ   0 e RC   0 e RC
i
RC
R
dt
i
Ignore - sign
V0
R
RC
t
How the charge on a capacitor varies with time as it is being charged
What about charging the capacitor?
CV0  Q0
Q
Q  CV0 (1  e

t
RC
t
)
V0  t
i
e
R
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
Instruments
Galvanometers:
a coil in a magnetic field that senses current.
Ammeters:
measures current.
Voltmeter:
measures voltage.
Ohmmeters:
measures resistance.
Multimeters:
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
Rg
Usually when R g  20
Ig  0  0.5milliAmp
Ohmmeter
i
V
R  Rs  Rg
Adjust Rs so when R=0 the
galvanometer read full scale.
Ammeter
10V
5A
Ig
Rg
2

I  5A
I  5A
Rs
Is
The idea is to find the value of RS that will give a full
scale reading in the galvanometer for 5A
I  Ig  Is  5 A
IgR g  IsR s
Rg  20 and Ig  0.5  103 A, So, Is  5A  .0005A  5A
Ig
0.5  10 3 A
So, R s  R g 
(20)  0.002
Is
5A
Very small
Ammeters have very low resistance when put in series in a circuit.
You need a very stable shunt resistor.
Voltmeter
Use the same galvanometer to construct a voltmeter for which full scale reading
in 10 Volts.
Rs
I
10V
2

What is the value of RS now?
We need
10 V  Ig (R s  R g )
10V
10 V
Rs  Rg 

Ig
5  10 4 A
R s  R g  20,000
Rs  19,980
Rg
Ig  0.5  10 3 A
Rg  20
So, the shunt resistor needs to
be about 20K.
Note: the voltmeter is in
parallel with the battery.