Transcript Document

Lecture 7 Circuits Chp. 28
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Cartoon -Kirchoff’s Laws
Opening Demo- transmission lines
Physlet
Topics
– Direct Current Circuits
– Kirchoff’s Two Rules
– Analysis of Circuits Examples
– Ammeter and voltmeter
– RC circuits
• Demos
– Ohms Law
– Power loss in transmission lines
– Resistivity of a pencil
– Blowing a fuse
• Warm-up problems
Transmission line demo
Direct Current Circuits
1.
The sum of the potential
charges 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  iR3  E1  ir1  0
i
E1  E 2
R1  R 2  R 3  r1  r 2
i
E1  E 2
R1  R 2  R 3  r1  r 2
Put in numbers.
Suppose: R1  R 2  R 3  10
r1  r 2  1
E1  10V
E 2  5V
10  5
5
i

amp
10  10  10  1  1 32
Suppose: E1  5V
E 2  10V
amp
5  10  5
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
Eeq. 10
I
 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)  i (1  5  5  1  1  3)  0
10 V
i
 0.625am ps 0.625A
16 
Example with numbers (continued)
Question: What are the terminal voltages of each battery?
12V:
2V:
4V:
V    ir 12V  0.625A 1 11.375V
V    ir  2V  0.625A 1 1.375V
V    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.
in any loop
V
i
2.
0
i
i  i
in
at any junction
out
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  26i1  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
For example solve for i
0
1
 12  4
1
0
 4  2 48  20  24 52


 2A
1 1
8  12  6
26
3 4 0
5
i
1
0
4 2
You try it for i1 and i2.
See Appendix in your book on how to use Cramer’s Rule.
Another example
Find all the currents including directions.
Loop 2
i
i
i2
i1
i
Loop 1
i
Loop 1
0  8V  4V  4V  3i  2i1
0  8  3i1  3i 2  2i1
0  8  5i1  3i 2
multiply by 2
i = i1+ i2
i2
Loop 2
 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.
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
Potential across capacitor = V =
Qo
C
just before you throw switch at time t = 0.
V(t)
Potential across Resistor = iR
Qo
Qo
 ioR  io 
C
RC
at t > 0.
What is the current I at time t?

Q(t)
i(t) 
RC
Time constant
= RC
What is the current?
Q  Q0e
t

RC
dQ
Q0
i

e
dt
RC
RC

t
RC
V0
 e
R

t
RC
Ignore - sign
How the charge on a capacitor varies with time
as it is being charged
Ohmmeter
Ammeter
Voltmeter
Warm up set 7
Warm up set 7 Due 8:00 am Tuesday
1.
HRW6 28.TB.05. [119859] In the context of the loop and junctions rules for electrical circuits a
junction is:
where
where
where
where
where
a wire is connected to a battery
three or more wires are joined
a wire is bent
a wire is connected to a resistor
only two wires are joined
2. HRW6 28.TB.18. [119872] Two wires made of the same material have the same length but
different diameter. They are connected in parallel to a battery. The quantity that is NOT the same
for the wires is:
the electric field
the electron drift velocity
the current
the current density
the end-to-end potential difference
3. HRW6 28.TB.26. [119880] The emf of a battery is equal to its terminal potential difference:
only when there is no current in the battery
only when a large current is in the battery
under all conditions
under no conditions
only when the battery is being charged