# Lecture 07 by ert554898

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```									PHY-2049

Current & Circuits
February ‘08
A closed circuit
Power in DC Circuit

In time t, a charge Q is pushed through
the resistor by the battery. The amount of work
done by the battery is :
W  VQ
Power :
W        Q
V        VI
t       t
Power  P  IV  I IR  I 2 R
E2
P  I 2 R  IV 
R
The figure below gives the electrical potential V(x) along a copper wire carrying a
uniform current, from a point at higher potential (x=0m) to a point at a lower
potential (x=3m). The wire has a radius of   2.45 mm.    What is the current in the
wire?

What does the graph tell us??

*The length of the wire is 3 meters.
*The potential difference across the
wire is 12 m volts.
*The wire is uniform.

Let’s get rid of the mm radius and
convert it to area in square meters:
A=pr2 = 3.14159 x 2.452 x 10-6 m2
or
copper                                A=1.9 x 10-5 m 2
12 volts                    0 volts
Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm meters
We have all we need….

8
L 1.69 x10 ohm - m  3.0 m
R                     5
 2.67m
A           1.9 x10
From Ohm's Law :
6
V  12 10 volts
i         3
 4.49 ma
R 2.67 10 ohms
SERIES Resistors
i                  i

Series Combinations
R1               R2

V1               V2
V
V1  iR1
V2  iR2
and
V  V1  V2  iR  iR1  iR2
R  R1  R2
general :
R( series )   Ri
i
The rod in the figure is made of two materials. The
figure is not drawn to scale. Each conductor has a
square cross section 3.00 mm on a side. The first
material has a resistivity of 4.00 × 10–3 Ω · m and is
25.0 cm long, while the second material has a
resistivity of 6.00 × 10–3 Ω · m and is 40.0 cm long.
What is the resistance between the ends of the rod?
Parallel Combination??

R1, I1
V  iR
V V V
i  i1  i2     
R1 R2 R

R2, I2
so..
1   1   1
   
R1 R2 R
general
1    1

V          R  i Ri
In Fig. 28-39, find the
equivalent resistance
What’s This???   between points
(a) F and H and [2.5]
(b) F and G. [3.13]
   (a) Find the equivalent resistance
between points a and b in Figure P28.6.
(b) A potential difference of 34.0 V is
applied between points a and b. Calculate
the current in each resistor.
Power Source in a Circuit

The ideal battery does work on charges moving
them (inside) from a lower potential to one that is
V higher.
A REAL Power Source
is NOT an ideal battery

Internal Resistance   V

E or Emf is an idealized device that does an amount of
work E to move a unit charge from one side to
another.

By the way …. this is called a circuit!
A Physical (Real) Battery

Emf
i
rR
Back to which is brighter?
Back to Potential

Change in potential as one circuits
this complete circuit is ZERO!

Represents a charge in space
Consider a “circuit”.

This trip around the circuit is the same as a path
through space.

THE CHANGE IN POTENTIAL FROM “a” AROUND
THE CIRCUIT AND BACK TO “a” is ZERO!!
To remember
   In a real circuit, we can neglect the
resistance of the wires compared to the
resistors.
   We can therefore consider a wire in a circuit to
be an equipotential – the change in
potential over its length is slight
compared to that in a resistor
   A resistor allows current to flow from a
high potential to a lower potential.
   The energy needed to do this is supplied
by the battery.
W  qV
NEW LAWS PASSED BY THIS SESSION
OF THE FLORIDUH LEGISLATURE.

   LOOP EQUATION
   The sum of the voltage drops (or rises)
as one completely travels through a
circuit loop is zero.
   Sometimes known as Kirchoff’s loop
equation.
   NODE EQUATION
   The sum of the currents entering (or
leaving) a node in a circuit is ZERO
TWO resistors again
i

R1               R2
V1               V2

V
V  iR  iR1  iR2
or
R  R1  R2
General for SERIES Resistors
R  Rj
j
A single “real” resistor can be modeled
as follows:

R
a                             b

V

position

ADD ENOUGH RESISTORS, MAKING THEM SMALLER
AND YOU MODEL A CONTINUOUS VOLTAGE DROP.
We start at a point in the circuit and travel
around until we get back to where we
started.

   If the potential rises … well it is a rise.
   If it falls it is a fall OR a negative rise.
   We can traverse the circuit adding each
rise or drop in potential.
   The sum of all the rises around the loop
is zero. A drop is a negative rise.
   The sum of all the drops around a circuit
is zero. A rise is a negative drop.
   Your choice … rises or drops. But you
must remain consistent.
Take a trip around this circuit.

Consider voltage DROPS:
-E +ir +iR = 0
or
E=ir + iR
Circuit Reduction

i=E/Req
Multiple Batteries
Reduction

Computes i
Another Reduction Example

1 1      1   50   1
           
R 20 30 600 12
R  12

PARALLEL
START by assuming a
DIRECTION for each Current

Let’s write the equations.
The Unthinkable ….
RC Circuit
   Initially, no current
through the circuit
   Close switch at (a)
and current begins to
flow until the
capacitor is fully
charged.
   If capacitor is charged
and switch is switched
to (b) discharge will
follow.
Close the Switch

I need to use E for E

Note RC = (Volts/Amp)(Coul/Volt)
= Coul/(Coul/sec) = (1/sec)
Really Close the Switch
Loop Equation
q
 E  iR   0
C
dq
since i 
dt
I need to use E for E
dq q
R  E
Note RC = (Volts/Amp)(Coul/Volt)        dt C
= Coul/(Coul/sec) = (1/sec)   or
dq q   E
  
dt RC R
differential
This is a

equation.
   To solve we need what is called a
particular solution as well as a
general solution.
   We often do this by creative
“guessing” and then matching the
guess to reality.
   You may or may not have studied
this topic … but you WILL!
General Solution
q  q p  Ke  at
Look at particular solution :
dq q       E
    
dt RC R
When the device is fully charged,dq/dt  0 and
q p  CE
When t  0, q  0 and from solution
0  CE  K
K  -CE
dq q          E
         and q  CE(1 - e -at )
dt RC R
CE (ae at )  CE(1 - e -at )  E / R
for t  0
CEa  0  E/R
E   1
a      
RCE RC
Time Constant

  RC
Result q=CE(1-e-t/RC)
q=CE(1-e-t/RC) and i=(CE/RC) e-t/RC

E t / RC
i e
R
Discharging a Capacitor
qinitial=CE BIG SURPRISE! (Q=CV)
i
iR+q/C=0
dq q
R      0
dt C
solution
q  q0 e t / RC
dq    q0 t / RC
i        e
dt    RC
In Fig. (a), a R = 21, Ohm a resistor is connected to a battery. Figure (b) shows the increase
of thermal energy Eth in the resistor as a function of time t.
(a)What is the electric potential across the battery? (60)
(b) If the resistance is doubled, what is the POWER dissipated by the circuit? (39)

Looking at the graph, we see that the
resistor dissipates 0.5 mJ in one second.
Therefore, the POWER =i2R=0.5 mW

P 0.5 mW
i2                2.38 10 5 amp 2
R     21 Ω
i  4.88 10 3 amp  4.88 ma      Voltage drop across the reisitor  iR or
V  iR  4.8810-3 amp  21  102mV
If the resistance is doubled what is the
power dissipated by the circuit?

R  42       V  102 mV
3
V 102 10
i               2.43ma
R        42
P  i R  0.248mJ
2

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