Transcript DC Circuits - UCF Physics

PHY-2049
Current & Circuits
February ‘08
News
• Quiz Today
• Examination #2 is on Wednesday of next week
(2/4/09)
• It covers potential, capacitors, resistors and any
material covered through Monday on DC
circuits.
• No review session on Wednesday – Exam Day!
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
2
E
P  I 2 R  IV 
R
Let’s add resistors …….
SERIES Resistorsi
i
R1
Series Combinations
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
V
so..
1
1
1


R1 R2 R
general
1
1

R
i Ri
What’s This???
In the figure, find the
equivalent resistance
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
Which is brighter?
Which is Brighter???
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
V1
V  iR  iR1  iR2
R2
V2
V
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
Reduction
Computes i
Another Reduction Example
1
1
1
50
1




R 20 30 600 12
R  12
PARALLEL
Battery
• A battery applies a potential difference between
its terminals.
• Whatever else is connected (circuits, etc.), the
potential between the points remains the same:
the battery potential.
Take a trip around this circuit.
Consider voltage DROPS:
-E +ir +iR = 0
or
E=ir + iR
Multiple Batteries
START by assuming a
DIRECTION for each Current
Let’s write the equations.
In the figure, all the resistors have a resistance of 4.0  and all the (ideal)
batteries have an emf of 4.0 V. What is the current through resistor R?
Consider the circuit shown in the figure. Find (a) the current in the
20.0-Ω resistor and (b) the potential difference between points a and b.
Using Kirchhoff’s rules, (a) find the current in each resistor in Figure P28.24.
(b) Find the potential difference between points c and f. Which point is at the
higher potential?
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
I need to use E for E
Note RC = (Volts/Amp)(Coul/Volt)
= Coul/(Coul/sec) = (1/sec)
Loop Equation
q
 E  iR   0
C
dq
since i 
dt
dq q
R
 E
dt C
or
dq
q
E


dt RC R
This is a
differential 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
a
E
1

RCE RC
Time Constant
  RC
Result q=CE(1-e-t/RC)
q=CE(1-e-t/RC) and i=(CE/RC) et/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
q0 t / RC
dq
i

e
dt
RC