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PHY-2049
Chapter 27
Circuits
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
#24 chapter 26: 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
current in the wire?
2.45 mm.
What is the
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
A=1.9 x 10-5 m 2
copper
12 uvolts
0 volts
Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm meters
We have all what we need….
8
L 1.69 x10 ohm - m  3.0 m
R 
 2.67 m
5
A
1.9 x10
From Ohm' s Law :
6
V
12  10 volts
i 
 4.49 mA
3
R 2.67  10 ohms
Let’s add resistors …….
SERIES Resistorsi
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
V
1
1

R
i Ri
#26 chapter 27:In Figure
What’s This???
below, find the equivalent
resistance between points
(a) F and H and [2.5]
(b) F and G. [3.13]
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
ε or Emf is an idealized device that does an amount
of work 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

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
Take a trip around this circuit.
Consider voltage DROPS:
ε-ir -iR = 0
or
ε=ir + iR
Circuit Reduction
i=ε/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.
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.
Really Close the Switch
Loop Equation
q
  iR   0
C
dq
since i 
dt
dq q
R
 
dt C
or
dq
q



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.
Result q=Cε(1-e-t/RC)
q=Cε(1-e-t/RC) and i=(Cε/RC) e-t/RC
i

R
e
 t / RC
Discharging a Capacitor
qinitial=Cε 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